Electricity and Magnetism: Physics 110AWritten by Roger L. Grifth @UCB fall 2007January 23, 2008Chapter 1Vector Analysis1.1 Vector OperationsThe Dot product of two vectors is dened asAB ABsin nwhere is the angle they form when placed tail-to-tail. Note that AB is itself a scalar (hence thealternative name scalar product). The dot product is commutative,A B = B Aand distributive,A (B+C) = A B+A Cif two vectors are parallel, then A B =AB. In particular, for any vector A,A A =A2if A and B are perpendicular, then AB =0.The cross product is dened asAB = ABsin n,where n is a unit vector pointing perpendicular to the plane of Aand B. The cross product is distributive,A(B+C) = (AB) +(AC)but not commutative. In fact,(BA) =(AB)geometrically, |AB| is the area of the parallelogram generated by A and B. If two vectors are parallel,their cross product is zero. In particular,AA = 011.2 Vector Algebra: Triple Products(i) Scalar triple product: A (BC). Geometrically, |A (BC)| is the volume of the parallelepipedgenerated by A, B, and C, since |BC| is the area of the base, and Acosis the altitude. Evidently,A (BC) = B (CA) = C (AB)for they all correspond to the same gure. Not that alphabetical order is preserved. The nonalpha-betical triple products,A (CB) = B (AC) = C (BA)have the opposite sign. In component form,A (BC) =

Ax Ay AzBx By BzCx Cy Cz

(ii) Vector Triple Product: A(BC). The vector triple product can be simplied by the so calledBAC-CAB rule,A(BC) = B(A C) C(A B)1.3 Position, Displacement, and seperation vectorsThe location of a point in three dimensions can be described by listing it Cartesian coordinates (x, y, z) .The vector to that point from the origin is called the position vector:r x x+y y+z zits magnitude is given asr =

x2+y2+z2is the distance to the origin, and r = rr = x x+y y+z z

x2+y2+z2is a unit vector pointing radially outward. The innitesimal displacement vector, is given asdl = dx x+dy y+dz zthe seperation vector from a source point to the eld is given as r rits magnitude is =

r r

and a unit vector in the direction from rto r is = = r r

r r

2in Cartesian coordinates, = (x x) x+(y y) y+(z z) z =

(x x)2+(y y)2+(z z)2 = (x x) x+(y y) y+(z z) z

(x x)2+(y y)2+(z z)21.4 Problems with SolutionsProblem # 1The Equation giving a family of ellipsoids isu = x2a2 + y2b2 + z2c2Find the unit vector normal to each point of the surface of these ellipsoids.Given a three-dimensional surface we can nd what the normal vector is by using u = u u and knowing thatu =

ux x + uy y + uz z

= 2xa2 x + 2yb2 y + 2zc2 zand for the magnitude of this vector we nd u = 2

x2a4 + y2b4 + z2c4thus the unit vector u is given as u =xa2 x + yb2 y + zc2 z

x2a4 + y2b4 + z2c4Problem # 2For each of the following vector elds A (1) determine whether A is constant, (2) nd the divergenceand the curl A, and (3) nd the components of A in cartesian (expressed in x, y, and z), cylindrical (ex-pressed using , and z) and spherical coordinates (expressed in r, , and ). In all cases a, b, and c areconstants.(a). A = a x +b y +c zA is constant because any point in the eld will give you the same unit vector at that point.The Divergence of A is given byA =

Ax + Ay + Az

= ax + by + cz = 03The curl of A is given byA =

x y zzzza b c

= 0in cartesian coordinates the components of A are given byAx = a Ay = b Az = cin cylindrical coordinates we must use a transformation matrix to go from one basis to another, i.e

z

=

cos sin 0sin cos 00 0 1

x y z

where the 3x3 matrix is the transformation matrix that lets you represent cylindrical coordinates incartesian coordinates. Thus we nd = cos x+sin y = sin x+cos y z = zthus the components are given byA = A = acos+bsinA = A =asin+bcosAz = z A = cIn spherical coordinates we must also use a transformation matrix.

r

=

sincos sinsin coscoscos cossin sinsin cos 0

x y z

where the 3x3 matrix is the transformation matrix that lets you represent the spherical coordinates incartesian coordinates, this yields r = sincos x+sinsin y+cos z = coscos x+cossin ysin z = sin x+cos ythus we nd the components to be given byAr = r A = asincos+bsinsin+ccos A = A = acoscos+bcossincsinA = A =asin+bcos\4(b). A = a +b+c zA is not constant because as your position changes, so do and . The component is the distancefrom the z axis and the direction is always in the direction.The Divergence of A is given byA = 1(a) + 1b + cz = aThe curl of A is given byA = b zin cartesian coordinates we must use

x y z

=

cos sin 0sin cos 00 0 1

z

where the 3x3 matrix is the transformation matrix that lets you represent cylindrical coordinates incartesian coordinates. Thus we nd x = cos sin y = sin +cos z = zthus the components are given byAx = x A = acosbsinAy = y A = asin+bcosAz = z A = cin cylindrical coordinates the components of A are given byA = a A = b Az = cIn spherical coordinates we must also use a transformation matrix.

r

=

sin 0 coscos 0 sin0 1 0

z

where the 3x3 matrix is the transformation matrix that lets you represent the spherical coordinates incylindrical coordinates, this yields r = sin +cos z = cos sin z = thus we nd the components to be given byAr = r A = asin+ccosA = A =acoscsinA = A = b5(c). A = a r +b+cA is not constant with the same argument as above except that r, , and all change according to wherethe point is.The Divergence of A is given by A = 1r2l(r2a) + 1r sin(sinb) + 1r sinc= 2ar + cr cot The curl of A is given byA = 1r sin (sinc) b

r + 1r 1sina r(rc)

+ 1r r(rb) a

= cr cot r cr+ brin cartesian coordinates the components of A are given by

x y z

=

sincos coscos sinsinsin cossin coscos sin 0

r

where the 3x3 matrix is the transformation matrix that lets you represent the spherical coordinates incartesian coordinates, this yields x = sincos r +cos cos+sin y = sinsin r +cossincos z = cos r +sinthus we nd the components to be given byAx = x A = asincos+coscos+csinAy = y A = asinsin+bcossinccosAz = z A = acosbsinin cylindrical coordinates the components of A are given by

z

=

sin cos 00 0 1cos sin 0

r

where the 3x3 matrix is the transformation matrix that lets you represent the spherical coordinates incylindrical coordinates, this yields = sin r +cos = z = cos r sin6thus we nd the components to be given byA = A = asin+bcosA = A = cAz = z A = acosbsinand in spherical coordinates they are given byAr = a A = b A = cProblem # 3Show that ZVf dV =IAf dAwhere A is the area of the closed surface bounding the volume V. Hint: multiply each side by a constantvector and use calculus theorems.If we multiply both sides by a constant vector eld A we getAZVf dV = AIAf daif we work on the left-hand side, along with using equation 5 from the front of Grifths we ndZV(A f )dV =ZV ( f A)dV ZVf ( A)dVwe know that the right hand term goes to zero because the gradient of a constant vector eld is 0.Invoking the Divergence theorem on the left hand term we getZV ( f A)dV =IAf A dawe nally ndAZVf dV =IAf A daAZVf dV =IAA ( f da)AZVf dV = AIAf dathus we haveZVf dV =IAf dAProblem # 4Show that the velocity (dr/dt = r) is expressed in cylindrical coordinates as + + z z and inspherical coordinates as r r +r+r sin.7We know that the velocity can also be written as r = dxdt x + dydt y + dzdt zand we know that in cylindrical coordinatesx = cos ddt(cos) = cossiny = sin ddt(sin) = sin+ cosz = z ddt(z) = zand we also know that in cylindrical cooridinates x = cos sin y = sin +cos z = zthus we can write the velocity as r = ( cossin)(cos sin) +( sin+ cos)(sin +cos) + z z= cos2 + sin2+ sin2 + cos2+ z z= ++ z zin Spherical coordinates we can apply the same technique.and we know that in spherical coordinatesx = r sincos ddt(r sincos) = r sincos+ r coscosr sinsiny = r sinsin ddt(r sinsin) = r sinsin+ r cossin+ r sincosz = r cos ddt(r cos) = r cossinand we also know that in cylindrical cooridinates x = cos sin y = sin +cos z = zthus we can write the velocity as being (minus some algebra) r = dxdt x + dydt y + dzdt z = r r +r+r sinProblem # 5Here we will practice the use of eld differential operators and visualize some elds.(a). Find the gradient of the scalar potential (x, y, z) = xy. Provide a clear sketch of the countourlines of in the x y plane and a representation of its gradient eld. Such a eld is known as a radial8quadrupole eld, and is used in the focusing of charged particle beams and of dipolar (electric or magnetic)particles.The gradient of is given by =

x(xy) + y(xy) + z(xy)

= (y x +x y)The sketch is given as(b). Provide a clear sketch (in the x y plane) of the vector eld expressed in cylindrical coordinates asE = where is a constant. Calculate its divergence. If E is an electric eld, what charge distributiongenerates it?The gradient of E is given byE =

1(2)

= 2 The divergence of E is given by E = 29which is the direction of most change. The charge distribution that produces this charge can be foundby using Gausss law E = qenc0= 2 qenc = 20The sketch is given by(c). Consider a large sheet placed on a turntable that is set rotating at angular velocity . What vectoreld v describes the local velocity on this sheet? Provide a clear sketch of this eld. What is the curl ofthis vector eld (now expressed as a vector in a three-dimensional space, i.e including a coordinate for thez axis)? If a magnetic vector potential A is dened with the functional form of v (just a different constantout front), what magnetic eld B =A does it represent.The vector describing this system is given byv =rwhere r is the distance from the origin and is the angle between any lines originating from the origin.The curl is given byv = 1rr(r2) z = 2 zThe magnetic eld is given byB =v = 2 zThe sketch is given by10(d). Now generate new vector eld by taking your vector eld v obtained in part (c) above, and adding(1) a constant eld, and (2) a radial quadrupole eld from part (a). Provide clear sketches of these twoelds and compare with that of v. Show explicitly that the curl of these three elds is the same. Note thatthese three elds all represent forms of a magnetic vector potential that describes the same magnetic eld.We can dene a constant vector eld ask = a x+b y+c zwhich can also be represented ask = (acos+bsin) +(asin+bcos)+c zso now we can add these two vectorsv+k = (acos+bsin) +(asin+bcos+r)+c zFrom part (a) we found = (y x+x y)= (ycos+xsin) +(xcosysin)adding up these vectors yieldv+ = (ycos+xsin) +[r+(xcosysin)To nd the curl we use(v+k) =v+kthe right hand term goes to 0 and the left hand term isv = 2 z11also(v+) =v+the right hand term is also 0, so we getv =2 zwe have just showed explicitly that all the curls are the same,. For the rs sketch we can dene aconstant vector eld ask = u yand then add these two vector elds, as show in the diagram part (a), the second diagram (b) is the sumof our original vector eld added to the quadrupole eld given by (a).Problem # 6(a) Show that Zsf (A) da =Zs[A(f )] da+Ipf A dlUsing equation 7 from the front cover of Grifthss we ndZsf (A) da =Zs[( f A)] da+Zs[A(f )] dainvoking Stokes theorem on the left hand term yieldZs[( f A)] da =Ipf A dlthus we ndZsf (A) da =Zs[A(f )] da+Ipf A dl12(b) Show that ZVB (A)d =ZVA(B)d +Is(AB) daUsing equation 6 from thr front cover of Grifths we ndZVB (A)d =ZV (AB)d +ZVA (B)dUsing Gausss theorem on the left hand term we ndZV (AB)d =Is(AB) daThus we haveZVB (A)d =ZVA(B)d +Is(AB) daProblem # 7(a) Let F1 =x2 z and F2 = x x+y y+z z. Calculate the divergence and curl of F1 and F2. Which one canbe written as a gradient of a scalar? Find a scalr potential that does the job. Which one can be written as acurl of a vector? Find a suitable vector potential.The divergence of F1 is given by F1 = zx2= 0and the curl is given byF1 =

x y zxyz0 0 x2

= xx2 y =2x yThe divergence of F2 is given by F2 = xx + zyy + zz = 3and the curl is given byF2 =

x y zxyzx y z

= 0We knowthat F1 can be written as the curl of a vector potential because it is divergen-less or solenoidal,to nd the vector potential we doF1 = VWe need to nd the vector potential V, so we can nd it usingF1 =V =

x y zxyzAx Ay Az

= Ayx z = x2 z13thus Ay is given byAy = x33and the vector potential is given asV = a x+ x33 y+b zwhere a and b are constants. We know that F2 can be written as the gradient of a scalar potentialbecause it is curl-less. we nd the scalar potential by usingF2 = V= x22 + y22 + z22

thus we know that the scalar potential isV = x22 + y22 + z22

(b) Show that F3 = yz x+zx y +xy z can be written both as the gradient of a scalar and as the curl of avector. Find scalar and vector potentials for this function.First we must show that F3 is is both divergen-less and curl-less, to show divergen-less and nd thevector potential we use F3 = x(yz) + y(xz) + z(xy) = 0thus we must useF3 = V =

x y zxyzAx Ay Az

=

Azy Ayz

x

Azx Axz

y+

Ayx Axy

zSolving for Ax,, Ay, and Az we ndyz x =

Azy Ayz

x Az = zy22zx y =

Axx Azz

y Ax = yz22xy z =

Ayx Axy

z Ay = xy22Thus the vector potential isV = zy22 x+ xy22 y+ zy22 z14For the scalar potential we doF3 =

x y zxyzyz zx xy

= 0by inspection. The scalar potential is given byF3 = yz x+zx y+xy z =Vand we ndV =xyzthus the vector can be written asF3 = (xyz)Problem # 8Check the divergence theorem for the functionv = r2sin r +4r2cos+r2tanusing the volume of the ice-cream cone shown on page 56 Fig. 1.52 (the top surface is spherical,with radius R and centered at the origin).The divergence theorem statesZV vdV =IAv dathe left hand side can be solved by taking a volume integral of v. v = 1r2r(r4sin) + 1r sin(4r2sincos) + 1r sin(r2tan)= 4r sin+ 4rsin

(cos)2(sin)2

= 4r sin+4r 1sin2(sin)

= 4r(cos)2sinand the integral is given byZV vdV =Z 20dZ R04r3drZ /60(cos)2dwhich yieldsZV vdV = 2R4 12 +38Now for the area integral must look at the the cone and the surface separatelyIAv da =Z 20R4dZ /60(sin)2d+ 12Z 20dZ R0sin(/6)cos(/6)4r3dr= 2R4 1238+R4

34

= 2R4 12 +3815Problem # 9A little more practice on vector analysis: consider the vector function F(r) = where is dened asconventional for spherical coordinates.(a). Calculate the line integralIcF dlwhere C is a circle of radius in the x y plane, centered at the origin, and where the integral isevaluated with dl oriented counter clockwise.from the diagram we can see thatdl =dgiven that F(r) = we ndIcF dl =Z 20 (d) =Z 20d = 2(b). Calculate the surface integralZH(F) dawhere the surface H is a hemisphere that is above and bounded by the curve C used for part (a). Thesurface integral is calculated with da oriented outward.from the diagram we can see thatda = 2sindd rand to nd what the curl of F we doF = 1r sin (sin)

r 1rr(r)

= cot r r 1rin our case r = thus we haveF = 1[cot r ]thus the integral is given byZH(F) da =ZH1[cot r ] 2sindd r = Z /20cosdZ 20d = 2(c). Calculate the surface integralZD(F) dawhere the surface D is now the disk in the x y plane that is bounded by the curve C used for part (a).The surface integral is calculated with da oriented upward.from the diagram we can see thatda =drd z = drd(cos r sin) = 2and to nd what the curl is we doF = 1r sin (sin)

r 1rr(r)

= cot r r 1r16in our case r = thus we haveF = 1[cot r ]thus the integral is given byZD(F) da =ZD1[cot r ] drd(cos r sin)=Z 20Z 0cos2+sin2sin

drd=Z 20dZ 0dr = 2(d). Verify that Stokes theorem holds for both surfaces H and D.Stokes theorem states I F dl =ZS(F) da = 2thus Stokes theorem holds.Problem # 10Let H(r) = x2y x+y2z y+z2x z. Find an irrotational function F(r) and a sonisoidal function G(r) suchthat H = F+G.We know that an irrotational function can be described asF = 0and a sonisoidal function is described as G = 0thus we know thatH =F+G = Gand we know thatH =

x y zxyzx2y y2z z2x

=y2 xz2 yx2 zthus we can nd G byG =

x y zxyzGx Gy Gz

=

Gzy Gyz

x

Gzx Gxz

y+

Gyx Gxy

zlooking at the following terms we ndGzy =y2 Gxz =z2 Gyx =x217thusGz =y33 Gx =z33 Gy =x33thus we nd G to beG =z33 xx33 yy33 zsince we know what H and G are we can easily nd FF = HG =

x2y + z33

x+

y2z + x33

y+

z2x + y33

zand it is seen thatF = 0 G = 0and by construction we know thatH = F+G18Chapter 2Electrostatics2.1 The Electric Field2.1.1 Coulombs LawThe fundamental problem electromagnetic theory hopes to solve is this: We have some electric chargesq1, q2, q3,... (call them source charges); what force do they exert on another charge, Q (call it the testcharge). The force on the test charge Q due to a single point charge q which is at rest a distance r awayis given by Coulombs LawF = 140qQr2 r (2.1)The cosntant 0 is called the permitivity of free space. In SI units, where force is in Newtons (N), distancein meters (m), and charge in Coulombs (C)0 = 8.851012 C2N m2In words, the force is proportional to the product of the charges and inversely proportional to the squareof the seperation distance. As alwaysr = r rwhere r is the seperation vector from the location of q to Q, r is the magnitude and r is the direction.The force points along the line from q to Q; it is repulsive if q and Q have the same sign, and attractive iftheir signs are different. Coulombs law and the principle of superposition constitute the physical input forelectrostatics- the rest, except for some special properties of matter, is mathematically elaborationof thesefundemental rules.2.1.2 The Electric FieldIf we have several point charges q1, q2, ..., qn at distances r1, r2, ..., rn from Q, the total force on Q isevidentlyF =F1 +F2 +... = 140

q1Qr21 r1 + q2Qr22 r2 +..

orF = QE19whereE(r) 140ni=1qir2i ri (2.2)E is called the electric eld of the source charges.2.1.3 Continuous Charge DistributionsOur denition of the electric eld Equation 2.2 assumes that the source of the eld is a collection ofdiscrete point charges qi. If, instead, the charge is distributed continuously over some region, the sumbecomes an integralE(r) = 140Z 1r2 rdqIf the charge is spread out along a line, with charge-per-unit length , then dq = dl (where dl is anelement of length along the line); if the charge is smeared out over a surface, with charge-per-unit are, the dq = da, (where da is an element of area on the surface) and if the charge lls a volume, withcharge-per-unit volume then dq = d (where d is an element of volume)dq dldadthus the electric eld of a line charge isE(r) = 140Z (r)r2 rdlfor a surface chargeE(r) = 140Z (r)r2 rdaand for a volume chargeE(r) = 140Z (r)r2 rd (2.3)Equation 2.3 itself is often referred to as Coulombs law, because it is such a short step from the original,and because a volume charge is in a sense the most general and realistic case.2.2 Divergence and Curl of Electrostatic Fields2.2.1 Field lines, Flux, and Gausss LawThe ux of E through a surface S is given asE ZSE dait is a measure of the number of eld lines passing through S. This suggests that the ux through anyclosed surface is a measure of the total charge inside. This is in essence Gausss Law. In the case of apoint charge q at the origin, the ux E through a sphere of radius r isI E da =Z 140

qr2 r

(r2sindd r = q020For any closed surface, thenI E da = 10Qencwhere Qenc is the total charge enclosed within the surface. As it stands, Gausss law is an integralequation, but we can readily turn it into a differential one, by applying the divergence theoremISE da =ZV( E)drewriting Qenc in terms of the charge density , we haveQenc =ZVdso Gausss law becomes ZV( E)d =ZV

0

dand since this holds for any volume the integrand must be equal E = 0(2.4)Equation 2.4 is Gausss law in differential form.2.2.2 The Curl of EIll calculate the curl of E, as I did the divergence by studying the simplest possible conguration, a pointcharge at the origin. In this caseE = 140qr2 rwhat if we calculate the line integral of this eld from some point a to some other point bZ baE dlin spherical coordinates, dl = dr r +rd+r sind, soE dl = 1401r2drthereforeZ baE dl = 140

qra qrb

where ra is the distance from the origin to point a and rb is the distance to point b. The integral around aclosed path is evidently zero (for then ra = rb)I E dl = 0and hence applying Stokes theoremE = 0 (2.5)Equation 2.5 holds for any static charge distribution.212.3 Electric PotentialThe electric eld E is not just any old vector; it is a special kind of vector function, one whose curl isalways zero. There is a theorem that asserts that any vector whose curl is zero is equal to the gradient ofsome scalar. Because the line integral is indepenent of path, we can dene a functionV(r) Z rOE dl (2.6)this is called the electric potential. This allows us to write the electric eld asE =V (2.7)Equation 2.7 is the differential version of Equation 2.6, it says that the electric eld is the gradient of ascalar potential. If you knowV, you can easily get E by just taking the gradient of V.2.3.1 Poissons Equation and Laplaces EquationWe found that the electric eld can be written as the gradient of a scalar potential. The question arises:What do the fundamental equations for E, E = 0E = 0look like, in terms of V? Well. E = (V) = 2V, so apart from the persisting minus sign, thedivergence of E is the Laplacian of V. Gausss law then says2V =0(2.8)this is known as Poissons equation. In regions where there is no charge, so that =0, Poissons equationreduces to Laplaces equation2V = 02.3.2 The Potential of a Localized Charge DistributionI dened V in terms of E. Ordinarily, though it is E that were looking for. The idea is that it might beeasier to get V rst, and the calculate E by taking the gradient. In general, the potential of a point chargeq isV(r) = 140qror for continuous distributionV(r) = 140Z 1rdqin particular, for a volume chargeV(r) = 140Z (r)r dthis is the equation we are looking for, telling us how to compute V when we know ; it is, if you like, thesolution to Poissons equation, for a localized charge distribution.222.3.3 Summary; Electrostatic Boundary ConditionsThe electric eld always undergoes a discontinuity when you cross a surface charge . In fact, it is asimple matter to nd the amount by which E changes at such a boundary. Gausss law states thatISE da = 10Qenc = 10Awhere A is the area of a pillbox used on the surface. Now the sides of the pillbox contribute nothing to theux, in the limit as the thickness goes to zero, so we are left withEaboveEbelow = 0Conclusion: The normal component of E is discontinuous by an amount /0 at any boundary. In par-ticular, where there is no surface charge, E is contnuous, as for instance at the surface of a uniformlycharged solid sphere. The tangential component of E, by contrast, is always contnuous.E

above = E

belowthe boundary conditions on E can be combined into a single formulaEaboveEbelow = 0 nwhere n is a unit vector perpendicular to the surface, pointing from below to above. The potential,meanwhile, is continuous across any boundary, sinceVaboveVbelow =Z baE dlas the path length shrinks to zero, so too does the integralVabove =VbelowHowever, the gradient of V inherits the discontinuity in E; since E =V, impliesVaboveVbelow =0 nor more convenientlyVaboven Vbelown =0whereVn = V ndenotes the normal derivative of V (that is, the rate of change in the direction perpendicular to thesurface).232.4 Work and Energy in Electrostatics2.4.1 The Work Done to Move a ChargeSuppose you have a stationary conguration of source charges, and you want to move a test charge Q frompoint a to point b. How much work will you have to do? The work is dened asW =Z baF dl =QZ baE dl = Q(V(b) V(a)]notice that the answer is independent of the path you take from a to b; in mechanics, then, we would callthe electric force conservative. Dividing through by Q, we haveWQ =V(b) V(a)in words, the potential difference between two points is equal to the work-per-unit charge required to carrythe particle between the points. In particular, if you want to bring the charge Q in from far away and stickit at a point r, the work you must do isW = Q[V(r) V()]so if you have set the reference point out at innityW = QV(r)2.4.2 The Energy of a Point Charge DistributionHow much work would it take to assemble an entire collection of point charges? Imagine bringing in thecharges, one by one, from far away. The work needed to assemble a conguration of point charges is givenbyW = 12ni=0qiV(ri) (2.9)this is also the amount of work youd get back out if you dismantled the system. In the mean time itrepresents the energy stored in the conguration.2.4.3 The Energy of a Continuous Charge DistributionFor a volume charge density , Equation 2.9 becomesW = 12Z Vd(The corresponding integrals for line and surface charges would be R Vdl and R Vda, respectively. Thiscan also be written asW = 02ZspaceE2dThe energy can be thought of as being stored in the eld or stored in the charge. We can also say02 E2= energy per unit volume24Because electrostatic energy is quadratic in the eld, it does not obey a superposition principle. The energyof a compound system is not the sum of the energies of its parts considered seperately, there are also crosstermsWtot =W1 +W2 +0Z E1 E2dfor example, if you double the charge everywhere, you quadruple the total energy.2.5 Conductors2.5.1 Basic PropertiesIn an insulator, such as glass or rubber, each electron is attached to a particular atom. In a metallicconductor, by contrats, one or more electrons per atom are free to roam about at will through the material.A perfect conductor would be a material containing an unlimited supply of completely free charges. Thereare some general rules when dealing with conductors E=0 inside a conductor = 0 inside a conductor any net charge resides on the surface A conductor is an equipotential E is perpendicular to the surface, just outside a conductor2.5.2 Surface Charge and the Force on a ConductorBecause the eld inside a conductor is zero, boundary conditions requires thatb the eld immediatelyoutside isE = 0 nconsistent with our earlier conclusion that the eld is normal to the surface. In terms of potential =0Vnthese equations enable you to calculate the surface charge on a conductor, if you can determine E or V. Inthe presence of an electric eld, a surface charge will, naturally experience a force; the force per unit area,f, is E. But theres a problem here, for the electric eld is discontinuous at a surface charge, so whichvalue are we supposed to use? The answer is that we should use the averagef = Eaverage = 12(Eabove +Ebelow)the force per unit area is also given asf = 1202 nthis amounts to an outward electrostatic pressure on the surface, tending to draw the conductor into theeld, regardless of the sign of . Expressing the pressure in terms of the eld just outside the surfaceP = 02 E2whic is identical to the energy-per-unit volume stored in the elds.252.5.3 CapacitorsSuppose we have two conductors, and we put charge +Q on one and Q the other. Since V is constantover a conductor, we can speak unambiguously of the potential difference between themV =V+V =Z +E dlWe dont know how the charge distributes itself over the two conductors, and calculating the eld wouldbe a mess, if their shapes are complicated, but this much we know: E is proprtional to Q. Since E is pro-portional to Q, so also is V. The constant of proportionality is called the capacitance of the arrangement:C = QVCapacitance is a purely geometrical quantity, determined by the sizes, shapes, and seperation of the twoconductors.2.6 Problems with SolutionsProblem # 1Electrostatics with cylindrical symmetry(a). Find the electric potential V(z) a distance z above the center of a circular loop of radius R whichcarries linear charge density . Obtain an expression correct for all values of z (not just z > 0).We know thatV = 140Z dqrsince we know that dq is given bydq = ds = Rd r =

R2+z2thus the above expression for the potential isV = 140Z 20RdR2+z2 = 20RR2+z2(b). Find the electric potential a distance z above the center of a circular disk of radius R, which carriesa uniform surface charge .26We know thatdq = da = 2rdrputting this into the expression for the potential we ndV = 140Z dqr =V = 140Z R02rz2+r2drusing a u substitution we nd the solution to the integral to which is given byu = z2+r2du = 2rdr (r = 0 : u z2, r = R : u z2+R2)which gives usV = 40Z z2+R2z2u1/2du = 20

z2+R2z

(b). Using the potential and the symmetry of the problem, determine the electric eld a distance zabove the center of the circular disk for part b. Check that your answer makes sense: What do you getfor |z| R and for z R: Show also that he magnitude of the eld at z is proprtional to the solid anglesubtended by the disk as seen from the point z.We know that the electric eld can be dened asE =V = 20z

z2+R2z2

z = 20 zz2+R21

zFor R z we can use a Taylor expansion(1+x)n1+nxso we ndE = 20 zz2+R21

z = 2

1+ R2z2

1/21 z 40R2z2 zand we know thatq =R2z = Rthus at R z we should see a point charge i.eE = q40R2 z27For R z we can use a Taylor expansion(1+x)n1+nxso we ndE = 20 zz2+R21

z =zR

1+ z2R2

1/21 2 z 2012z3R3 zR +1

zthis is only for R z.To show also that the magnitude of the eld at z is proprtional to the solid angle subtended by the diskas seen from the point z. We will say=Z 20dZ 0sind =2cosbut the limits of are given by0 and = cos1

zz2+R2

which gives us = 2z zz2+R21

andE = 40= constantthusE = kProblem # 2Two spheres, each of radius R carrying uniform charge densitys + and , respectively, are placedso that they partialy overlap. Call the vector from the positive center to the negative center d. Show thatthe eld in the region of overlap is constant, and nd its value.We can solve this problem by nding the electric elds for 1 sphere and adding the spheres usingsuperposition. It is well known that the electric eld of a uniform sphere is given byI E da = qtot0qtot =(r)V = 43(r)r3since the electric eld is constant we ndEA = 43(r)r3 10E = (r)r30 rso we nd that the two spheres are given byE+ = (r)r+30 r E =(r)r30 r28and so the total electric eld is just the sum of these two (superposition), we also know that the vectorsare pointing towards each other and that the vertical components of the vectors cancel by symmetry. Wecan nd the total electric eld asEt = E+ +E =(r)r+30(r)r30

r = (r)30[r+r] rbut since we know that the total distance between this two vectors will always be the same we canwrite the r vector as rwhich would give you just give you d (which is constant) and thus your totalelectric eld would be constant. i.ed =r+ +r = r+rProblem # 3Consider the so-called screened Coulomb potential of a point charge q that arises, for example, inplasma physicsV(r) = q40er/rwhere is a constant (called the screening length)(a). Determine the charge distribution (r) that produces this potential. Sketch this function in amanner that clearly describes all of its characteristics (i.e whats the best way of representing this three-dimensional charge distribution? Use it and explain what your potting.We can solve this problemtwo ways, you can take the Laplacian of a Laplacian or just use the followingidentities2V = (( f g)) = ( f g+gf )and we also know the following relationshipsE =V E = (V) = (r)0thus we know that(r) =0( E) =0( (V))so we need to nd what the divergence of the electric eld is, by nding the gradient of the potentialrst and taking the divergence of thisV = r

q40er/r

= q40r

er/r

=er/r2

1+ r

rthus we ndE = er/r2

1+ r

rlettingf = er/

1+ r

and g = rr2we nd E = er/

1+ r

rr2 + rr2er/

1+ r

29but we know from Grifth page 50 that

rr2

= 43(r)so we nd E = er/

1+ r

43(r) + rr2er/

1+ r

= er/

1+ r

43(r) + 1r2r

er/

1+ r

= q40er/

1+ r

43(r) er/r2thus the charge density is given by(r) = q4er/

1+ r

43(r) er/r2The plot is given bywhere the plot is a countour map that describes the distribution with respect to the radius.(b). Show by explicit calculation over (r) that the net charge represented by this distribution is zero.Explain how this observation is easily derived using the integral form of Gauss law.the total charge Q can be found byQnet =ZV(r)d = q4Zspaceer/

1+ r

43(r)d Zspaceer/r2 dwhich gives usZV(r)d = q444Z 0rer/2 drwhere the term on the right hand side is just the Gamma function and the delta function integral is just4 by denition. So we nd that the total charge is zeroZ 0rer/2 dr = 130Q(tot) =Zspace(r)d = q4[44] = 0this tedious calculation could have been solved by using the divergence theorem and Gausss law. i.eZV( E)dV =IsE da = qenc0but we know tZV( E)dV =Z

(r)0

dV = 0 =I E da = qenc0Since we now know that we have a +q charge due to the delta function and a -q charge from the integralof the potential contained within a surface S then the surface has the same number of eld lines comingin that are going out and thus the surface integral goes to zero because the divergence goes to zero thusimplying that qenc is zero.Problem # 4(a). Show that the electrostatic energy of a uniformly charged solid sphere, with total charge Q andradius R is 3Q2/(200R).It is assumed that we know what the electric eld is in the inside of the sphere and on the outside ofthe sphere is. i.eEin = q40R3r r Eout = q40r2 rand using the following expression for the electrostatic energyW = 02ZspaceE2dthus we must consider two integrals, the contribution from the inside and the contribution from theoutside. Knowing the limits of integration we ndW = q280Z R0r4R6dr +Z R1r2dr

= q280 15R + 1R

W = 3q2200R(b). Use the result above to compute the electrostatic energy of an atomic nucleus with Z protons and atotal of A nucleons, using an approximation for the nuclear radius as R = (1.21015m)A1/3. Give yourresults in units of MeV times Z2/A1/3this is just a plug and chug problem, gure out your constants and plug away, i.eW = 3(Zq)2200(1.21015m)A1/3 0.720MeV Z2A1/3(c). Calculate the change of electrostatic energy when a uranium nucleus undergoes ssion.31We know that U235decays asU2359236 Kr +14156 Ba+3nthus we know thatW =WreacWu =|(Wkr +WB) WuthusW = 0.720MeV 362921/3 + 5621411/3

9222351/3

=347.02MeVwhich is a large amount of liberated energy.Problem # 5Evaluate the following integrals(a). Zspace(r2+r a+a2)3(r a)dwhere a is a xed vector and a is its magnitude.We know that the delta function is zero everywhere except at r = a thus we must evaluate this integralat a i.e Z f (r)3(r a)d = f (a)thus this yieldsZspace(r2+r a+a2)3(r a)d = 3a2(b). ZV|r b|23(5r)dwhere V is a cube of side 2, centered on the origin, and b = 4 y+3 z.We know that we must evaluate this function at r = 0, which yieldsZV|r b|23(5r)d =| b|2= 25(c). ZV(r4+r2(r c) +c4)3(r c)dwhere V is a sphere of radius 6 about the origin, and c= 5 x+3 y+2 z.We know that the delta function is zero everywhere excpet for where it is dened, so this means if thelimits of the integral are not within the specied value at which the delta function is equal to 1 then theintegral is zero.c =

52+32+22=38 = 6.16 6.16 > 6.0thus this implies that the magnitude of c is outside of the range and soZV(r4+r2(r c) +c4)3(r c)d = 032(d). ZVr(dr)3(e r)dwhere d = (1, 2, 3),and e = (3, 2, 1), and V is a sphere of radius 1.5 centered at (2,2,2).From our experience with delta function we now know that this will be given byZVr(dr)3(e r)d = e (de) = 4Problem # 6The farad is actually an enormous unit of capacitance. To illustrate this, treat the Earth as a conductingsphere and nd its capacitance.++QR2R1EEEQ+The above gure describes the earth capacitance problem, Where R1 is the radius of the earth and alsoR2R1 and also a positive charge Q on the outside surface and a negative charge Q for the earth.We know that the capacitance is dened asC = QVwhere Q is the charge and V is the potential. The potential is given asV =Z R2R1E dr = 140Z R2R1Qr2(cos)dr = 140 1R2 1R1

because the electric eld points in the r direction we must consider the cos term. Thus we ndV = Q40 1R2 1R1

and so the capacitance is given byC =R1R240R2R1R2R1 R1 = 6, 378 km33so we ndC = R140 = 700 FProblem # 7Consider a system of n conductors that are prepared with total charges of Qi , with i{1, 2, 3, . . .n} labeling the conductors. The surface charge densities on each of the conductors are described by thefunctions i , for which, clearlyQi =ZSiidaiwith the integral being over the surface Si of conductor i.(a). Find an expression for the potential on conductor i. This you may obtain as usual by summingover the contributions to the potential from all the surface charges on all the conductors (including thei th conductor).Since we know that there ith and jth conductors, to nd the potential of the ith conducter from all thejth conductors, we must sum over all the jth conducters contribution to the potential, i.eVi = 140ZSjQjri j= j=1ZSjjdajri jwhere ri j is the distance from the ith conducter to the jth conductor.(b). Let us dene a coefcient pi j aspi j = 140SjZSjfidajri jHere, ri j denotes the distance from one point on conductor i to the point on conductor j at which theintegrand is evaluated (as we integrate over the surface of j ). In this integral, we make us of a relationi =j fj = QjSjfjin which Sj is now the area of the jth conductor. The dimensionless function fj describes how the actualcharge density differs from the average. With this denition, show that the following set of equations hold:V1 = p11Q1 + p12Q2 +... + p1nQnV2 = p21Q1 + p22Q2 +... +P2nQn..Vn = pn1Q1 + pn2Q2 +... +PnnQnThe coefcients pi j are known as coefcients of potential. They express how the potential on conductori will vary (linearly) upon changing the charge Qj on conductor j . We note here that the coefcients nolonger contain any reference to the specic potentials or charges placed on the conductors - they simplyreect geometric properties of the conductors and their placements.(c). By considering the quantity pi jQiQj , show that pi j = pji. This nice relation expresses the fol-lowing fact: if a charge Q on conductor i brings that conductor to a potential V , then the same charge Qplaced on conductor i would bring conductor j to the same potential V.34using the fact that these matrix elements are symmetric, nn, and diagonalizable, we can use a linearalgebra by invoking the theory of matrix multiplication, i.eni=1(AB)ii =ni=1nj=1Ai jBji =nj=1ni=1BjiAi j =nj=1(BA)j jthis can be shown bypi jQiQj = QiQj40SjZSjfjdajri j= QjQi40SiZSifidairji= pjiQjQithis comes from the symmetry of the problem, i.e whatever conductor i sees will be exactly the sameif conductor i now became conductor j.(d). Consider a situation with only two conductors that are used as a capacitor. What is the capacitancefor this system expressed in terms of the coefcients of potential p11 , p12 and p22 ?if we know thatV1 = p11Q1 + p12Q2V2 = p21Q1 + p22Q2we also know thatp12 = p21thus we can nd the potential by usingV =V2V1 = (p12p11)Q1 +(p22p12)Q2and we we assume that Q1 =Q2 we ndC = Q1V = 1p11 + p12 + p22Problem # 8Suppose that a parallel plate ca-pacitor has rectangular plates but the plates are not exactly parallel.The separation at one edge is d a and d +a at the other edge, where a d. Show that the capacitance isgiven approximately byC 0Ad

1+ a23d2

where A is the area of the plate. A gure is given by35Thus we know from the diagramA = 4lwand knowing from Gausss lawI E da = dA20which give usEdA = dA20E = 20( z)to nd the potential we must useV =Z E dland knowing thatdl = zdswhere ds is in the direction of the eld. Thus we nd the potential to be given byV =Z 20( z) ( zds) = 20Swhich S is the the distance between the plane that is a function of x which is given by the equation ofthe line in the formS = d + awxthus we now know that the capacitance is given by and qenc = dAC =Z ww20dA

d + awx =Z ww20lZ wwdx

d + aw

using a u substitution, in which we will just qoute the solution. We know how to do u substitutions atthis levelC = 20lwa

ln

d

1+ ad

ln

d

1ad

which if we do a bit of algebra with a knowledge of how the natural log functions work along with theTaylor expansion for the natural log function which isln(x +1) x x22 + x33 ...we ndC = 20lwaa2 a22d2 + a33d3ad a22d2 a33d3

the even terms cancel and thus to third order we nd, with a bit of algebraC = 2lw01a2ad + a33d3

= 4lw01d1+ a23d2

= A0d1+ a23d2

Problem # 9(a) Two grounded conducting planes meet at an angle of = 600at the origin. A point charge q isa distance r0 from the origin along their angular bisector. What image charges are needed to satisfy theboundary conditions? (Hint: their arrangement is going to be highly symmetric)36We can see from the diagram will require 6 total charges and 5 image charges. This is necessary tomeet the boundary condition implied by the use of images. We can see that there is strong dependence inthe angle between the plates. You must balance the charges. where the total charge is zero.What is the r depedence of the potential along the angular bisector for r r0?We can nd the r dependence is given byV = q40 1r1 1r2 1r3+ 1r4+ 1r5 1r6

if we pick an arbitrary point (0, 0, z) we can nd a relationship for the distance between these chargesand that point. The diagram is given byknowing that the distance between each pair of charges is the same because it is a equilatiral triangleand by symmetry we ndr4 = r5 and also r2 = r3and also from the diagram we can seer1 = r R0 r6 = r +R0Thus we nd that the potential is given byV = q40 1r R0 2r2+ 2r4 1r +R0

37combining the rst term and the last term gives usV = q40 2R0r2+R20 2r2+ 2r4

From the diagram we can also infer thatr22 = 34R20 +

r R02

2and by symmetry we know thatr24 = 34R20 +

r + R02

2thus we nd that the r dependence on the potential is given byV = q40

2R0r2+R20 2

34R20 +

r R02

2 + 2

34R20 +

r + R02

2(b) Repeating the above problem but for different angles between the grounded plates. Show that if = 2/n, where n is any even integer, then n 1 image charges are required to ground both sides of thewedge. Show that if, however, n is an odd integer (e.g. n = 3) then an image solution cannot be obtained.We can see that there is a dependence on n by36090 = 4 (41) = 3 360180 = 2 (21) = 1 36060 = 6 (61) = 5from these three examples we can see that2 = n total number of chargesandn1 image chargeswe can also see that must be even to allow an even number of charges that to be used in order tosatisfy the boundary conditions.Problem # 10A conductor is formed with a grounded conducting plane spanning the x y axis (z = 0). Upon it isattached a hemispherical conductor of radius R0 that extends into the half-plane z > 0 and whose center isat the origin. A charge q is placed at the point (x, y, z) = (0, 0, z) with z > R0 .38(a). What is the electrostatic potential V(r, ) with r and being conventionally dened sphericalcoordinates (by symmetry, the potential must be independent of the azimuthal angle )? Hint: there willbe three image charges. Answer:1 z2R20z

z20 +R20A diagram is given byThus we know thatV(r, ) = q40 1r1 1r2+ 1r3 1r4

Using geometry and symmetry, along with the the example in Grifthss we nd that the relationshipsfor the all the radius are given asr12= r2sin2+(r cosz)2r22= r2sin2+

r cosR20z

2and by symmetry we nd that the other two radius are given byr32= r2sin2+

r cos+ R20z

2r42= r2sin2+(r cos+z)2so we know that the potential is given by after cleaning it up a bitV(r, ) = q40

1r2+z22rzcos R0z

r22R20r cos z + R40z2+ R0z

r2+ 2R20r cos z + R40z2 1r2+z2+2rzcosand we know that the total charge distribution induced on the plane is given byp =01rdVd39and so we nd that the derivative of the potential with respect to is given asdVd = q

rR30sinz2

r22R20r cos z + R40z2

3/2 + rR30sinz2

r2+ 2R20r cosz + R40z2

3/2 rzsin(r2+z22rzcos)3/2 q rzsin(r2+z22rzcos)3/2thus we know thatp = q4

R30sinz2

r22R20r cos z + R40z2

3/2 + R30sinz2

r2+ 2R20r cos z + R40z2

3/2 zsin(r2+z22rzcos)3/2

q4

zsin(r2+z22rzcos)3/2and since we are integrating over the plane this means = 2 and da = r sinddr andqp =Zspacedawhich will give usqp = q4Z 20dZ R0r

R30z2

r2+ R40z2

3/2 + R30z2

r2+ R40z2

3/2 z(r2+z2)3/2 z(r2+z2)3/2which becomesqp =q2Z R0R30rdrz2

r2+ R40z2

3/2 + q2Z R0zrdr(r2+z2)3/2solving this integral using u substitution yieldsqp =q

z2(2) 1

R20 +z2R30(2)2z21

R20 + R40z2simpyfying this term yieldsqp =q

z

R20 +z2R20z2z

z2+R20=q z2+R20z

z2+R20and knowing thatqp +qh =q qhq = 1 qpqthusqhq = 1 z2+R20z

z2+R2040Chapter 3Special Techniques3.1 Laplaces EquationThe primary task of electrostatics is to nd the electric eld of a given stationary charge distribution. Inprinciple, this purpose is accomplished by Coulombs law, in the form of Equation 2.3E(r) = 140Z rr2(r)dUnfortunately, integrals of this type can be difcult to calculate for any but the simplest charge congura-tions. We can attempt to solve for the potential usingV(r) = 140Z 1r(r)d (3.1)Still, even this integral is often too tough to handle analytically. In such cases it is fruitful to recast theproblem in differential, using Poissons equation2V = 10which, together with appropriate boundary conditions, is equivalent to Equation 3.1. Very often in fact, weare interested in nding the potential in a region where = 0. There may be plenty of charge elsewhere,but were conning our attention to places where there is no charge. In this case Poissons equation reducesto Laplaces equation2V = 0or, written in cartesian coordinates2Vx2 + 2Vy2 + 2Vz2 = 0this formula is so fundamental to the subject that one might almost say electrostatics is the study ofLaplaces equation. At the same time, it is a ubiquitous equation, appearing in such diverse branches ofphysics as gravitation and magnetism, the theory of heat, and the study of soap bubbles.3.1.1 Boundary Conditions and Uniqueness TheoremsLaplaces equation does not by itself determine V; in addition, a suitable set of boundary conditions mustbe supplied. The proof that a proposedset of boundary conditions will sufce is usually presented in the41form of a uniqueness theorem. There are many such theorems for electrostatics, all sharing the samebasic format-Ill show you the two most useful onesFirst uniqueness theorem: The solution to Laplaces equation in some volume V is uniquely deter-mined if V is specied on the boundary surface S.Proof: Imagine a region and its boundary, there could be islands inside, so long as V is given on alltheir surfaces; also the outer boundary could be at innity, where V is ordinarily taken to be zero. Supposethere were two solutions to Laplaces equation2V1 = 0 2V2 = 0both of which assume the specied value on the surface. I want to prove that they must be equal. The trickis to look at the differenceV3 =V1V2This obeys Laplaces equation2V3 = 2V12V2and it takes the value zero on all the boundaries (since V1 and V2 are equal there). But Laplaces equationallows no maxima or minima-all extrema occur on the boundaries. So the maximum and minimum of V3are both zero. Therefore V3 must be zero, and henceV1 =V2The uniqueness theorem is a license to your imagination. It doesnt matter how you come to your solution;if a) it satises Laplaces equation and (b) it has the correct values on the boundaries, then its right.3.1.2 Conductors and the Second Uniqueness TheoremThe simplest way to set boundaries for an electrostatic problem is to specify the value of V on all thesurfaces surrounding the region of interest. And this situation often occurs in practice. In the laboratory,we have conductors connected to batteries, which maintain a given potebtial, or groundm which is theexperimentalists word for V = 0.Second uniqueness theorem: In a volume V surrounded by conductors and containing a speciedcharge density , the electric eld is uniquely determined if the total charge on each conductor is given.(The region as a whole can be bounded by another conductor, or else unbounded.)3.2 Multipole Expansion3.2.1 Approximate Potentials at Large DistancesIf you are very far away froma localized charge distribution, it looks like a point charge, and the potentialis, to good approximation, (1/40)Q/r, where Q is the total charge. We can expand the potential asV(r) = 140n=01r(n+1)Z (r)nPn(cos)(r)d (3.2)or more explicitelyV(r) = 1401rZ (r)d+ 1r2Z rcos(r)d+ 1r3Z (r)2

32 cos212

(r)d+...

42this is the desired result- the multipole expansion of V in powers of 1/r. The rst term n = 0 is themonopole contribution. The second (n = 1) is the dipole term; the third is the quadrupole term; thefourth is the octopole term and so on. As it stands Equation 3.2 is exact, but it is useful primarily as anapproximation scheme: The lowest nonzero term in the expansion provides the approximate potential atlarge r, and succesive terms tell us how to improve the approximation if greater precision is required.3.2.2 The Monopole and Dipole Terms3.3 Problems and SolutionsProblem # 1Consider a cube made of 6 conducting plates of size aa , that encloses the space a/2 x, y, z a/2.The plates at z =a/2 are held at potentials V0 , respectively, while the other four sides are all grounded(V = 0).(a). Solve the boundary value problem for V(x, y, z) inside the cube.First we must solve the Laplace equation2V = 2Vx2 + 2Vy2 + 2Vz2 = 0As always, we look for solutions that are productsV(x, y, z) = X(x)Y(y)Z(z)Plugging this into the Laplace equation we nd1Xd2Xdx2 + 1Yd2Ydy2 + 1Zd2Zdz2 = 0thus we know that1Xd2Xdx2 = c11Yd2Ydy2 = c21Zd2Zdz2 = c3which implies thatc1 +c2 +c3 = 0we can now solve this second order differential equation. One of these coefcients must be negative,this can only be c3 because we know that this must be an oscillatory solution. In this case we ndc3 =(c1 +c2) c1 = l2c2 = k2c3 =(l2+k2)and sod2Xdx2 = l2X d2Ydy2 = k2Y d2Zdz2 =(l2+k2)Z43the solutions to these differential equations areX(x) = Asin(lx) +Bcos(lx)Y(y) = Csin(ky) +Dsin(ky)Z(z) = Eel2+k2z+Fel2+k2zBoundary conditionsour boundary conditions are given as(i) V = 0 when y =a/2(ii) V = 0 when x =a/2(iii) V = V0 when z =a/2from (i) we can see thatY(a/2) =Csin

la2

+Dcos

la2

= 0 Y(a/2) =Csin

la2

+Dcos

la2

= 0and from (ii) we can see thatX(a/2) = Asin

ka2

+Bcos

ka2

= 0 X(a/2) =Asin

ka2

+Bcos

ka2

= 0from the rst and second boundary condition we ndC = 0 A = 0and alsol = na k = ma l2= n22a2 k2= m22a2where n and m are positive integers. Since the given boundary conditions imply an odd potential wecan see thatZ(z) =Z(z) =

El2+k2z+Fel2+k2z

=

Eel2+k2z+Fel2+k2z

and so we nd thatF =Eand this now givesZ(z) = E

el2+k2zel2+k2z

= E2sinh(

l2+k2z)combining the remaining constants we are left withV(x, y, z) =Ccos

na x

cos

ma y

sinh(

l2+k2z)where C has absorbed the other constants, we can also write this asV(x, y, z) =Ccos

na x

cos

ma y

sinh

a

n2+m2z

44we can now write this as a double sum over the integers n and mV(x, y, z) =n=1m=1Cn,msinh

a

n2+m2z

cos(nx/a)cos(my/a)and nally the last boundary conditions (iii) gives usV(x, y, a/2) =n=1m=1Cn,msinh

2

n2+m2

cos(nx/a)cos(my/a) =V0 (3.3)to determine the constant Cn,m we multiply both sides by cos(nx/a)cos(my/a) where n and m arepositive integers and integrate, this is known as Fouriers trick. For the left hand side we ndnmsinh

2

n2+m2

Cn,mZ a/2a/2cos

na x

cos

na x

dxZ a/2a/2cos

ma y

cos

ma y

dywe also know from Fouriers trick and orthogonality that n = n m = m and so we getsinh

2

n2+m2

Cn,mZ a/2a/2cos2

na x

dxZ a/2a/2cos2

ma y

dy =Cn,msinh

2

n2+m2

a24and now for the right hand side of equation 1 we ndV0Z a/2a/2cos

na x

dxZ a/2a/2cos

ma y

dy =V04a2nm2 sin

na

sin

ma

and nally we can put these two solutions together to getCn,msinh

2

n2+m2

a24 =V04a2nm2 sin

na

sin

ma

this yieldsCn,m = 16V0nm21sinh

2n2+m2 for n and m oddif n and m are even then Cn,m is zero. We can now write the potential asV(x, y, z) = 16V02 n,m=1,3,51nm cos

nxa

cos

ma y sinh

an2+m2z

sinh

2n2+m2

(b). The potential is obviously zero at the center of the cube. What is the electric eld there?To nd the electric eld we use

E =V = Vx + Vy + Vz45and since we are evaluating the electric eld at the origin we know thatVx = Vy = 0this leaves us with

E(x, y, z) = Vz =16aV0 n,m=1,3,51nm cos

nxa

cos

ma y cosh

an2+m2z

sinh

2n2+m2

n2+m2 zand so the electric eld at the origin is given by

E(0, 0, 0) = 16V0a n,m=1,3,51nmn2+m2sinh

2n2+m2 zProblem # 2(a). Suppose that on the surface of a sphere of radius R there is a surface charge density () =0cos2, where 0 is a constant. What is V(r, ) inside and outside the sphere? What is the electric eldE(r, ) inside and outside the sphere? Conrm that the eld is discontinuous at the surface of the spherein accord with the given charge distribution.From equation 3.65 in Grifths we know thatV(r, ) =l=0

Alrl+ Blrl+1

Pl(cos) (3.4)this is the general form of a potential in which we assumed azimuthal symmetry in spherical coordi-nates, this was derived in Grifths on page 137-139. Looking at the inside of the sphere when r < R wend that Bl = 0 for all l otherwise the potential will diverge. Looking at the potential outside the spherewe can see that Al =0 otherwise the potential would not satisfy the boundery condition that V 0 r .From this we can see thatV(r < R, ) =l=0AlrlPl(cos)V(r R, ) =l=0Blrl+1Pl(cos)we know that the potentials are the same at r = R and continous so we can set these expression equalto each otherl=0AlRlPl(cos) =l=0BlRl+1Pl(cos)and we can see thatBl = AlR2l+1since there is a discontinuity in the radial derivative we can use equation 2.36

Voutr Vinr

r=R=100()46from this we ndl=0(l +1) BlRl+2Pl(cos) l=0lAlRl1Pl(cos) =100substituting the above expression for Bl and some algebra yieldsl=0(2l +1)AlRl1Pl(cos) = 100cos2we can use our knowledge of Legendre polynomials to solve this problemcos2 =

P2(cos) + 12

23 =

P2(cos) + P0(cos)2

23looking at only the l = 0, 2 terms in the sum and using this on the above expression yieldsA0R1P0(cos) +5A2RP2(cos) = 1300P0(cos) + 2300P2(cos)this allows us to solve for A0 and A2 asA0 = 0R30A2 = 20150R1so the potentials for inside and outside the spheres are given asV(r < R, ) = 0R30+ 20r2150RP2(cos)V(r R, ) = 0R230r + 20R4150r3P2(cos)We can nd the electric eld by

E =V =Vr r + 1rV

for inside the sphere we nd

Ein =Vin = 415r0R0

3cos212

r + r03sin(2)5R0knowing that the derivative of the term isdd32 cos2 = 32dd12[1+cos2] = 32 sin2which can be simplied as

Ein =Vin = 215r0R032 sin(2)[3cos21] r

47for the outside we nd

Eout =Vout = 030R2r2 15R40r40(3cos21)

r + 15R40r40(sin2)

to showthat these elds are discontinous at the surface, one would just have to showthat at r =R Ein=

Eout and this can be seen from the above expressions.(b). Now repeat the above exercise for the surface charge density () = 0sin2. Note that in thiscaseV(r, ) =

2R030 2R0150

rR

2P2(cos) r < R2R030 2R0150

Rr

3P2(cos) r R

For this problem we can just start withl=0(2l +1)AlRl1Pl(cos) = 0sin20= 00(1cos2)and only looking at the l = 0, 2 case in the sum we ndA0R1Po(cos) +5A2RP2(cos) = 00

P0(cos) 23P2(cos) 13P0(cos)

setting like terms equal to each other givesA0 = 20R30A2 = 2015R0and so nally we nd that the potentials are given byV(r < R, ) = 2R0302R0150

rR

2P2(cos)and alsoV(r R, ) = 2R2030r 2R0150

Rr

3P2(cos)and the electric eld is given as

Ein =Vin = 2r0R150(3cos21) r + 150r20(sin2)and for the outside we have

Eout =Vout = 2R2030r2 + 60R4150r4(cos21)

r + 150R40r4 sin2

48Problem # 3Solutions to Laplaces equation in two dimensions using complex functions: One sometimes encoun-ters a boundary-value problem that is essentially two-dimensional, in that, by the nature of the problem,we can assume the electrostatic potential is constant along one Cartesian direction. In this case, it may bepossible to nd the solution to our boundary-value problem among the many solutions obtained by anal-ysis of analytic functions of a complex variable. For this, lets consider the complex number = x +iy,where x and y are the remaining two Cartesian coordinates for which we wish to determine V(x, y). We letF() be an analytic function of - meaning that it is a well-behaved function with well dened derivatives.(a). Show that F() satises Laplaces equation, i.e. that2F = 2Fx2 + 2Fy2 = 0To show this we begin withF() = F(x +iy) = x +iywe also know thatFx = Fzzx = Fz 12Fx2 = xFz

= zFx

= 2Fz2and now for y we ndFy = Fzzy = Fz i2Fy2 = i yFz

= zFy

= i22Fz2 =2Fz2because we know that the differential operator is commutative and we can say thatxy = yxso we have shown that all nice well behaved complex functions satises Laplaces equation.2F = 2Fx2 + 2Fy2 = 2Fz2 2Fz2 = 0Given this fact, we have indeed found two solutions to Laplaces equation in two dimensions. Express-ing F = g(x, y) +ih(x, y), then the real-valued functions g and h are both such solutions.Deriving from a single complex analytic function, the functions g(x, y) and h(x, y), known as conjugatefunctions, possess an important relation to one another. By considering rst derivatives of F() one candemonstrate the validity of the Cauchy-Riemann relations:gx = hygy =hx49(b). Show that the electric eld obtained by setting V(x, y) = g(x, y) is everywhere orthogonal to thatobtained for V(x, y) =h(x, y). Based on this observation, one can state that the curves h(x, y) =h0 describethe eld lines for the potential V(x, y) = g(x, y), and visa versa.To show that these two solutions are orthogonal we must take the dot products of the two solutions

E(g) = V =g(x, y) =gx x+ gy y

E(h) = V =h(x, y) =hx x+ hy y

the dot product must be zero for these two solutions to be orthogonal.

E(g)

E(h) =gx x+ gy y

hx x+ hy y

= gxhx + gyhy = 0but since we know that the Cauchy-Reimann equations aregx = hyhx =gyusing this we ndgxhxgxhx = 0(c). One application of this method: Find the potential V(x, y) in the wedge-shaped region betweentwo plates (innite half-planes) that intersect at the z axis at an angle > R, ) = k +d lns +=1[Cs+Ds][Asin() +Bcos()] =E0scos53where the term Dsbecomes negligable for s R, from conditions (i) and (ii) we can see thatk = 0 d = 0 A = 0and also that = 1. This gives usV(s, ) =Cs +Ds1)Bcos()we can now absorb B into C and D we getV(s, ) = (Cs +Ds1)cos()applying the boundary conditions again gives us(i) CR+ DR = 0 D =CR2(ii) Cscos() = E0scos C =E0this works because we can ignore the D/s term because it is negligable.V(s, ) =

E0s + E0R2s

cos()To nd the induced charge we can use =0Vswe know thaVs = cosE0E0R2s2

= 0cosE0E0R2s2

s = R = 20E0cosProblem # 6A spherical surface of radius R has charge uniformly distributed over its surface with a density Q/4R2, except for a spherical cap at the north pole, dened by the bounding cone = .54(a). Before we get started here, show the following recursion relation for Legendre polynomials to betrue:dPl+1(x)dx dPl1(x)dx (2l +1)Pl(x) = 0 (3.6)(for l > 0). This can be obatined by considering the Rodriguez formula. (Grifths 3.62)The Rodriguez formula (Equation 3.62 in Grifths) is given asPl(x) = 12ll!

ddx

l(x21)ltaking the rst derivative with respect to x on Pl+1(x) yieldsddxPl+1(x) = 12l2(l +1)!

ddx

ld2dx2(x21)l+1= 12ll!

ddx

l[(x21)l+2lx2(x21)l1]in the above calculation we used the following relationships(l +1)! = (l +1)l! 1(l 1)! = ll!taking the rst derivative with respect to x on Pl1(x) yieldsddxPl1(x) = 22l(l 1)!

ddx

l(x21)l1= 2l2ll!

ddx

l(x21)l155and nally expanding the last term gives(2l +1)Pl(x) = (2l +1) 12ll!

ddx

l(x21)l= (2l +1) 12ll!

ddx

l(x21)l1(x21)putting all this together yields

ddx

l (x21)l+2lx2(x21)l1(x21)l1[(2l +1)(x21)l

= 0

ddx

l 2lx2(x21)l1(x21)l12lx2(x21)l1+(x21)l1

= 00 = 0(b). Show that the potential inside the spherical surface can be expressed asV = Q2l012l +1[Pl+1(cos) Pl1(cos) rlRl+1Pl(cos)where, for l = 0, we should substitute Pl1(cos) =1.Integrating every term in Equation 1 givesZ Pl(x)dx = 12l +1 [Pl+1(x) Pl1(x)]Knowing that the surface charge density and the general expression of the potential inside a sphere aregiven byV(r, ) =l=0AlrlPl(cos) = Q4R2motivated by example 3.9 in Grifths we know that the coefcients Al are given by equation 3.84Al = 12Rl1Z 0()Pl(cos)sindwe also know from Fouriers trickZ baPlcossind =Z baPl(x)dx =Z abPl(x)dxWe ndAl = 12Rl1Z 0()Pl(cos)sind+ 12Rl1Z ()Pl(cos)sindthe rst integral is just zero because the surface charge density is zero there and the second integral isgiven byAl = 12Rl1Z ()Pl(cos)sind = ()2Rl1Z cos 1Pl(x)dxwhere the integration limits are given by the change of variable of the integral. From the relationshipfrom part (a) we ndAl = 2(2l +1) [Pl+1cosPl1cos] = Q8(2l +1)Rl+1 [Pl+1cosPl1cos]56where Pl+1(1) Pl1(1) = 0 , putting this into the general expression for the potential inside asphere yieldsV(r < R, ) = Q8l=0rlPl(cos)(2l +1)Rl+1[Pl+1cosPl1cos](c). What is the potential outside the sphere?We know that the general expression for the potential outside of a sphere is given byV(r > R, ) =l=0Blrl+1Pl(cos)but we know that the coefcients Bl (Equation 3.81) can be expressed asBl = AlR2l+1thusV(r > R, ) =l=0AlR2l+1rl+1 Pl(cos) = Q8l=0Rl(2l +1)rl+1[Pl+1cosPl1cos](d). Find the magnitude and the direction of the electric eld at the origin.We know that the electric eld is given byE =V =VrSince is not well dened at the origin and we know that the electric eld points in the z direction andwe do not compute the derivative with respect to . The electric eld is given byEin = Q8l=1lrl1Plcos(2l +1)Rl+1[Pl+1cosPl1cos] rand so we know r = z, and = 0. We can also see that the above expression tells us the l = 1 in orderfor the electric eld to not blow up at the origin or go to zero.E = Q8R213[P2cosP0cos] z= Q8R21332 cos212

1

z= Q4R214[cos21] zE = Q16R2[1cos2] zand the magnitude is given by|E| = Q16R2[1cos2]57(e). Discuss the limiting forms of the potential (b,c) and the electric eld (d) as the spherical capbecomes either (1) very small or (2) so large that the entire remaining charged area becomes a very smallcap at the south pole.We can see from the gure that if gets very small the sphere would be almost represented as aspherical conductor and so 1 very smallthe potentials and the electric eld would behave asV(r < R, ) V0V(r > R, ) 1rE(r < R, ) 0and if the cap becomes very large the sphere would be represented by a point charge at the south poleof the sphere. The potentials and electric eld would behave asV(r < R, ) 1rV(r > R, ) 1rE(r < R, ) 1r2Problem # 7Many nuclei have electric quadrupole moments. The quadrupole moment of a given state of a givennucleus is typically quantied as Q = Q33/e where Qi j is dened as in Grifths Problem 3.45, and theindex 3 refers to the z axis. Now, consider a nucleus centered at the origin nding itself in a cylindricallysymmetric electric eld with a gradient Ez/z along the z axis at the position of the nucleus.(a). Show that the inhomogeneous eld modies the energy of the quadrupole byW =e4Q

Ezz

For this, you might want to nd a suitable form of the electrostatic potential near the position of thenucleus. Be sure that this potential satises the Laplace equation. You should let V = 0 at the origin.We know that energy of a continuous charge distribution can be written asW =Z (r)(r)dwhere (r) is the externel potential energy, where the general form can be expressed as(r) = 14l=01rn+1Z (r)nPn(cos)(r)dwe can expend this using Taylor expansion to nd(0) = (0) +r (0) + 12i, jrirjij(0)58where we know that the last term is the quadrupole term, which is the one we are concerned about.Plugging this into the work function yieldsW = 12Z (r)i, jrirjij(0)dand since we are only concerned with Q33 we know that i = j = 3 , thus we ndW = 12Z (r)d(r3)2 2z2(0)

but we know that2z2(0) =Ezzputting this into the above expression yieldsW =12EzzZ (r3)2(r)dusing the relationship from problem 3.45Q33 =Z [3(r3)2(r3)2](r)d Q332 =Z (r)d(r3)2using this expression we nd that the work is given byW =Q334Ezzsince we know thatQ33 = eQwe ndW = eQ4Ezz(b) Nuclear charge distributions can be approximated by a constant charge density throughout aspheroidal (egg-shaped) volume of semimajor axis a and semimajor axis b. Calculate the quadrupolemoment of such a nucleus assuming that the total charge is Ze. Given that the 153Eu(Z = 63) nucleus hasa quadrupole moment of Q = 2.5 1025cm2and a mean radius R = a+b2 = 7 1013cm, determine thefractional difference in radius (ab)/R.59Using the denition of the quadrupole momentQi j =Z [3rirj(r)2i j](r)dand since we are only looking for Qzz we ndQzz =(r)Z [2z2x2y2]dxdydzThis gives us the quadrupole moment in cartesian coordinates, it would be easier to solve this integralin spherical coordinates, we can do a linear transformation to do this, the equation that models our spheroidis given byx2b2 + y2b2 + z2a2 = 1lettingu = xb v = yb w = zawhere the derivatives are given bydx = bdu dy = bdv dz = adwand so we knowdxdydz = ab2dudvdwwe also know that the cartesian components in spherical coordiantes are given byu = r sincosv = r sinsinw = r cos60using the above relationships we getQzz =(r)Z Z Z [2(wa)2(bv)2(bu)2]ab2dwhich gives(r)Z 20Z 0Z 10[2a2r2cos2b2r2sin2cos2b2r2sin2sin2]ab2r2sindddrsimplifying this givesQzz =(r)Z 20Z 0Z 10[2a2r2cos2b2r2sin2]ab2r2sindddrand using Mathematica yieldsQzz =(r)815ab2(a2b2)but we know that the charge density is given as(r) = qV = 3q4ab2 Vell = 43ab2subsituting this expression givesQzz = 25q((ab)(a+b))solving for ab/R givesabR = 5Qzz2Ze(a+b)R = 5Qzz4ZeR2since we know thatQzz = Qewe nd that the fractional difference is given byabR = 5Q4ZR2 0.01Problem # 8Consider a dipole with moment p at a distance z0 above a grounded conducting plane that is takento dene the x y plane. p is taken to lie in the x z plane, oriented at an angle to the normal (z)of the plane. Find the torque N on the dipole. What are the equilibrium values of ? [Answer: N=p2sincos y/320z30 )]61We know that the torque from an electric eld on a dipole is given asN = pEand we also know that the energy of a dipole is given asEdip = 141r3[3(p r) r p]we can solve this problem by using the image of the dipole, we know that the components of the dipoleand the image are given bypdip = psin, 0, pcospim = psin, 0, pcospim = pdipwe also know that r points in the z direction, thusp r =p z = pcos r = 2z0putting this into the electric dipole equation yieldsEim = 141(2z0)3[3pcos z +[psin xpcos z]] = 1321z30[psin x+2pcos z]and thus we know that the torque isN = pE = p232z30

x y zsin 0 cossin 0 2cos

which yieldsN = p2sincos32z30 y62The equilibrium values of are given when the torque goes to zero, i.e = n2 n 0 integerProblem # 9Using Eq. 3.103, calculate the average electric eld of a dipole, over a spherical volume of radius R,centered at the origin. Do the angular intervals rst. [Note : You must express r and in terms of x, y and z before integrating. Compare your answer with the general theorem Eq. 3.105. The discrepency here isrelated to the fact that the eld of the dipole blows up at r = 0. The angular integral is zero, but th radialintegral is innite, so we really dont know what to make of the answer. To resolve this dilemma, lets sayEq. 3.103 applies outside a tiny sphere of radius - its contribution to Eave is then unambiguously zero,and the whole answer has to come from the eld inside the sphere.We know that the average electric eld and the electric eld from the dipole can be expressed asEave =RspaceEdipdV Edip = p4(2cos r +sin)we need to change to cartesian coordiantes so that we can evaluate this integral, i.e r = sincos x+sinsin y+cos z = coscos x+cossin ysin zthus Edip in cartesian coordinates givesEdip = p4[3sincoscos x+3sincossin y+(2cos2sin2) z]and so we nd that the integral can be expressed asZspaceEdipd = p4Z r0Z 20Z 03sin2coscosr drdd x+ p4Z r0Z 20Z 03sin2cossinr drdd y+ p4Z r0Z 20Z 0(2cos2sin2r drdd yusing Mathematica yields (for the angular integrals)ZangEdipd = 0and we know that the integral for the radial component (at the origin) yieldsZradEdipd = thus we nd thatEave = 0 which seems very weird? There must be a delta function in the center of the sphere.(b). What must the eld inside the sphere be, in order for the general theorem (3.105) to hold?[Hint : since is arbitrarily small, were talking about something that is innite at r = 0 and whoseintegral over an innitesimal volume is nite.][Answer : (p/30)3(r)]63[Evidently, the true eld of a dipole isEdip(r) = 1401r3[3(p r) rp] 130p3(r)You may well wonder how we missed the delta-function term when we calculated the eld back inSect. 3.4.4. The answer is that the differentiation leading to Eq. 3.103 is perfectly valid except at r = 0 ,but we should have known (from our experience in Sect 1.5.1) that the point r = 0 is problematic.Since we know that the average electric eld inside the sphere is not zero we can write the electricdipole asEdip = 14r3[3(p r) rp] +K3(r)where K is a non-zero coefcient that we need to solve for. we can solve for K by solving the averageelectric eld in terms of a delta functionEave =R K3(r)dV = 3K4R3setting this expression to equation 3.105 we nd3K4R3 = p40R3 K = p30thus the correct expression for the electric dipole is given byEdip = 14r3[3(p r) rp] p303(r)64Chapter 4Electric Fields in Matter4.1 Problems and SolutionsProblem # 1 Forces on dipoles(a). A dipole p with xed magnitude is placed in an external electric eld E(r) that varies with position.Show that there is a force on the dipole, and nd how this force depends on the orientation of the dipoleand on the variation of the eld. [Answer: F = (p )E]So we know thatF+ = qE+ F =qEand we also know that the total force is given byFtot = F+ +F = q(E+E) = qEwhere the components of E are given byEx = (Ex) d Ey = (Ey) d Ez = (Ez) dwhere d is the displacement vector. Thus we knowE = (d )Ethus the force is given byF = (p )E(b). Find the force and torque on a dipole in the eld of a point charge. Let the charge q be at theorigin, and the dipole p = p(sin0 x +cos 0 z) be at the point (0, 0, z0 ). Also, nd the force on q, andverify Newtons third law. [Answer: the force on q isF = p0q40z30(sin0 x+2cos0 z)We know that the potential of a point charge can be written as (in cartesian coordinates)Vc = q41

x2+y2+z265and we know that the electric eld can be written asE =V = q4 x(x2+y2+z2)3/2 x+ y(x2+y2+z2)3/2 y+ z(x2+y2+z2)3/2 z

now that we have the electric eld we can nd the forceF =

psin0xE+ pcos 0zE

which yieldsF(0, 0, z0) = pq4[sin0 x2cos0 z]and the torque is given byN(0, 0, z0) = pE =

x y zpsin0 0 pcos 00 0 q4z20

= pqsin04z20 yTo verify Newtons third law we need to show the force on q from the dipole is equal and opposite ofthe force on the dipole from q, thus we can start with the defenition for the electric eld of a dipoleEdip(r) = 14r3[3(p r) r p]and we know that the force is given byF = qEdipwe also know that r = z p = p(sin0 x+cos0 z) (p r) r = pcos0 zthus we can see that the force isF = pq4z30[3cos0 z sin0 xcos0 z] = pq4z30[2cos0 z sin0 x]thus we can see thatF = qEdip =qEcWe know that the equilibrium values for are given when = n2 n 0Problem # 2 Capacitors with dielectric slabsThe space between the plates of a parallel capacitor (Figure 1) is lled with two slabs of linear dialectricmaterial. Each slab has thickness a, so the total distance between the plates is 2a. Slab 1 has a dialectricconstant of 2, and Slab 2 has a dialectric constant of 1.5. The free charge density on the top plate is andonthe bottom plate .66Figure 1a). Find the electric displacement D in each slab.Since we know what the free charge is on both the top plate of the capacitor and the bottom plate ofthe capacitor we can apply the integral form of Gausss law to nd the displacement, usingItopD da = Qfwe can nd what the displacement if we place a Gaussian pillbox encompassing the top plate of thecapacitor and Slab 1Qf =fAItopD1 da = fA da = A( z)where z points in the direction of the electric eld, which is downward. So for Slab 1 we nd thedisplacement to beD1 = f( z) = ( z)and for the displacement of Slab 2 we can place a Gaussian pillbox encompassing the bottom plate ofthe capacitor and Slab 2Qf =fAIbotD2 da = fA da = A( z)thusD2 = f( z) =( z)and the magnitudes are given byD1 = D2 =b). Find the electric eld E in each slab.given that we know what D is in each slab we can easily nd EE1 = D111 = 20 E1 = 20 zandE2 = D222 = 320 E2 =230 zand the magnitudes are given byE1 = 20E2 = 23067c). Find the polarization P in each slab.We know that the polarization is dened asP =0eEwhere e is dened as0= r = 0(1+e) e =r1 1 = 1 2 = 12for Slab 1 and 2 we ndP1 = 2 z P2 =3 zand the magnitudes are given byP1 = 2 P2 = 3d). Find the potential difference between the plates.we know that the potential difference is given by the following integralV =|V1V2| =|Z a0E1 dl Z 2aaE2 dl| =

a20+ 2a30

given that dl = a z we ndV = 7a60e). Find the location and amount of all bound charge.given that we know the polarization we can nd what the bound charge is by usingb = P nwhere n points outward with respect to the dialectric, so for Slab 1 we ndb = 2 Slab 1 top surfaceb = 2 Slab 1 bottom surfaceand for Slab 2 we ndb = 3 Slab 2 top surfaceb = 3 Slab 2 bottom surfacethis is illustrated in Figure 2.68Figure 2f). Now that you know all the charge (free and bound), recalculate the eld in each slab, and conrmyour answer to b.Using Figure 2 we can nd the total charge t for Slab 1 and Slab 2. For the top of Slab 1 we add upall the charge in the dashed box labeled total-top(1) which yieldst =2 = 2 Top of Slab 1for the bottom of Slab 1 we add up all the charge in the box labeled total-bottom(1) which yieldst = 23 + 3 =2 bottom of Slab 1and now for the bottomof Slab 2 we add up all the charge in the dashed-dot box labeled total-bottom(2)which yeildst = 3 =23 bottom of Slab 2and nally for the top of Slab 2 we add up all the charge in the dashed-dot box labeled total-top(2)which yieldst =3 + 22 + = 23 Top of Slab 2and the magnitude of the electric eld is given byE1 = t0= 20E2 = t0= 230which is the same thing we found in part b.part 2: problem 2Suppose you have enough linear dialectric material, of dialectric constant r , to hal f -ll a parallel-plate capacitor (Figure 3). By what fraction is the capacitance increased when you distribute the materialas in Fig.3 (a)? How about Fig.3 (b)?Figure 3For Figure 3 (a) we can see that the displacement is given byI D da = Qf = A D = this is the displacement inside the dialectric, with this we can now nd what the electric eld is insidethe dialectricE = D Edia = 69we also know that the electric eld between the dialectric and the top plate of the capacitor is given byEair = 0given that the defenition of capacitance is given byC1/2 = QV1/2we are now in a position to nd what the potential is between the capacitorV1/2 =Z d/20Eair dl +Z dd/2Edia dl = d20+ d2 = d2

10+ 1

= d2(r +1)thus the capacitance is given byC1/2 = QV1/2= AV1/2= 2Ad(r +1)and since we know that a capacitor with no dialectric has a capacitance given byC = A0dthus the ratio of the capacitance with dialectric to capacitance without dialectric for Figure 3 (a) isgiven byC1/2C = 20(r +1) = 2r(r +1)From Figure 3 (b) we can see that the area has been decreased by 1/2 and the dialectric now lls thisentire 1/2-space, thus we can treat this as two separate capacitors one in which the 1/2 area is lled withthe dialectric and another 1/2 area that has no dialectric. We nd that the total capacitance is given byC1/2 = A2d + A02d = A2d( +0)and we know that the capacitance with no dialectric is given byC = A0dthus the ratio givesC1/2C = +020= r +12Problem # 3A spherical conductor, of radius a, carries a charge Q. It is sorrounded by linear dialectric material ofsusceptibility e, out to radius b. Find the energy of this conguration.We know that the total work done, as we build up the free charge from zero to its nal conuguartionis given byW = 12Z D Ed70We can nd D by usingI D da = Qf D = Q4r2 r r > awe know that P=D=E=0 inside the sphere. Now that we know D we can easily nd E,E = D = Q4r2 r a < r < bandE = Q40r2 r r > bnow that we have D and E we can now nd what the energy of this conguration isW = Q22Z ba1(4)2r2drd+Z b1(4)20r2drd

knowing that d= 4 when integrating over all solid angles allow us to symplify this asW = Q281Z ba1r2dr + 10Z b1r2dr

= Q2811a1b

+ 101b

we can also write this asW = Q281a1b

+r1b

= Q281a +(r1)1b

but since we knowr = 1+e = 0(1+e)we ndW = Q280(1+e)1a +e1b

Problem # 4The dielectric constant of air is 1.00059. From this, determine the mean polarizability of atmosphericmolecules. Compare this result to the atomic polarizabilities listed, for example, in Grifths Table 4.1.We can write the polarization asP =e0EThe polarization in a dialectric is also dened asP = npwhere p is the dipole moment and n is the particle density. We also know that the dipole moment canbe dened asp =Eotherwhere is the polarizability and Eother is the total electric eld, the applied eld and the enduced eldEother = EEsel f71we need to nd out what the self induced eld is. The external eld produced by the sphere itself is apure dipole. The total eld inside the conducting sphere is 0, because a conductor is an equipotential, thusthe induced eld inside must be E0, this eld is needed to cancel the applied uniform eld, so we ndEsel f =E0If we treat the sphere as a conducting sphere we nd that the dipole moment goes asp = 40a3E0where a is the radius, thus Esel f can be written asEsel f = p40a3substituting this in Eother we ndEother = E+ p40a3we also know thatEsel f =np30= P30thus we ndP = np = n

E+ P30

there is a linear relationship between P and E, and the susceptibility which is given bye = n/01n/(30)if we solve for the atomic polarizability we nd = 0ne1+e/3 = 30n1+2and usin the ideal gas law we ndn = NV = kTPwe nd for /4040= 3kT4P1+2 1.901030 1m3using T = 293 K and 1105Pa for the pressure, along with the dialectric constant given.Problem # 5A hollow dielectric sphere, with dielectric constant /0 = , inner radius a and outer radius b, isplaced in a uniform applied electric eld E0 z. The presence of the sphere changes the eld. Find the eldin the three dened regions, i.e. r < a, a < r < b and r > b. What is the eld at the center of the sphericalshell? What is the dipole moment of the dielectric medium? [for the last part, youll nd:pz = 40(a3b3)b3E0(1+2)(+1)2a3(1)2b3(2+)(1+2)72Using the general solution for the potential in spherical coordinates we nd thet the potentials are givenbyV1 = l=0AlrlPlcos r < aV2 = l=0

Blrl+ Clrl+1

Plcos a < r < bV3 = l=0Dlrl+1PlcosE0r cosand the boundary conditions are given by(i) V1 = V2 r = a(ii) V2 = V3 r = b(iii) V2r = 0V1r r = a(iv) V2r = 0V3r r = bthus we have four unknown constants and four boundary conditions. Lets gure out what these con-stants are using the boundary conditions. From 1 we ndl=0AlalPlcos =l=0

Blal+ Clal+1

Plcoswhich simplies toAlal= Blal+ Clal+1 l = 1 A = B+ Ca3 l = 1ifl = 1 Al = Bl =Cl = 0from this and from the next boundary condition we know that l = 1 for all cases. From the secondboundary condition we nd (using l = 1)Bb3+C = DE0b373from boundary condition number 3 we nd0A = B2Ca3

= 0which simplies toA = B2Ca3and from our last boundary condition we nd2Db3 E0 = B2Cb3so now we have four equation and four unknowns, we can do this by brute force or just use useMathematica. The four equations are summarized asA = B+ Ca3Bb3+C = DE0b3A = B2Ca32Db3 E0 = B2Cb3Mathematica yieldsA = 9b3E02a3(1)2b3(2+)(1+2)B = 3b3E0(1+2)2a3(1)2b3(2+)(1+2)C = 3a3b3E0(1)2a3(1)2b3(2+)(1+2)D = (a3b3)b3E0(1+2)(1)2a3(1)2b3(2+)(1+2)plugging these constants into the general expression for all potentials yieldV1 = 9b3E02a3(1)2b3(2+)(1+2)r cos = 9b3E02a3(1)2b3(2+)(1+2)z r cosand for the other two we ndV2 = 3b3E0(1+2)2a3(1)2b3(2+)(1+2)z + 3a3b3E0(1)2a3(1)2b3(2+)(1+2)z(x2+y2+z2)3/2and the nal potential is given byV3 = (a3b3)b3E0(1+2)(1)2a3(1)2b3(2+)(1+2)z(x2+y2+z2)3/2E0z74The electric eld in all three regions is given byE =VsoE1 = 9b3E02a3(1)2b3(2+)(1+2) z r < athe electric eld inside the dialectric is given byE2 = 3b3E0(1+2)2a3(1)2b3(2+)(1+2) z + 3a3b3E0(1)2a3(1)2b3(2+)(1+2) 3xz(x2+y2+z2)5/2 x 3yz(x2+y2+z2)5/2 y 3z2(x2+y2+z2)5/2 z + 1(x2+y2+z2)3/2 z

and nally for the electric eld on the outside we nd, usingK = (a3b3)b3E0(1+2)(1)2a3(1)2b3(2+)(1+2)E3 = K 3xz(x2+y2+z2)5/2 x 3yz(x2+y2+z2)5/2 y 3z2(x2+y2+z2)5/2 z + 1(x2+y2+z2)3/2 z

E0 zThe eld in the center of the dialectric is given byEcenter = 9b3E02a3(1)2b3(2+)(1+2) zWe know that the potential in terms of the dipole moment is given byV = p r40r2 = pzcos40r2and looking at the potential on the outside we can see that the potential from the dipole moment is justthe rst term in the potential V3, the E0 is from the applied eld, we are only interested in the potentialfrom the dipole, i.epz = 40(a3b3)b3E0(1+2)(1)2a3(1)2b3(2+)(1+2)Problem # 6Consider two insulating media with dielectric constants 1/0 = 1 and 2/0 = 2 , placed togetherwith a planar interface between them. In the region of ?1 there is an innite line charge with linear chargedensity 1, parallel to the interface at perpendicular distance75a. Use the method of images to nd the electric eld in both media.We can see from diagram (2) that the potential in the top region is given byVt = 21ln|z a| + 21ln|z +a|and the potential at the bottom is given byVb = 22ln|z a|we also know that the boundary conditions are given byVt =Vb z = 0from continuity. Our othert boundary condition is given as1Vtz =2VbzFrom boundary condition number 1 we can see that+ = 12and from condition 2 we nd =plugging in from the last equation into the previous one yields+ = 12()thus =

212 +1

plugging this into either equation yields =1+ 212 +1

=

222 +1

76plugging these solutions into our potentials we ndVt = 21ln|z a| +

212 +1

121ln|z +a|and the electric eld at the top is given byEt = 21|z a|

212 +1

1211|z +a|and the potential at the bottom is given byVb = 122

222 +1

ln|z a|and the electric eld is given asEb = 1

12 +1

1|z a|Show that the force per unit length on the line charge isdFdL = 2(12)41a(12)We know thatF = qE dF = dqEbutdq = dland since we know that there is no self induced force on we must use the term as the one con-tributing to the force, so we nddFdL =E1 = 2 (12)41(1 +2)Note that the sign of the force depends on the difference 12 . Explain physically whats going onhere.We can see from the expression for the force that if 1 is greater than 2 the force will be pointing inthe z direction and if we have the inverse it will point in the - z direction.Problem # 777Two long cylindrical sheets of metal (radii r1 and r2 with r2 > r1 ) are arranged coaxially. The platesconnected to a battery that maintains a potential difference V between the sheets. The region between theconductors is lled with a material of conductivity and permittivity .a) Determine the capacitance per unit length of this system.We know that the capacitance and the potential are given byC = QV V =Z E dlwe can nd the electric eld by using Gausss law, i.eZ E da = Q da = 2rL rwe nd the electric eld to be given asE = Q2rL rknowing the electric eld allows us to nd the potentialV =Z r1r2Q2rLdr = Q2Lln

r2r1

knowing this allows us to nd the capacitance per unit lengthCL = 2ln

r2r1

b) Use Ohms law to calculate the electric current per unit length between the conducting shells.We know that Ohms law can be expressed asI =Z J da = Z E dausing Gausss law we ndZ E da = Q Q = Lthus we nd that the current per unit length is given byIL = = 2ln(r2/r1)Vc) Suppose the battery that maintains the potential difference V is suddenly disconnected from thecircuit. Show that charge will leak off the two plates of this capacitor as an exponential function of time.What is the exponential time constant? You should obtain a simple function of and .We know that once the battery is disconnected the current will be given asI =dQdt = Q78and this is a simple integral of the formZ QfQidQQ =Z t0dtwhich yieldsQf = Qietand the exponential time constant is given byk = d) Show that the total energy dissipated by Joule heating as the capacitor discharges completely equalsthe electrostatic energy that was originally stored in the capacitor.We know that the work can be expressed asW =Z Pdtwe know that the power given for Joule heating can be expressed asP = I2Rthus we ndW =Z I2Rdtbut we also know thatI =dQdt =QietI = Qietwe also know thatR = VI =ln

r2r1

2Lthus we nd that the work is given byW =

Qi

2ln

r2r1

12LZ 0e2tdt = 2

Qi

2ln

r2r1

12Lthis simplies intoW = Q24Lln

r2r1

= 12 Q2Lln

r2r1

Q = 12QVwe know that the work from a capacitor is given byW = 12CV2= 12QVthus we have just shown that these two expressions are equivalent.79Chapter 5Magnetostatics5.1 The Lorentz Force LawThe basic problem in classical electrodynamics we have a collection of charges q1, q2, q3... which areknown as the source charges, and we want to know the force they exert on a charge Q. Up till now wewere dealing with charges that were at rest electrostatics and the time has come to consider the forcesbetween charges in motion. Imagine that we have two wires seperated by a few centimeters and when Iturn on a current so that it passes up one wire and down the other and the wires jump apart. How do youexplain this? Whatever force accounts for the attraction of parallel currents and repulsion of antiparallelones is not electrostatic in nature. The force responsible for the observed phenomenom is the magneticforce, we can understand this using Figure 1From Figure 1 we can see that a cross product will allow us to solve for the force. The magnetic forcein a charge Q , moving with a