The charges all have the same sign so the forces are repulsive. For the force on q3 due to q2,

12m

9m 2 2

23r (9m) (12m) 15m= + =

2

292 3 N m

23 2 2C23

q q (400 C)(500 C)F k (9 10 ) 8.0Nr (15m)

⋅ µ µ= = × =

θθ

Now x 12m 9mcos & sinr 15m 15m

θ = = θ =

And 23ˆ ˆF 8.0N(cos i sin j)= θ − θ

So, 2312 9ˆ ˆ ˆ ˆF 8.0N( i j) 6.4Ni 4.8Nj15 15

= − = −

For the force on q3 due to q1 , which is entirely along x,

2

291 3 N m

13 2 2C13

q q (300 C)(500 C)F k (9 10 ) 9.38Nr (12m)

⋅ µ µ= = × =

13ˆF 9.38Ni=

Then

net 23 13ˆ ˆ ˆF F F 6.4Ni 4.8Nj 9.38Ni= + = − +

netˆ ˆF 15.78Ni 4.8Nj= −

& 2 2netF (15.78N) (4.8N) 16.5N= + =

(– sign from sketch)

23F

2e p

e 2 2

q q eF m a k kr r

= = =

2

2 2

2 19 29 22N m m

2 31 11 2C se

e (1.6 10 C)a k (9 10 ) 9.0 10m r (9.1 10 kg)(5.3 10 m)

−⋅

− −

×= = × = ×

× ×

Configuration C has all the + charge, repelling Q, on one side and all – charge, attracting Q, on the other so it will have the greatest net force on Q. Config. B has the same sign charges equidistant on either side of Q resulting in forces that cancel each other (net force on Q is zero) so it will be the smallest. In config. D the two outermost like charges cancel each other leaving only the net force from the two oppositely signed nearer charges. Config. A has the same situation for the nearer charges giving the same net force magnitude as in D from these but it has opposite sign outer charges oriented to oppose the force from the inner charges making the magnitude of the force in config. D greater than that in A so: C, D, A, B.

+3Q q3 –9Q

0m 2m

3 1 3 22 23,1 3,2

q q q qk k 0

r r− =

3,1 3,2F F 0+ =

1 22 23,1 3,2

q q0

r r− =

1 22 23,1 3,2

q qr r

=

The sketch shows approximately where the charge must lie.

3,2 2

3,1 1

r q 9Q3

r q 3Q−

= = =

3,2 3,1r 3r=

2 3 1 3x x 3 (x x )− = −

3 32m x 3 (0m x )− = −

3 32m x 3x− = −

3 3 32m x 3x (1 3)x= − = −

32m x

(1 3)=

−

3x 2.73m= −

For equilibrium,

Draw equal length arrows representing the force on a unit positive test charge (by definition the electric field) at the center point, due to the charges at the corners for each case. For clarity the arrows are color coded to go with the charge that produced it. The net field at the center is zero in cases (3) & (4). In case (1) the x components of the individual forces cancel while their y components sum to make the magnitude proportional to 4cos(45o)=2.83. In case (2) the magnitude is proportional to 2, which is smaller.

mg

qEθ TTcosθ

Tsinθ

Equilibrium requires that,

xF 0 qE Tsin= → = θ∑yF 0 mg Tcos= → = θ∑

Divide the first eqn by the second,

qE Tsin tanmg Tcos

θ= = θ

θ

2m

os NC6

(0.00075kg)(9.8 )mgE tan tan(28 ) 122q 32 10 C−= θ = =

×

Solve for E:

F qE ma= =

qEam

= 21o o 2x x v t at= + +

0 (starts from rest)

21o2 at x x= −

2 o2(x x )ta−

=

319o o

19 4 NC

2(x x ) 2m(x x ) 2(9.1 10 kg)(0.025m 0)t 4.4 10 sa qE (1.6 10 C)(1.5 10 )

−−

−

− − × −= = = = ×

× ×

enc oq E dA= ε ⋅∫

Since there is no electric field along the z direction there is no electric flux through the faces perpendicular to the z axis. There is electric flux passing through the faces perpendicular to the y axis but the field is in the same direction through each face, the areas are the same but their normal vectors are in opposite directions so they contribute to the integral with equal magnitudes but opposite sign and so they vanish. This leaves only the integral over the faces perpendicular to the x axis. For the one at x = 0 the electric field is zero leaving only,

2

212 2 12C N

enc o x 1m o x 1m o x 1m CN mq E dA E dA E dA 8.85 10 ( 2 )(1m ) 17.7 10 C− −

= = = ⋅= ε ⋅ = ε = ε = × − = ×∫ ∫ ∫

The electric field inside a uniform shell of charge due to the charge on the shell is zero (draw a Gaussian surface just inside the shell. It encloses zero charge, making the field inside, from the shell, zero). But there is field there due to the infinite sheet of charge given by, (see section 23-5 of your textbook for the proof)

o

E2σ

=ε

To make the field inside the metal of the conducting shell zero (as it must be) the charge on the inner surface of the conducting shell must have the same magnitude but opposite sign as the charge inside the shell (to make the net charge inside a Gaussian surface drawn entirely within the conductor, see sketch, zero).

+Q

–Q

All the charge is at the same distance r = R from the point of interest so,

( ) ( )1 2 11 2 1 1

o o o o

1 Q Q 1 1 5QV Q Q Q 6Q4 r r 4 r 4 r 4 r

− = + = + = − = πε πε πε πε

( )2

2

121

12 Co N m

5Q 5(8.58 10 C)V 3.27 V4 r 4 8.85 10 (0.118 m)

−

−⋅

− − ×= = = −

πε π ×

The integral of f(x)dx is the signed area between the function and f(x)=0 between the limits of the integral. So,

f iV V E ds− = − ⋅∫

f iV V E ds= − ⋅∫

4

f0

V 30V E(x)dx= − ∫

N1f 2 CV 30V (4m)(20 ) 30V 40V 10V= − = − = −

f

f ii

V V E ds− = − ⋅∫

Where the conservative nature of the electric force let’s us go from O to P by any path to get the same value. In that case we chose the path shown below. The electric field for the negative sheet of charge is uniform and points in, towards the sheet with magnitude (everywhere) given by

o

E2σ

=ε

E

P T P

P OO O T

V V E ds E ds E ds− = − ⋅ = − ⋅ − ⋅∫ ∫ ∫

Then

T

From O to T while from T to P oE ds Edscos(90 ) 0⋅ = =

dsoE ds Edscos(180 ) Eds⋅ = = −

P P

P OoT T

V V 0 Eds E ds Ed d2σ

− = − − − = = =ε∫ ∫

So,

Vx m

V (xyz)E yz 1x x

∂ ∂= − = − = − = −

∂ ∂(at 1,1,1)

Vy m

V (xyz)E xz 1y y

∂ ∂= − = − = − = −

∂ ∂(at 1,1,1)

Vz m

V (xyz)E xy 1z z

∂ ∂= − = − = − = −

∂ ∂(at 1,1,1)

The magnitude is then: 2 2 2x y yE E E E 1 1 1 3 1.7= + + = + + = =

6i 1 2A

i i 1 2

q q q 6 C 9 C 48 C 45 C 93 CV k k k k k 2.093 10 Vr r r 0.05m 0.08m 0.40m 0.40m 0.40m

µ µ µ µ µ = = + = + = + = = ×

∑

6i 1 2B

i i 1 2

q q q 6 C 9 C 30 C 72 C 102 CV k k k k k 2.295 10 Vr r r 0.08m 0.05m 0.40m 0.40m 0.40m

µ µ µ µ µ = = + = + = + = = ×

∑

app f i B AW U q V q(V V ) q(V V )= ∆ = ∆ = − = −

6 6 6app B AW q(V V ) (4 10 C)(2.295 10 V 2.093 10 V) 0.81J−= − = × × − × =

2o

f i o oo

U 1 q 1W U U U U U 1 12C

= ∆ = − = − = − = − κ κ κ

2

2

2 9 26

12 2Co N m

q d 1 (4 10 C) (0.003m) 1W 1 1 2.26 10 J2 A 1.52(8.85 10 )(0.02m)

−−

−⋅

× = − = − = × ε κ ×

Where oo

ACdε

=

eqC 20 | ((12 | 24) 12)= + With all values in µF

eq12 24C 20 | ( 12) 20 | (8 12) 20 | 20 10 F12 24

⋅ = + = + = = µ +

The equivalent capacitance for the right side of the circuit (all in µF) is

eq15 10C ((10 |10) 10) |10 (5 10) |10 15 |10 6 F15 10

⋅= + = + = = = µ

+

That means the total charge on that network is eqQ C V 6 F 10V 60 C= = µ ⋅ = µ

That’s also the charge on the cap we’ve labeled C3. The voltage drop across that is .

C3

33

3

Q 60 CV 6VC 10 F

µ= = =

µThe voltage drop across the upper part of that network is then 4V. Since that splits evenly between the series combination of C2 and the equal cap below it, C2 has 2V across it. Then,

2 2 2Q C V 10 F 2V 20 C= = µ ⋅ = µ

Since we had not yet covered resistors in series at the time of this exam the word successive here was meant to convey its temporal meaning i.e. that the resistors each got 12V across them, connecting them (successively in time) one at a time. In that case, the resistor with the smallest resistance has the greatest current flowing through it and using with the voltages all the same (12V) we see that the smaller the resistor the greater power dissipated (as heat). In that case the resistor values are such that C>A>B.

P IV=

Many of you, however, interpreted successive as the resistors in series. In that case the same current flows through each resistor (limited by the series resistance of the 3 series resistors). In that case using the form for the power, , since the same current flows through each resistor the power scales with the resistance and the order of the resistances would be, B>A>C.

2P I R=

ii

i

LRA

= ρff

f

LRA

= ρ

Let f refer to the final case and i to the initial, then, with Lf = 4Li , for constant volume, we must have,

f f i iL A L A=

i f i i4L A L A=

if

AA4

=

Now, and divide the first equation by the 2nd,

f

f f i i if

i ii f ii

i

LR L A 4L AA 16L AR A L L

A 4

ρ= = = =ρ

f iR 16R=