1
However, in some cases, a SYMMETRIC CHARGE DISTRIBUTION allows us to guess the orientation of the corresponding electric field. That is the case, for example, when we consider a INFINITELY-LONG line of uniform charge distribution (λ= charge per unit length.)
Asymmetric charge distribution
Symmetric charge distribution
Andres La Rosa Lecture Notes Portland State University PH-212
THE GAUSS' LAW
Q
+ + + + + + + + + + + + + + + + + + + + +
The electric field at any point away from the line (A and B, for instance) turns out to be perpendicular to the line.
2
It turns out, the Gauss's law will allow calculating the magnitude, as we will see below
+ + + + + + +
+ + + + + + +
Since the points P, Q, R and S are at the same distance from the charged-line, the corresponding electric fields should have the same magnitude:
Vertical line
Hypothetical ring on a horizontal plane
3 The Gauss's law (to be described below) is a tool that allows to calculate (in a very simplified way) the electric field produced by symmetrically distributed charges Note: But keep in mind that Gauss's law is valid for both, symmetric or asymmetric charge distributions.
5
Definition of the solid angle ΔΩ
ΔΩ
ΔΩ
ΔΩ ΔS ΔS
ΔS
CASE: ΔS is parallel to r
ΔΩ
Magnitude of ΔS
Total solid angleenclosed by aspherical surface
Sphere ofradius R
11
E ΔS Cos θ
θ
Electric flux through a closed surface
CASE: The electric field is produced by one point-charge located inside the closed surface
Flux depends only on the solid angle, independent on the distance of ΔS to the point-charge !!
E
ΔΩ
= 4π
13
whereφ is the electrical flux
crossing the mathematicalsurface S ,
andq is the point charge
inside the surface S
q
q1
q
CASE: More than one point-charge are located inside the mathematical closed surface
q2
IN GENERALnet charge insideS
Gauss' Law
14
CONCLUSION: Charges located outside the surface S do not contribute to the electric flux
Exercise
charge inside
q1q4
q2
q3
q2 + q3
q1, q2, q3 , q4 , q5
Exercise: Evaluate the electric flux across the surface S produced by the give discrete charges indicated in the figure
This result is valid for any arbitrary surface S
q5
18
Applying Gauss' Law to problems that present planar symmetry
See also textbook,page 617
S cylindrical surface(it has a circular base of area A)
S
22
σ1 σ2E1 E1 E1
E2 E2E2
E2
E1
EA = E1 + E2
Example
EC = E1 + E2 =
σ1 σ2E1 E1 E1
E2 E2 E2
+ E2 = E1 =
Example
This arrangement ofcharges is used todescribe(approximately) theworking principle of aCAPACITOR
E
E
X
X
-
23
Applying Gauss' Law to problems that present spherical symmetry
E (r) =
E
E
E
CASE 1: Spherical shell uniformly charged.
Question:What is the electric field inside the sphere?
Answer
S: Gaussian surfaceA spherical surface of radius r
E
E
Q is the total charge on the sphere
24
Question:What is the electric field outside the sphereof radius R?
S: Gaussian surfaceA spherical surface of radius r
E
26
ECASE 2: Compact sphere of radius R uniformly charged.
S: Gaussian surfaceA spherical surface of radius r
29
We have just determined the electric field E for positions r<R (that is, fields inside the charged dielectric sphere). Let's procedd calculating E for positions outside the sphere, that is for r>R
33 A compact conductor has a total charge Q. Under electrostatic conditions, where are those charges located?
There is not net charge inside the conductor. All the charges distribute at the outer surface of the conductor
r
Would there be charge inside the conducting sphere?
R
r < R
Gaussian surface
Let's use Gauss' law
Therefore, charge inside = 0