Transcript Gauss’ Law

Some Measures of Electric Charge
Name
Symbol
SI Unit
Charge
q
C
Linear Charge
Density

C/m
Surface Charge
Density

C/m2
Volume Charge
Density

C/m3
Electric Field Due to a Ring of Charge
The ring of uniform charge has a linear charge density of 
(in coulomb/m). Find the electric field strength at point P
which is located a distance x from the center of the ring as
shown.
You must consider the contribution each differential
piece (dQ) has on the electric field at point P.
dQ =  ds
Solving for E
1 dQ
1 ds
dE 

2
40 r
40 r 2
dE 
1
 ds
40 ( x 2  a 2 )
x
x
cos    2
r ( x  a 2 )1 / 2
The vertical component of E will cancel out,
while the vertical component has a
magnitude of dE cos.
1
x
dE cos  
ds
2
2 3/ 2
40 ( x  a )
The Integral
dE cos  
1
x
ds
2
2 3/ 2
40 ( x  a )
1
x
E  dE cos  
40 ( x 2  a 2 )3 / 2

1
x
E
40 ( x 2  a 2 )3 / 2
2R
 ds
 ds
0
1
( 2R)x
E
40 ( x 2  R2 )3 / 2
1
qx
E
40 ( x 2  R2 )3 / 2
Electric Field Due to a
Charged Disk
The disk of uniform charge has a surface charge
density of  (in C/m2). Find the electric field
strength at point P which is located a distance z
from the center of the disk as shown.
z
The plan is to divide the disk into concentric flat
rings, and to add up the contributions each ring
has to the electric field at point P.
dr
dq   dA  ( 2r dr )
P
r
R
Solving for E
P
1
qz
E
40 (z 2  R2 )3 / 2
1
z 2r dr
dE 
40 (z 2  r 2 )3 / 2
z
E  dE 
4 0
Electric field due to
a ring of charge.
r
dr

R
0
2 3 / 2
(z  r )
2
z
( 2r)dr
R
Evaluating the Integral
z
E  dE 
4 0

R
(z 2  r 2 )3 / 2 ( 2r)dr
P
0
Let U = (z2 +r2) and dU = 2r dr.
z
E  dE 
4 0

R
0
U
3 / 2
z
dU 
4 0
R
z
E
4 0
 (z 2  R )

 21




0

E
2 0

z
1 

z 2  R2





2  12
z
R
 U-1/2 
 1 
  2  0
r
R
Infinite Sheet of Charge
 
z
1 
E
2 0 
z 2  R2

R  , E 
2 0




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The electric field strength at any distance z
from an infinite sheet of charge is constant.
Calculate the Electric Field
Strength at Point P
Gauss’ Law
  qencl
 E  dA 
0
What is it and how can it help me?
What is Gauss’ Law???
Electric Flux Through a
Closed Surface

  q
encl
E  dA 
0
Charge
Enclosed
Gauss’s Law, like Coulomb’s Law can help you to
evaluate the electric field by relating the electric
field at a location to the charge(s) creating it.
(o = the permittivity of free space 8.854x10-12 C2/(N m2) = 1/4k)
Electric Flux
The Electric Flux FE is the
product of component of the
electric field strength E that
is perpendicular to a surface
times the surface area.
 
FE  E  A  EA cos   E A

ˆ
A  An
(SI: V.m = N.m2/C)
Unit vector
perpendicular to A
The Surface Integral
 
FE  E  dA

Gaussian Surface
Visualizing Electric Flux
What Makes Electric Flux??
You can think of electric flux as the number of electric field
lines that pass through a surface. You can imagine that the
more dense the the electric field lines, the greater the
electric field flux.
  qencl
More Importantly…  E  dA 
0
The net flux through a closed surface is proportional to the
charge enclosed by that surface.
AND, Therefore…
  qencl
 E  dA 
0
If there is no net charge inside some volume of
space then the electric flux over the surface of
that volume is always equal to zero.
If the electric field is parallel to the Gaussian
surface, the electric flux is through that surface
is zero.
Using Gauss’ Law
Gauss’ Law can be used to evaluate the electric field
strength in an electric field, by taking advantage of the
symmetry of a charge distribution.
  qencl
 E  dA 
0
The skill in using Gauss’ Law comes in picking a
Gaussian surface such that E everywhere on that
surface is constant, so that you can pull E out of
the integral, and evaluate (1) the area of the
Gaussian surface and (2) the enclosed charge
(qencl).
A Derivation of Coulomb’s Law
For a point charge q, the electric field strength E at distance r
from the charge may be found using Gauss’s Law:

r
+
  q
E  d A  enc 
0

E dA 
EA 
E4r 2 
E 
q enc
40r 2
By using a spherical Gaussian surface, we recognize that the electric
field strength E at distance r will be constant, and therefore, E falls
out of the integral. That’s what we want to happen. The surface
integral for a spherical Gaussian surface is the surface area of a
sphere.
Recipe for Success
1. Carefully draw a figure showing the location of all
charges, the direction of the resulting electric field
vectors E and the given dimensions.
2. Draw an imaginary closed Gaussian surface so that the
value of the magnitude of the electric field, E, is constant
on the surface… you’ll be dealing with spheres, cylinders
or other shapes that have some symmetry.
3. Write Gauss’ Law and perform the surface integral, which
should be a friendly integral if you choose the right
Gaussian surface.
4. You can factor the electric field out of the integral, and
simply evaluate the surface integral.
5. Write an expression for qencl, and then solve the Gauss’
Law for the magnitude of the electric field.
Electric Field Due to an Infinite
Line of Charge

  q
E  d A  enc 
0

E dA  E( 2rl)  0  0
q enc
l
E( 2rl) 

0
0
qenc= l
l

E

 0 ( 2rl)  0 ( 2r )
Electric Field Due to an
Infinite Sheet of Charge
  q
E  d A  enc 
0

E dA  2E( r
2
) 0
2
q

r
2E( r 2 )  enc 
0
0
r 2
qenc= A= r2

E

2
2 0 ( r ) 2 0
Electric Field Due to a Sphere
of Charge Inside the sphere (r < R):

R

E dA  E( 4r 2 )
r
qenc= V= 4r3/3

3Q
4R3
  q
E  d A  enc 
0
3
q

4

r
2
E( 4r )  enc 
0
3 0
4r
3
Q
r
E

2
3 0 ( 4r ) 4R3  0
Electric Field Due to a Sphere
of Charge
Outside the sphere (r > R):
R

  q
enc
E  dA 

0

E dA  E( 4r 2 )
r
q enc
Q
E( 4r ) 

0
0
2
E
Q
40r 2