Baixe Griffiths Problema 2.27-28 e outras Exercícios em PDF para Eletromagnetismo, somente na Docsity!
ELECTRIC POTENTIAL FROM CHARGES - EXAMPLES 2
Link to: physicspages home page. To leave a comment or report an error, please use the auxiliary blog. Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall - Problems 2.27 - 2.28. Post date: 12 Oct 2011. Here are a couple more examples of calculating the potential from the charge distribution. Example 1. Given a uniformly charged solid sphere of radius R and total charge q, find the potential from the formula
V (r) =
4 π 0
|r − r′|
ρ(r′)d^3 r′^ (1)
In this case, the charge density is constant, and can be expressed in terms of q by
ρ =
3 q 4 πR^3
Using the cosine law, we have
|r − r′| =
r^2 + r′^2 − 2 rr′^ cos θ (3)
so in spherical coordinates, we get
V (r) =
3 q 16 π^2 0 R^3
ˆ (^2) π
0
ˆ R
0
ˆ (^) π
0
r′^2 sin θdθdr′dφ √ r^2 + r′^2 − 2 rr′^ cos θ
q 8 π 0 R
r^2 R^2
which agrees with the result in Example 1 in an earlier post. Example 2. Given a uniformly charged solid cylinder of length L, radius R and charge density ρ with its axis along the z axis and centre at the origin, find the potential at location zV on the z axis, where we’re assuming that zV > L/2. We can use the result of Example 3 in an earlier post. The potential of a charged disk is 1
ELECTRIC POTENTIAL FROM CHARGES - EXAMPLES 2 2
V =
σ 2 0
[√
z^2 + R^2 − z
]
where z is the distance above the centre of the disk. In the case of the cylinder, we can slice the cylinder into a number of disks, each of thickness dz. For a slice at height z the distance from the centre of the slice to location zV is then zV − z, so the potential of the entire cylinder will be
V =
ρ 2 0
ˆ L/ 2
−L/ 2
[√
(zV − z)^2 + R^2 − zV + z
]
dz (7)
This is an unpleasant integral, but can be done with software. Maple gives (after condensing a few of the terms):
ρ
V =
( 2 zV + L)^2 + ( 2 R)^2
L
zV 4
( 2 zV − L)^2 + ( 2 R)^2
L
zV 4
R^2
ln
[√
( 2 zV − L)^2 + ( 2 R)^2 − 2 zV + L √ ( 2 zV + L)^2 + ( 2 R)^2 − 2 zV − L
]
− zV L (9)
Since the system has symmetry about the z axis, we can calculate the electric field along the axis by taking the gradient. After using Maple and doing some simplification we get (remember that we take the derivative w.r.t. zV in the gradient):
E = −∇V (10)
ρ 2 0
[
L −
( 2 zV + L)^2 + ( 2 R)^2 +
( 2 zV − L)^2 + ( 2 R)^2
]
z ˆ (11)
PINGBACKS Pingback: Work and energy - continuous charge Pingback: Laplace’s equation - average values of solutions Pingback: Laplace’s equation - charged disk
