Electric field

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Electric field
The problem:
Given a thin sphere, radius R, charge Q (divided evenly across the sphere), calculate the electric
field inside and outside of the sphere:
1. using direct integration (the Coulomb’s law).
2. using the Gauss’s law.
Using the result, calculate the electric field of a homogeneously charged solid sphere, radius R,
charge Q.
The solution:
1.According to the Coulomb’s law
~0
~ = kdq(~r − r )
dE
|~r − r~0 |3
In our case:
Q
Q
dq
=
=
σ =
da
A
4πR2
2
da = R sin θdθdφ
Q sin θ
dq =
dθdφ
4π
For simplicity let us choose
(1)
(2)
(3)
(4)
~r = (0, 0, z)
(5)
(any other point can be reached by rotating the coordinates) and
r~0 = (R sin θ cos φ, R sin θ sin φ, R cos θ)
(6)
then
|~r − r~0 | = (R2 + z 2 − 2Rz cos θ)1/2
(7)
Therefore,
E=
kQ
4π
Z
π
Z
sin θdθ
0
2π
dφ
0
1
(−R sin θ cos φ, −R sin θ sin φ, z − R cos θ) (8)
(R2 + z 2 − 2Rz cos θ)3/2
Ex and Ey both equal 0 (since sin φ and cos φ are both periodic for 2π).
Z
kq π
sin θ(z − R cos θ)
Ez =
dθ
2 0 (R2 + z 2 − 2Rz cos θ)3/2
substituting cos θ for t we get:
Z
kQ 1
z − Rt
Ez =
dt
2
2 −1 (R + z 2 − 2Rzt)3/2
Z 1
∂
dt
kQ
=
· (−1)
2
∂z −1 (R2 + z 2 − 2Rzt)1/2
kQ ∂
1
2
2
1/2 1
=
(R + z − 2Rzt) |−1
2 ∂z Rz
1
kQ ∂
=
(|R − z| − |R + z|)
2 ∂z Rz
(
kq/z 2 , |z| > R
Ez =
0,
|z| < R
1
(9)
(10)
(11)
(12)
(13)
(14)
2. According to the Gauss’s law:
Z
~ · d~a = 4πkq
Φ= E
(15)
Let us find the electric field at the point r1 (spherical coordinates). We build a sphere with a
radius r1 . Due to the symmetry, the magnitude of the field on the sphere is the same, and in the r̂
direction. Then
~ ·A
~ = E · 4πr2 = 4πkq
Φ=E
1
(16)
where Q being the charge inside the sphere r1 .
If r1 > R then all of charge is in the sphere, and we get :
E=
kQ
r2
(17)
and if r1 < R, then no charge is inside the sphere r1 , and so:
E=0
(18)
which is, of course, the same result as we got before.
3. Let us think of the solid sphere as a collection of thin spheres. Each thin sphere at the radius
r0 < R and of thickness dr0 is charged with the charge
dq = ρdV =
Q
dr0
4/3πR3
(19)
For r > R each thin sphere induces dEr = kdq
. Therefore, Er = kQ
.
r2
r2
For r < R one has to consider only the thin spheres with the radii r0 < r (since the outer spheres
3
contribute zero). Therefore, the total contributing charge is q = Q Rr 3 , and the field is Er = kQ
r.
R3
Summarising,
(
kQ
r, r < R
R3
Er = kQ
(20)
, r>R
r2
2
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