Surface Charge on Conducting Sphere

Electricity and Magnetism Level 4

A conducting sphere of radius R R with a layer of charge Q Q distributed on its surface has the electric potential everywhere in space:

V = { 1 4 π ϵ 0 Q R 2 r 3 sin ⁡ θ cos ⁡ θ cos ⁡ ϕ , r > R 1 4 π ϵ 0 Q r 2 R 3 sin ⁡ θ cos ⁡ θ cos ⁡ ϕ , r < R . V = \begin{cases} \dfrac{1}{4\pi \epsilon_0} \dfrac{QR^2}{r^3} \sin \theta \cos \theta \cos \phi, \ \ r>R \\ \dfrac{1}{4\pi \epsilon_0} \dfrac{Qr^2}{R^3} \sin \theta \cos \theta \cos \phi, \ \ r<R. \end{cases}

Which of the following gives the surface charge density on the surface of the sphere?


Note: Recall that the change in electric field across either side of a conductor is equal to σ ϵ 0 , \frac{\sigma}{\epsilon_0}, where σ \sigma is the surface charge density.

− Q 4 π R 2 sin ⁡ θ cos ⁡ θ cos ⁡ ϕ -\frac{Q}{4\pi R^2} \sin \theta \cos \theta \cos \phi Q 4 π R 2 sin ⁡ θ cos ⁡ θ cos ⁡ ϕ \frac{Q}{4\pi R^2} \sin \theta \cos \theta \cos \phi Q 4 π R 2 3 sin ⁡ θ cos ⁡ θ cos ⁡ ϕ \frac{Q}{4\pi R^2} 3 \sin \theta \cos \theta \cos \phi Q 4 π R 2 5 sin ⁡ θ cos ⁡ θ cos ⁡ ϕ \frac{Q}{4\pi R^2} 5 \sin \theta \cos \theta \cos \phi

Matt DeCross by Matt DeCross , 27, USA

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2 solutions

Matt DeCross
May 10, 2016

One can use spherical harmonics for this problem, but it is overkill. Simply apply E = − ∇ ϕ E = -\nabla \phi using spherical symmetry and the hint. The electric field is thus:

E = { 3 4 π ϵ 0 Q R 2 r 4 sin ⁡ θ cos ⁡ θ cos ⁡ ϕ , r > R − 1 2 π ϵ 0 Q r R 3 sin ⁡ θ cos ⁡ θ cos ⁡ ϕ , r < R . E = \begin{cases} \dfrac{3}{4\pi \epsilon_0} \dfrac{QR^2}{r^4} \sin \theta \cos \theta \cos \phi, \qquad &r>R \\ \dfrac{-1}{2\pi \epsilon_0} \dfrac{Qr}{R^3} \sin \theta \cos \theta \cos \phi, \qquad &r<R \end{cases}.

The difference on either side of r = R r=R is:

σ ϵ 0 = 3 4 π ϵ 0 Q R 2 R 4 sin ⁡ θ cos ⁡ θ cos ⁡ ϕ + 1 2 π ϵ 0 Q R R 3 sin ⁡ θ cos ⁡ θ cos ⁡ ϕ = Q 4 π R 2 ϵ 0 5 sin ⁡ θ cos ⁡ θ cos ⁡ ϕ . \frac{\sigma}{\epsilon_0} = \dfrac{3}{4\pi \epsilon_0} \dfrac{QR^2}{R^4} \sin \theta \cos \theta \cos \phi + \dfrac{1}{2\pi \epsilon_0} \dfrac{QR}{R^3} \sin \theta \cos \theta \cos \phi = \frac{Q}{4\pi R^2 \epsilon_0} 5 \sin \theta \cos \theta \cos \phi .

Multilying both sides by ϵ 0 \epsilon_0 gives the answer.

4 Helpful 0 Interesting 0 Brilliant 0 Confused
Aryan Goyat
Apr 23, 2016

it would have been a master piece if you would have not given "THE HINT".

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Very true! it was a kind of hard to think of this !

Prakhar Bindal - 5 years, 1 month ago

Exactly.😀😁

rajdeep brahma - 3 years, 2 months ago

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