#2015_27 How well you know about electric field?

Electricity and Magnetism Level 3

A non-conducting solid sphere has volume charge density that varies as ρ = ρ 0 r \rho ={ \rho }_{ 0 }r where ρ 0 { \rho }_{ 0 } is a constant and r r is distance from center. Find out electric field intensities at r < R r<R , where R R is the radius of sphere.

Given,

R R = 4 meter

r r = 2 meter

ρ 0 { \rho }_{ 0 } = 8.854 × 10 − 11 8.854\times { 10 }^{ -11 }

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The answer is 10.

Anandhu Raj by Anandhu Raj , 23, India

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1 solution

In this problem first we find the charge enclosed by the sphere and then apply Gauss' law.

Q = ∫ 0 2 4 π r 2 ρ 0 r d r Q=\int^2_0 4\pi r^2\rho_0 r \ dr

Q = 16 π ρ 0 Q=16\pi\rho_0

Since the Electric field is equal at all points on the outer surface of the sphere, E is constant.

Φ = ∮ S E → ⋅ d A → . \Phi = \oint_{\mathcal{S}} \overrightarrow{E} \cdot d \overrightarrow{A}.

Φ = E → ⋅ ∮ S d A → . \Phi = \overrightarrow{E} \cdot\oint_{\mathcal{S}} d \overrightarrow{A}.

Φ = E → ⋅ 4 π r 2 \Phi = \overrightarrow{E}\cdot 4\pi r^2

So by Gauss' law, Φ = Q ϵ 0 \Phi=\frac{Q}{\epsilon_0}

Putting the known values we get E = ρ ϵ 0 = 10 E=\frac{\rho}{\epsilon_0} =\boxed{10}

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