Question

An insulating sphere of radius a, centered at the origin, has a uniform volume charge density ρ. part a find the electric field e⃗ (r⃗ ) inside the sphere (for r<
a.in terms of the position vector r⃗ . express your answer in terms of r⃗ , ρ (greek letter rho), and ϵ0. hints

Answer 1

As per Gauss law we can say

$\int E.dA = \frac{q_{en}}{\epsilon_0}$

now here in order to find the enclosed charge

$q_{en} = \rho * \frac{4}{3}\pi r^3$

now from the above equation

$E.\int dA = \frac{\rho *\frac{4}{3} \pi r^3}{\epsilon_0}$

$E.4\pi r^2 = \frac{\rho *\frac{4}{3} \pi r^3}{\epsilon_0}$

$E = \frac{\rho r}{3\epsilon_0}$

so above is the electric field in this inner position

Answer 2

The electric field is in the $E= \dfrac{ \rho \pi r}{3\epsilon_0}$ inner area.

Given to us

Insulating sphere radius = a,

Uniform volume charge density = ρ,

What is Gauss law?

We know that according to the Gauss Law,

$\int E.da= \dfrac{q_{en}}{\epsilon_0}$,

We know that to know that the charge is enclosed or not,

$q_{en} = \rho \times \dfrac{4}{3}\pi r^3$

Substitute the above equation against each other,

$\int E.da= \dfrac{ \rho \times \dfrac{4}{3}\pi r^3}{\epsilon_0}$

$E.4\pi r^2= \dfrac{ \rho \times \dfrac{4}{3}\pi r^3}{\epsilon_0}$

$E= \dfrac{ \rho \pi r}{3\epsilon_0}$

Hence, the electric field is in the $E= \dfrac{ \rho \pi r}{3\epsilon_0}$ inner area.

Learn more about Gauss law:

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