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

Given,

$R$
=
**
4 meter
**

$r$
=
**
2 meter
**

${ \rho }_{ 0 }$ = $8.854\times { 10 }^{ -11 }$

The answer is 10.

**
This section requires Javascript.
**

You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.

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

$Q=\int^2_0 4\pi r^2\rho_0 r \ dr$

$Q=16\pi\rho_0$

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

$\Phi = \oint_{\mathcal{S}} \overrightarrow{E} \cdot d \overrightarrow{A}.$

$\Phi = \overrightarrow{E} \cdot\oint_{\mathcal{S}} d \overrightarrow{A}.$

$\Phi = \overrightarrow{E}\cdot 4\pi r^2$

So by Gauss' law, $\Phi=\frac{Q}{\epsilon_0}$

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