Question

A sphere of radius $R$ have volume charge density given as
$\rho \left(\right.r\left.\right)=Kr$ for $r\leq R$
$=0$ for $r>R$
If electric field at distance $\frac{R}{2}$ from centre is $\frac{KR^{2}}{N \epsilon _{0}}$ Then $N$ will be.

Answer 1

$\oint\overset{ \rightarrow }{E}\cdot d\overset{ \rightarrow }{s}=\frac{q_{in}}{\epsilon _{0}}$
$E4\pi \left(\frac{R}{2}\right)^{2}=\frac{\displaystyle \int _{0}^{R / 2} \rho 4 \pi r^{2} dr}{\left(\epsilon \right)_{0}}$
$\Rightarrow E\pi R^{2}=\frac{\displaystyle \int _{0}^{R / 2} Kr 4 \pi r^{2} dr}{\epsilon _{0}}=\frac{K R^{2}}{16 \epsilon _{0}}$