Question

A solid sphere of radius $R$ carries a uniform volume charge density $p$. The magnitude of electric field inside the sphere at a distance $r$ from the centre is :

• A. $\frac{ rp }{3 \varepsilon_{0}}$
• B. $\frac{R \rho}{3 \varepsilon_0}$
• C. $\frac{ R ^{2} p }{ r \varepsilon_{0}}$
• D. $\frac{ R ^{3} p }{ r ^{2} \varepsilon_{0}}$

Answer 1

Answer: (A)

Given, solid sphere,
Radius $=R$
Volume charge density $=p$
charge $=q$
distance from centre $=r$
To find Electric Field, $E$.
We know,
Electric field ( $E$ ),
$E=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q r}{R^{3}}\,\,\, ...(1)$
(above is formula for $E$ inside the sphere)
$\therefore E=\frac{q}{\frac{4}{3} \pi R^{3}} \times \frac{r}{3 \varepsilon_{0}}$
$\left[\right.$ Reurranging $\left.e q^{n}(1)\right]$
We know, volume $\Rightarrow V=\frac{4}{3} \pi R^{3}\,\,\, ...(2)$
charge density $\Rightarrow p=\frac{q}{V}\,\,\,...(3)$
$\therefore P=\frac{q}{\frac{4}{3} \pi R^{3}} \,\,\,[$ from (2) $f(3)]$
$\therefore E=\frac{p r}{3 \varepsilon_{0}}$