Question: Q. 4. A charge $+Q$, is uniformly distributed within a sphere of radius $R$. Find the electric field, due to this charge distribution, at a distant point $r$ from the centre of the sphere where :
(i) $0<r<R$
(ii) $r$
A [Foreign, 2016]
Show Answer
Solution:
Ans. We have
$$ E=\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r^{2}}, \text { for a point charge } $$
Now (volume) charge density
$$ \rho=\frac{Q}{\left(\frac{4 \pi}{3} R^{3}\right)} $$
(i) $\therefore$ Charge contained within a sphere of radius $r$ $(0<r<R)$
$$ Q^{\prime}=\rho \cdot \frac{4 \pi}{3} \cdot r^{3}=Q\left(\frac{r^{3}}{R^{3}}\right) $$
$\therefore$ Electric Field,
$$ E=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q^{\prime}}{r^{2}}=\left(\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{R^{3}} \cdot r\right) \quad 1 / 2 $$
(ii) For $r>R$
Electric field $=($ Electric field due to a point charge $Q$ at the centre) 1
$$ E=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{r^{2}} $$
[CBSE Marking Scheme, 2016]
