Electrostatics Ques 17
- A solid sphere of radius $R$ has a charge $Q$ distributed in its volume with a charge density $\rho=k r^{a}$, where $k$ and $a$ are constants and $r$ is the distance from its centre. If the electric field at $r=\frac{R}{2}$ is $\frac{1}{8}$ times that at $r=R$, find the value of $\underset{(2009)}{a}$
(2009)
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Answer:
Correct Answer: 17.2
Solution:
Formula:
- From Gauss’s theorem,
$$ \begin{array}{rlrl} & E & \propto \frac{q}{r^{2}} \quad(q \text { charge enclosed }) \\ \therefore \quad & \frac{E_{2}}{E_{1}}= & \frac{q_{2}}{q_{1}}=\frac{r_{1}^{2}}{r_{2}^{2}} \\ \text { or } \quad & =\frac{\int_{0}^{R / 2}\left(4 \pi r^{2}\right) k r^{a} d r}{R / 2} \times \frac{(R / 2)^{2}}{(R)^{2}} \\ & \int_{0}^{R}\left(4 \pi r^{2}\right) k r^{a} d r \end{array} $$
Solving this equation, we obtain,
$$ a=2 \text {. } $$
BETA
