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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 $a$.
Solution:
1798 Upvotes
Verified Answer
The correct answer is:
2
From Gauss theorem,
$$
\begin{aligned}
& E \propto \frac{q}{r^2} \quad(q=\text { charge enclosed }) \\
& \therefore \quad \frac{E_2}{E_1}=\frac{q_2}{q_1}=\frac{r_1^2}{r_2^2} \\
& 8=\frac{\int_0^R\left(4 \pi r^2\right) k r^a d r}{\int_0^{R / 2}\left(4 \pi r^2\right) k r^a d r} \times \frac{(R / 2)^2}{(R)^2} \\
&
\end{aligned}
$$
Solving this equation we get, $a=2$
$$
\begin{aligned}
& E \propto \frac{q}{r^2} \quad(q=\text { charge enclosed }) \\
& \therefore \quad \frac{E_2}{E_1}=\frac{q_2}{q_1}=\frac{r_1^2}{r_2^2} \\
& 8=\frac{\int_0^R\left(4 \pi r^2\right) k r^a d r}{\int_0^{R / 2}\left(4 \pi r^2\right) k r^a d r} \times \frac{(R / 2)^2}{(R)^2} \\
&
\end{aligned}
$$
Solving this equation we get, $a=2$
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