JEE Advanced · Physics · 17. Electrostatics
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 \(a\).
- A 1
- B 2
- C 3
- D 4
Answer & Solution
Correct Answer
(B) 2
Step-by-step Solution
Detailed explanation
From Gauss theorem,
\(E \propto \frac{q}{r^2} (q=\text { charge enclosed }) \)
\( \therefore \frac{E_2}{E_1}=\frac{q_2}{q_1}=\frac{r_1^2}{r_2^2} \)
\( 8=\frac{\int_0^R\left(4 \pi r^2\right) k r^a d r}{\int_0^{R / 2}\left(4 \pi r^2\right) k r^a d r} \times \frac{(R / 2)^2}{(R)^2}\)
Solving this equation we get, \(a=2\)
\(E \propto \frac{q}{r^2} (q=\text { charge enclosed }) \)
\( \therefore \frac{E_2}{E_1}=\frac{q_2}{q_1}=\frac{r_1^2}{r_2^2} \)
\( 8=\frac{\int_0^R\left(4 \pi r^2\right) k r^a d r}{\int_0^{R / 2}\left(4 \pi r^2\right) k r^a d r} \times \frac{(R / 2)^2}{(R)^2}\)
Solving this equation we get, \(a=2\)
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