A solid sphere of radius R has a charge Q distributed in its volume with a charge density =k r^a , where k and a are constants and r is the distance from its centre. If the electric field at r=(R)/(2) is (1)/(8) times

Correct answer: 0 From Gauss theorem, aligned & E (q)/(r^2) (q= charge enclosed ) & (E_2)/(E_1)=(q_2)/(q_1)=(r_1^2)/(r_2^2) & 8= ∫_0^R(4 π r^2) k r^a d r ∫_0^ R / 2 (4 π r^2) k r^a d r × ((R / 2)^2)/((R)^2) & aligned Solving this equation we get, a=2

JEE Advanced 2009Paper 2

Exam: JEE Advanced
Year: 2009
Paper: Paper 2
Subject: Physics
Chapter: Electrostatics

JEE Advanced 2009 · Paper 2 · Physics · Electrostatics

A solid sphere of radius R has a charge Q distributed in its volume...

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}

\frac{1}{8}

times that at

r=R

, find the value of

a

Correct answer

Step-by-step solution

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

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