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BITSAT 2015 Paper

Section: Physics

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Question : 6 of 150

Marks: +1, -0

A solid sphere of radius R has a charge Q distributed in its volume with a charge density r =

kr^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, the value of a is

3
5
2
7

Solution:

Using Gauss's law, we have

∮

\overset{\rightarrow}{E}.\overset{\rightarrow}{dA}

=

\frac{1}{\epsilon_0}

∫ (ρdv) =

\frac{1}{\epsilon_0} \int\limits_{0}^{R} kr^a 8 4\pi r^2 dr

or E ×

4\pi R^2

=

\frac{4\pi k}{\epsilon_0} \frac{R^{a+3}}{a+3}

∴

E_1

=

\frac{k R^{a-1}}{\epsilon_0 (a+3)}

For , r =

\frac{R}{2}

,

E_2

=

\frac{k \left(\frac{R}{2}\right)^{a+1}}{\epsilon_0 (a+3)}

Given,

E_2

=

\frac{E_1}{8}

or

\frac{k \left(\frac{R}{2}\right)^{a+1}}{\epsilon_0 (a+3)}

=

\frac{1}{8} \frac{k R^{a+1}}{\epsilon_0 (a+3)}

∴

\frac{1}{2^{a+1}}

=

\frac{1}{8}

or a = 2

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