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BITSAT 2015
Exam
BITSAT
Year
2015
Subject
Physics
Chapter
Electrostatics

BITSAT 2015 · Physics · Electrostatics

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

A solid sphere of radius RR has a charge QQ distributed in its volume with a charge density ρ=kra\rho=k r^{a}, where kk and aa are constants and rr is the distance from its centre. If the electric field at r=R2r=\frac{R}{2} is 18\frac{1}{8} times that at r=R,r=R, the value of aa is

Options

  1. A3
  2. B5
  3. C2
  4. D7

Correct answer

C. 2

Step-by-step solution

Using Gauss's law, we have \oint E⃗⋅\vec{E} \cdot d A⃗=1ϵ0∫(ρ\vec{A}=\frac{1}{\epsilon_{0}} \int(\rho d v)=1ϵ0∫0Rkra×4πv)=\frac{1}{\epsilon_{0}} \int_{0}^{R} k r^{a} \times4 \pi r^{2} d r or E×4πR2=(4πkϵ0)R(a+3)(a+3)∴E1=kR(a+1)ϵ0(a+3)E \times 4 \pi R^{2}=\left(\frac{4 \pi k}{\epsilon_{0}}\right) \frac{R^{(a+3)}}{(a+3)} \therefore \quad E_{1}=\frac{k R^{(a+1)}}{\epsilon_{0}(a+3)} For r=R2⋅E2=k(R2)a+1ϵ0(a+3)r=\frac{R}{2} \cdot E_{2}=\frac{k\left(\frac{R}{2}\right)^{a+1}}{\epsilon_{0}(a+3)} Given, E2=E18E_{2}=\frac{E_{1}}{8} or k(R2)a+1ϵ0(a+3)=18kR(a+1)ϵ0(a+3)\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)}∴1a+1=18 or a=2.\begin{array}{ll} \therefore& \quad \frac{1}{a+1}=\frac{1}{8}\\ \text{ or }& a=2. \end{array}$$
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