Electrostatics Question 734
Question: A solid sphere of radius R has a charge Q distributed in its volume with a charge density$ \rho =kr^{a} $ , where k and an are constants and r is the distance from its center. If the electric field at $ r=\frac{R}{2} $ is $ \frac{1}{8} $ times that at $ r=R $ , the value of a is.
Options:
A) 3
B) 5
C) 2
D) both [a] and [b]
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Using Gauss’s law, we have $ \int _{\vec{E}.d\vec{A}}=\frac{1}{{\in _{0}}}\int _{{}}^{{}}{( \rho dv )=\frac{1}{{\in _{0}}}\int _{0}^{R}{kr^{a}\times 4\pi r^{2}dr}} $
$ \text{or }E\times 4\pi r^{2}=( \frac{4\pi k}{{\in _{0}}} )\frac{{{R}^{( a+3 )}}}{( a+3 )}\text{ }$
$\therefore E _{1}=\frac{k{{R}^{( a+1 )}}}{{\in _{0}}( a+3 )} $
$ \text{For } r = \frac{R}{2}\cdot E _{2}$
=$\frac{k{(\frac{R}{2} )}^{a+1}}{\in _{0}}( a+3 )=\frac{1}{8}\frac{k{{R}^{( a+1 )}}}{{\in _{0}}( a+3 )} $
$ \therefore {{2}^{\frac{1}{a+1}}}=\frac{1}{8}\text{ or }a=2. $
