Electrostatics Question 737
Question: A sphere of radius R carries charge density $ \rho $ proportional to the square of the distance from the center such that $ \rho =CR^{2} $ , where C is a positive constant. At a distance R/2 from the center, the magnitude of the electric field is
Options:
A) $ \frac{CR^{3}}{20{\in _{0}}} $
B) $ \frac{CR^{3}}{10{\in _{0}}} $
C) $ \frac{CR^{3}}{5{\in _{0}}} $
D) $ \frac{CR^{3}}{40{\in _{0}}} $
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Answer:
Correct Answer: D
Solution:
[d] For, $ r=R/2 $ Using Gauss’s law, we have
$ \int _{\vec{E}.d\vec{A}=\frac{q _{in}}{{\varepsilon _{0}}}\text{ or }E\times 4\pi r^{2}=\int\limits _{0}^{R/2}{\frac{\rho 4\pi r^{2}dr}{\varepsilon { _{0}}}}} $
$ \text{or }E=\frac{Cr^{3}}{5{\in _{0}}}=\frac{CR^{3}}{40{\in _{0}}}. $
