A) \[\frac{\rho R}{3{{\varepsilon }_{0}}}\]
B) \[\frac{\rho r}{{{\varepsilon }_{0}}}\]
C) \[\frac{\rho r}{3{{\varepsilon }_{0}}}\]
D) \[\frac{3\rho R}{{{\varepsilon }_{0}}}\]
Correct Answer: C
Solution :
Charge which is enclosed in the surface is given by\[\sigma =\frac{4}{3}\pi {{r}^{3}}\rho \]? Now, according to Gaussian law \[\varepsilon \oint{Eds}=\sigma \] \[{{\varepsilon }_{0}}E4\pi {{r}^{2}}=\frac{4}{3}\pi {{r}^{3}}\rho \] \[E=\frac{\rho r}{3{{\varepsilon }_{0}}}\]
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