question_answer

The electric field in a certain region is acting radially outward and is given by \[E=Ar.\text{ }A\]. charge contained in a sphere of radius \['a'\] centred at the origin of the field' will be given by [NEET 2015]

A) \[4\pi {{\varepsilon }_{0}}A{{a}^{2}}\]

B) \[A{{\varepsilon }_{0}}{{a}^{2}}\]

C) \[A\pi {{\varepsilon }_{0}}A{{a}^{3}}\]

D) \[{{\varepsilon }_{0}}A{{a}^{3}}\]

Correct Answer: A

Solution :

[a] Given,

Here,     \[r=a\]

\[E=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{q}{{{a}^{2}}}\]

From Eq. (i),

\[\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{q}{{{a}^{2}}}=Aa\]

\[\Rightarrow \]   \[q=4\pi {{\varepsilon }_{0}}A{{a}^{3}}\]