A) \[4\pi {{\varepsilon }_{0}}A{{a}^{2}}\]
B) \[4{{\varepsilon }_{0}}{{a}^{2}}\]
C) \[4\pi {{\varepsilon }_{0}}A{{a}^{3}}\]
D) \[{{\varepsilon }_{0}}A{{a}^{2}}\]
Correct Answer: C
Solution :
Let charge enclosed in the sphere of radius is q. According to Gauss theorem, \[\oint{\vec{E}.\vec{d}s}=\frac{q}{{{\varepsilon }_{0}}}\Rightarrow E.4\pi {{r}^{2}}=\frac{q}{{{\varepsilon }_{0}}}\] We are given E = Ar, substituting this value in the above equation we get \[4\pi A{{r}^{3}}=\frac{q}{{{\varepsilon }_{0}}}\] (i) If\[r=a,\]then substituting the value in equation (i), we get\[q=4\pi {{\varepsilon }_{0}}A{{a}^{3}}\]
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