A) \[4\pi {{\varepsilon }_{0}}Q\times {{10}^{22}}\,V/m\]
B) \[12\pi {{\varepsilon }_{0}}Q\times {{10}^{20}}\,V/m\]
C) \[4\pi {{\varepsilon }_{0}}Q\times {{10}^{20}}\,V/m\]
D) \[12\pi {{\varepsilon }_{0}}Q\times {{10}^{22}}\,V/m\]
Correct Answer: A
Solution :
The electric potential due to charge Q is \[V=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{Q}{r}\] where r is the distance of observation point from the charge. At the same point, electric field is \[E=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{Q}{r}\] Combining Eqs. (i) and (ii), we have \[E=\frac{4\pi {{\varepsilon }_{0}}{{V}^{2}}}{Q}=\frac{4\pi {{\varepsilon }_{0}}\times {{(Q\times {{10}^{11}})}^{2}}}{Q}\] \[=4\pi {{\varepsilon }_{0}}Q\times {{10}^{22}}\,V/m\]
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