A) \[4\pi {{\varepsilon }_{0}} Q \times 1{{0}^{22}}\,V/m\]
B) \[4\pi {{\varepsilon }_{0}} Q \times 1{{0}^{20}}\,V/m\]
C) \[12\pi {{\varepsilon }_{0}} Q \times 1{{0}^{22}}\,V/m\]
D) \[12\pi {{\varepsilon }_{0}} Q \times 1{{0}^{20}}\,V/m\]
Correct Answer: A
Solution :
As electric potential \[V=\frac{1}{4\pi {{\varepsilon }_{0}}}\,\frac{Q}{r}\] Given, \[V=Q\times {{10}^{11}}\,V\] So, \[\frac{1}{4\pi {{\varepsilon }_{0}}}\,\,\frac{Q}{r}=Q\times {{10}^{11}}\,V\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{1}{r}=4\pi {{\varepsilon }_{0}}\,\,{{10}^{11}}\,\,V\] Now, electric field \[E=\frac{V}{r}=Q\times {{10}^{11}}\times 4\pi \,{{\varepsilon }_{0}}\,{{10}^{11}}\] \[=\,\,\,4\pi \,{{\varepsilon }_{0}}Q\times {{10}^{22}}\]
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