A) \[64\,J\]
B) \[41\,J\]
C) \[16\,J\]
D) \[10\,J\]
Correct Answer: A
Solution :
Potential at the centre of square \[V=4\,\times \,\left( \frac{9\times {{10}^{9}}\times 50\times {{10}^{-6}}}{2/\sqrt{2}} \right)=90\sqrt{2}\times {{10}^{4}}V\] Work done in bringing a charge (q = 50 mC) from ¥ to centre (O) of the square is \[W=q\,({{V}_{0}}-{{V}_{\infty }})=q{{V}_{0}}\] Þ \[W=50\times {{10}^{-6}}\times 90\sqrt{2}\times {{10}^{4}}=64J\]You need to login to perform this action.
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