A) 1.8 V
B) \[1.8\times {{10}^{6}}\] V
C) \[1.8\times {{10}^{5}}\]V
D) \[1.8\times {{10}^{4}}\]V
Correct Answer: C
Solution :
Length of each side of square is so distance of it?s centre from each corner is \[\frac{\sqrt{2}}{\sqrt{2}}=1m.\] Potential at the centre \[V=9\times {{10}^{9}}\left[ \frac{10\,\times {{10}^{-6}}}{1}+\frac{5\times {{10}^{-6}}}{1}-\frac{3\times {{10}^{-6}}}{1}+\frac{8\times {{10}^{-6}}}{1} \right]\] \[=\text{ }1.8\times {{10}^{5}}V\]You need to login to perform this action.
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