A) \[2.7\times {{10}^{4}}V\]
B) \[1.5\times {{10}^{3}}V\]
C) \[3\times {{10}^{2}}V\]
D) \[5\times {{10}^{3}}V\]
E) \[3\times {{10}^{3}}V\]
Correct Answer: A
Solution :
The potential \[V=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{1}{r}({{q}_{1}}+{{q}_{2}}+{{q}_{3}}+{{q}_{4}})\] \[V=(9\times {{10}^{9}})\left( \frac{1}{1m} \right)\{(3-3-4+7)\times {{10}^{-6}}C\}\] \[=2.7\times {{10}^{4}}\frac{N-m}{C}=2.7\times {{10}^{4}}\frac{J}{C}\] \[=2.7\times {{10}^{4}}V\]
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