A) 9 V
B) zero
C) 2 V
D) 4.5 V
Correct Answer: B
Solution :
Potential at A due to charge at O \[{{V}_{A}}=\frac{1}{4\pi {{\varepsilon }_{0}}}=\frac{({{10}^{-3}})}{OA}\] \[=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{({{10}^{-3}})}{\sqrt{{{(\sqrt{2})}^{2}}}+{{(\sqrt{2})}^{2}}}\] \[=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{10}^{-3}}}{2}\] Potential at B due to charge at O \[{{V}_{B}}=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{({{10}^{-3}})}{OB}\] \[=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{({{10}^{-3}})}{2}\] So, \[{{V}_{A}}-{{V}_{B}}=0\]You need to login to perform this action.
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