A) 1.65 V
B) 0.165 V
C) 16.5 V
D) 2.65 V
Correct Answer: A
Solution :
The situation is summarised in figure. \[BC=AD=3cm,\] \[AB=DC=4\text{ }cm,~\]so \[AC=5cm.\] Now, potential at A \[{{V}_{A}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{B}}}{AB}+\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{{{q}_{C}}}{AC}+\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{{{q}_{P}}}{AD}\] \[=\frac{1}{4\pi {{\varepsilon }_{0}}}\left[ \frac{10\times {{10}^{-12}}}{4\times {{10}^{-2}}}-\frac{20\times {{10}^{-12}}}{5\times {{10}^{-2}}}+\frac{10\times {{10}^{-12}}}{3\times {{10}^{-2}}} \right]\] \[=9\times {{10}^{9}}\times {{10}^{-10}}\left[ \frac{10}{4}-\frac{20}{5}+\frac{10}{3} \right]\] \[=\frac{9\times {{10}^{-1}}\times 11}{6}\] \[=16.5\times {{10}^{-1}}=1.65V\]You need to login to perform this action.
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