question_answer

 ABCD is a rectangle. At comers B, C and D of the rectangle are placed charges \[+10\,\times \,{{10}^{-10}}\,C,-20\,\times \,{{10}^{-12}}\,and\,10\,\times \,{{10}^{-12}}C,\]respectively. Calculate the potential at the fourth comer. (The side \[AB=4\text{ }cm\] and \[BC=3\text{ }cm\])

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 \[{{V}_{A}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{B}}}{AB}+\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{{{q}_{C}}}{AC}+\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{{{q}_{P}}}{AD}\] \[=\frac{1}{4\pi {{\varepsilon }_{0}}}\left[ \frac{10\times {{10}^{-12}}}{4\times {{10}^{-2}}}-\frac{20\times {{10}^{-12}}}{5\times {{10}^{-2}}}+\frac{10\times {{10}^{-12}}}{3\times {{10}^{-2}}} \right]\] \[=9\times {{10}^{9}}\times {{10}^{-10}}\left[ \frac{10}{4}-\frac{20}{5}+\frac{10}{3} \right]\] \[=\frac{9\times {{10}^{-1}}\times 11}{6}\] \[=16.5\times {{10}^{-1}}=1.65V\]