question_answer

Three point charges \[3\,nC\], \[6\,nC\] and \[9\,nC\] are placed  at the corners of an equilateral triangle of side 0.1m. The potential energy of the system is

A)  \[89100J\]          

B)  \[99100J\]

C)  \[8910\text{ }J\]            

D)  \[9910J\]

E)  None of the Above

Correct Answer: E

Solution :

: The potential energy of the system of three charges is \[U=\frac{1}{4\pi {{\varepsilon }_{0}}}\left[ \frac{{{q}_{1}}{{q}_{2}}}{{{r}_{12}}}+\frac{{{q}_{1}}{{q}_{3}}}{{{r}_{13}}}+\frac{{{q}_{2}}{{q}_{3}}}{{{r}_{23}}} \right]\] Here,             \[{{q}_{1}}=3nC=3\times {{10}^{-9}}C\] \[{{q}_{2}}=6nC=6\times {{10}^{-9}}C\] \[{{q}_{3}}=9nC=9\times {{10}^{-9}}C\] \[{{r}_{12}}={{r}_{13}}={{r}_{23}}=0.1m\] \[\therefore \]  \[U=(9\times {{10}^{9}}N\,{{m}^{2}}{{C}^{-2}})\left[ \frac{(3\times {{10}^{-9}}C)(6\times {{10}^{-9}}C}{0.1} \right.\]\[+\frac{(3\times {{10}^{-9}}C)(9\times {{10}^{-9}}C)}{0.1m}+\left. \frac{(6\times {{10}^{-9}}C)(9\times {{10}^{-9}}C)}{0.1m} \right]\] \[U=\frac{(9\times {{10}^{9}}N\,{{m}^{2}}\,{{C}^{-2}})}{(0.1m)}[(18\times {{10}^{-18}}{{C}^{2}})+(27\times {{10}^{18}}{{C}^{2}})\]        \[+(54\times {{10}^{-18}}{{C}^{2}})]\] \[=\frac{(9\times {{10}^{9}}\,N\,{{m}^{2}}\,{{C}^{-2}})(99\times {{10}^{-18}}{{C}^{2}})}{(0.1m)}\] \[=8910\times {{10}^{-9}}J\] * None of the given options is correct.