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.You need to login to perform this action.
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