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Two equal charges q are placed at a distance of 2a and a third charge -2 q is placed at the midpoint. The potential energy of the system is-

Option: 1

\frac{q^{2}}{8 \pi \epsilon_{0} a}


Option: 2

\frac{6 q^{2}}{8 \pi \epsilon_{0} a}


Option: 3

-\frac{7 q^{2}}{8 \pi \epsilon_{0} a}


Option: 4

\frac{9 q^{2}}{8 \pi \epsilon_{0} a}


Answers (1)

best_answer



Potential energy -

U =\frac{1}{4 \pi \epsilon_{0}} \frac{q(-2 a)}{a}+\frac{1}{4 \pi \epsilon_{0}} \frac{q(-2 q)}{a}+\frac{1}{4 \pi \epsilon_{0}} \frac{q a}{2 q}
    =\frac{1}{4 \pi \epsilon _{0}} \frac{q^{2}}{a}\left[-2-2+\frac{1}{2}\right]

U =\frac{1}{4 \pi \epsilon_{0}} \frac{q^{2}}{a}\left(-\frac{7}{2}\right)=-\frac{7 q^{2}}{8 \pi \epsilon_{0} a} Ans.

Posted by

Sanket Gandhi

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