Question

Three charges $-q, +Q$ and $-q$ are placed at equal distances along a straight line. If the total $P.E.$ of the system is zero, the ratio $ \frac{Q}{q} $ becomes

• A. $ \frac{1}{8}$
• B. $ \frac{1}{6}$
• C. $ \frac{1}{4}$
• D. $ \frac{1}{2}$

Answer 1

The Correct Option is C

Total potential energy of the system is $U = \frac{1}{4\pi e_{0}} \left[ \frac{\left(-q\right)\left(+Q\right)}{x} + \frac{\left(+Q\right)\left(-q\right)}{x} + \frac{\left(-q\right)\left(-q\right)}{2x}\right] $ $ = \frac{1}{4\pi e_{0}}\left[-\frac{qQ}{x} - \frac{qQ}{x} + \frac{q^{2}}{2x}\right]$ $ = \frac{1}{4\pi e_{0}} \left[ -\frac{2qQ}{x} + \frac{q^{2}}{2x}\right]$ As per question , $ U = 0 $ $ - \frac{2qQ}{x}+ \frac{q^{2}}{2x} = 0$ $ \frac{2qQ}{x} =\frac{q^{2}}{2x} or \frac{Q}{q} = \frac{1}{4}$