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Physics Question on Electrostatic potential

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

A

18 \frac{1}{8}

B

16 \frac{1}{6}

C

14 \frac{1}{4}

D

12 \frac{1}{2}

Answer

14 \frac{1}{4}

Explanation

Solution

Total potential energy of the system is U=14πe0[(−q)(+Q)x+(+Q)(−q)x+(−q)(−q)2x]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] =14πe0[−qQx−qQx+q22x] = \frac{1}{4\pi e_{0}}\left[-\frac{qQ}{x} - \frac{qQ}{x} + \frac{q^{2}}{2x}\right] =14πe0[−2qQx+q22x] = \frac{1}{4\pi e_{0}} \left[ -\frac{2qQ}{x} + \frac{q^{2}}{2x}\right] As per question , U=0U = 0 −2qQx+q22x=0 - \frac{2qQ}{x}+ \frac{q^{2}}{2x} = 0 2qQx=q22xorQq=14 \frac{2qQ}{x} =\frac{q^{2}}{2x} or \frac{Q}{q} = \frac{1}{4}