Q117 / 150 · MHT CET 2023 (11 May Shift 1) · Physics

Two charges of equal magnitude ' $q$ ' are placed in air at a distance ' $2 r$ ' apart and third charge ' $-2 \mathrm{q}$ ' is placed at mid point. The potential energy of the system is $\left(\varepsilon_0=\right.$ permittivity of free space)

  • A
    $-\frac{\mathrm{q}^2}{8 \pi \varepsilon_0 \mathrm{r}}$
  • B
    $-\frac{3 q^2}{8 \pi \varepsilon_0 r}$
  • C
    $-\frac{5 q^2}{8 \pi \varepsilon_0 r}$
  • Correct answer
    D
    $-\frac{7 q^2}{8 \pi \varepsilon_0 r}$

Solution

Potential energy of ' $n$ ' point charges, $$ \mathrm{U}=\frac{1}{4 \pi \varepsilon_0} \sum_{\text {all pairs }} \frac{\mathrm{q}_{\mathrm{j}} \mathrm{q}_{\mathrm{k}}}{\mathrm{r}_{\mathrm{jk}}} $$ For 3 point charges, $$ \begin{aligned} & U=-\frac{q(2 q)}{4 \pi \varepsilon_0 r}-\frac{q(2 q)}{4 \pi \varepsilon_0 r}+\frac{q(q)}{4 \pi \varepsilon_0(2 r)} \\ & U=-\frac{2 q^2}{4 \pi \varepsilon_0 r}-\frac{2 q^2}{4 \pi \varepsilon_0 r}+\frac{q^2}{4 \pi \varepsilon_0(2 r)} \\ & U=-\frac{4 q^2}{8 \pi \varepsilon_0 r}-\frac{4 q^2}{8 \pi \varepsilon_0 r}+\frac{q^2}{8 \pi \varepsilon_0 r} \\ & U=-\frac{7 q^2}{8 \pi \varepsilon_0 r} \end{aligned} $$

Hence, the correct answer is option: 'D'.

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