MHT CET · Physics · Electrostatics
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 \(-\frac{5 q^2}{8 \pi \varepsilon_0 r}\)
- D \(-\frac{7 q^2}{8 \pi \varepsilon_0 r}\)
Answer & Solution
Correct Answer
(D) \(-\frac{7 q^2}{8 \pi \varepsilon_0 r}\)
Step-by-step Solution
Detailed explanation
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,
\(
\begin{aligned}
& U=-\frac{q(2 q)}{4 \pi \varepsilon_0 r}-\frac{q(2 q)}{4 \pi \varepsilon_0 r}+\frac{q(q)}{4 \pi \varepsilon_0(2 r)} \\
& U=-\frac{2 q^2}{4 \pi \varepsilon_0 r}-\frac{2 q^2}{4 \pi \varepsilon_0 r}+\frac{q^2}{4 \pi \varepsilon_0(2 r)} \\
& U=-\frac{4 q^2}{8 \pi \varepsilon_0 r}-\frac{4 q^2}{8 \pi \varepsilon_0 r}+\frac{q^2}{8 \pi \varepsilon_0 r} \\
& U=-\frac{7 q^2}{8 \pi \varepsilon_0 r}
\end{aligned}
\)
\(
\mathrm{U}=\frac{1}{4 \pi \varepsilon_0} \sum_{\text {all pairs }} \frac{\mathrm{q}_{\mathrm{j}} \mathrm{q}_{\mathrm{k}}}{\mathrm{r}_{\mathrm{jk}}}
\)
For 3 point charges,
\(
\begin{aligned}
& U=-\frac{q(2 q)}{4 \pi \varepsilon_0 r}-\frac{q(2 q)}{4 \pi \varepsilon_0 r}+\frac{q(q)}{4 \pi \varepsilon_0(2 r)} \\
& U=-\frac{2 q^2}{4 \pi \varepsilon_0 r}-\frac{2 q^2}{4 \pi \varepsilon_0 r}+\frac{q^2}{4 \pi \varepsilon_0(2 r)} \\
& U=-\frac{4 q^2}{8 \pi \varepsilon_0 r}-\frac{4 q^2}{8 \pi \varepsilon_0 r}+\frac{q^2}{8 \pi \varepsilon_0 r} \\
& U=-\frac{7 q^2}{8 \pi \varepsilon_0 r}
\end{aligned}
\)
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