
Charges $+q$ and $-q$ are placed at points $\mathrm{A}$ and $\mathrm{B}$ respectively which are a distance…

- $\frac{q Q}{2 \pi \varepsilon_0 L}$
- $-\frac{q Q}{6 \pi \varepsilon_0 L}$
- $\frac{q Q}{3 \pi \varepsilon_0 L}$
- $\frac{q Q}{4 \pi \varepsilon_0 L}$
Solution
Potential at \(\mathrm{C}=\mathrm{V}_{\mathrm{C}}=0\)
Potential at \(\mathrm{D}=\mathrm{V}_{\mathrm{D}}\)
\(=k\left(\frac{-q}{L}\right)+\frac{k q}{3 L}=-\frac{2}{3} \frac{k q}{L}\)
Potential difference
\(\mathrm{V}_{\mathrm{D}}-\mathrm{V}_{\mathrm{C}}=-\frac{2}{3} \frac{k q}{L}=\frac{1}{4 \pi \varepsilon_0}\left(-\frac{2}{3} \cdot \frac{q}{L}\right)\)
\(\Rightarrow\) Work done \(=\mathrm{Q}\left(\mathrm{V}_{\mathrm{D}}-\mathrm{V}_{\mathrm{C}}\right)\)
\(=-\frac{2}{3} \times \frac{1}{4 \pi \varepsilon_0} \frac{q Q}{L}=\frac{-q Q}{6 \pi \varepsilon_0 L}\)Asked in: NEET 2007