Two opposite and equal charges each of magnitude \(4 \times 10^{-8} \mathrm{C}\) form a dipole. Their…
- \(64 \times 10^{-4} \mathrm{Nm}\) and \(64 \times 10^{-4} \mathrm{~J}\)
- \(32 \times 10^{-4} \mathrm{Nm}\) and \(32 \times 10^{-4} \mathrm{~J}\)
- \(64 \times 10^{-4} \mathrm{Nm}\) and \(32 \times 10^{-4} \mathrm{~J}\)
- \(32 \times 10^{-4} \mathrm{Nm}\) and \(64 \times 10^{-4} \mathrm{~J}\)
Solution

\(\therefore\) Electric dipole moment, \(p=q \times 2 a\) \(=4 \times 10^{-8} \times 2 \times 10^{-4}=8 \times 10^{-12} \mathrm{C}-\mathrm{m}\) \(\therefore\) Maximum torque \(\left(\theta=90^{\circ}\right)\) is given as, \(\begin{aligned} \tau_{\max } & =p E \sin 90^{\circ}=p E \\ & =8 \times 10^{-12} \times 4 \times 10^8=32 \times 10^{-4} \mathrm{~N}-\mathrm{m} \end{aligned}\) Work done in rotating through \(180^{\circ}\) is given as \(\begin{aligned} W & =p E\left(\cos \theta_1-\cos \theta_2\right) \\ & =8 \times 10^{-12} \times 4 \times 10^8\left(\cos 0^{\circ}-\cos 180^{\circ}\right) \\ & =64 \times 10^{-4} \mathrm{~J} \end{aligned}\)
Asked in: AP EAMCET 2020 (18 Sep Shift 2)