Explanation:
Here,\(E = \sqrt 3 \times {10^5}N{C^{ - 1}},\,l = 4\,cm,\,\tau = 6\,Nm\)
Torque, \(\vec \tau = \vec p \times \vec E;\tau = pE\,\,\sin \theta \)
\(\therefore 6 = p \times \sqrt 3 \times {10^5} \times \sin 60^\circ \) or \(p = 4 \times {10^{ - 5}}Cm\)
\(\therefore \,\,\,{\rm{Charge}},\,\,q = \frac{p}{l} = \frac{{4 \times {{10}^{ - 5}}Cm}}{{0.04\,\,m}}\)
\( = 1 \times {10^{ - 3}}C = 1\,\,mC\)