An electric dipole is placed at an angle of $${30^ \circ }$$ with an electric field intensity $$2 \times {10^5}\,N/C.$$    It experiences a torque equal to $$4\,Nm.$$  The charge on the dipole, if the dipole length is $$2\,cm,$$  is

Solution :
$$\because $$ Torque on an electric dipole in an electric field,
$$\tau = p \times E \Rightarrow \left| \tau \right| = pE\sin \theta $$
where, $$\theta $$ is angle between $$E$$ and $$p$$
$$\eqalign{ & \Rightarrow 4 = p \times 2 \times {10^5} \times \sin 30 \cr & \Rightarrow p = 4 \times {10^{ - 5}}cm \cr & \Rightarrow p = q2l \cr & \therefore q2l = 4 \times {10^{ - 5}} \cr} $$
where, $$2l = 2\,cm = 2 \times {10^{ - 4}}m$$
$$\therefore q = \frac{{4 \times {{10}^{ - 5}}}}{{2 \times {{10}^{ - 2}}}} \Rightarrow 2 \times {10^{ - 3}}C = 2\,mC$$