question_answer

An electric dipole of dipole moment p is placed in a uniform electric field E. Write the expression for the torque \[\tau \] experienced by the dipole. Show diagrammatically the orientation of the dipole in the field for which the torque is half of the maximum value.

Or

Two insulated charged copper spheres A and B have identical sizes and charge \[6.5\times {{10}^{-7}}\] C on each, and their centres are separated by distance of 50 cm.

A third sphere C of same size but uncharged is brought in contact with first, then brought in contact with the second and finally removed from both, find the new force of interaction, between spheres A and B.

Answer:

Consider an electric dipole consisting of two charges \[-\,q\] and \[+\,q\] placed in a uniform external electric field of intensity E. The length of the electric dipole is \[2l\].The dipole moment p makes an angle q with the direction of the electric field. Two forces F and \[-\,\mathbf{F}\]which are equal in magnitude and opposite in directions act on the dipole. \[|F|\,\,=\,\,|-F|\,\,=\,\,qE\] The net force is zero. Since, the two forces are equal in magnitude and opposite in direction and act at different points, therefore they constitute a couple. A net torque t acts on the dipole about an axis passing through the mid-point of the dipole. Now, \[\tau =\]perpendicular distance BC between the parallel \[\text{forces}\times \text{either force}=2lsin\,\theta .qE\] \[\tau =(q\times 2l)\,E\,\sin q\,\,\text{or}\,\,t=pE\,\sin \,\theta \] If \[q=30{}^\circ ,\] then \[\tau =pE\,sin\,\,30{}^\circ =(1/2)\,pE\] Or Since, all three spheres are of same size, their capacities are equal and they equally share their charges when brought in contact. On bringing 3rd sphere C in contact with A, \[{{q}_{c}}=q{{'}_{A}}=\frac{{{q}_{A}}}{2}=\frac{6.5\times {{10}^{-7}}}{2}\,\,C\] And then bringing C in contact with B, we have \[q{{'}_{C}}=q{{'}_{B}}=\frac{{{q}_{C}}+{{q}_{B}}}{2}\]             \[=\frac{1}{2}\,\,\left[ \frac{6.5\times {{10}^{-7}}}{2}+6.5\times {{10}^{-7}} \right]=\frac{9.75\times {{10}^{-7}}}{2}\,\,C\] \[\therefore \] New force of repulsion between A and B, \[F'=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{q{{'}_{A}}.q{{'}_{B}}}{{{r}^{2}}}\] \[=9\times {{10}^{9}}\times \frac{\frac{6.5\times {{10}^{-7}}}{2}\times \frac{9.75\times {{10}^{-7}}}{2}}{{{(0.5)}^{2}}}=5.7\times {{10}^{-\,3}}N\]