(a) Distinguish, with a help of a suitable diagram, the difference in the behavior of a conductor and a dielectric placed in an external electric field. How does polarized dielectric modify the original external field?
(b) A capacitor of capacitance C is charged fully by connecting it to a battery of emf, E. it is then disconnected from the battery. If the separation between the plates of the capacitor is now doubled, how will the following change?
(i) charge stored by the capacitor.
(ii) field strength between the plates.
(iii) Energy stored by the capacitor.
Justify your answer in each case.
OR
(a) Explain why, for any charge configuration, the equipotential surface through a point is normal to the electric field at that point.
Draw a sketch of equipotential surfaces due to a single charge (-q), depicting the electric field lines due to the charge.
(b) Obtain an expression for the work done to dissociate the system of three charges placed at the vertices of an equilateral triangle of side ‘a’ as shown below.

(a) When a conductor is placed in an external field, free charge carriers move and themselves such that induced electric field opposes the external field so that the net electrostatic field is zero.

In a dielectric, free movement is not possible. The dipole is induced by stretching or re-orienting the molecules such that the net effect of dipoles opposes the external field. This causes polarization. Thus the polarized dielectric is equivalent to two charged surfaces with induced surface charge densities.

(b) For the capacitor:
(i) The charge stored does not change because of the law of conservation of charges.
(ii) The field strength between the plates is
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As field strength is independent of the distance between the plates. Field strength remains same.
(iii) The energy stored in the capacitor is
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When distance between the plates is doubled, capacitance becomes half. Hence, energy stored will also double.
OR
(a) For any charge configuration, the equipotential surface through a point is normal to the electric field at that point. If field is not normal, it would have a component along the surface. To move a unit test charge against this component, work would have to be done.
This is contradiction to the definition of equipotential surface that there is no potential difference between any two points on the surface and no work is required to move a test charge on the surface. So, the equipotential surface through a point is normal to the electric field at that point.
Equipotential surface for a negative charge:
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(b) Work done to keep the system bound is
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So, work done to dissociate the system is
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Couldn't generate an explanation.
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