(a) Distinguish, with the help of a suitable diagram, the difference in the behaviour of a conductor and a dielectric placed in an external electric field. How does polarised 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.
Solution
(a) A conductor placed in an external field develops induced surface charges whose field cancels the external field inside, so Einside=0. A dielectric only partially cancels it: the polarised bound charges set up an opposing field, reducing the net internal field to E=KE0. (b) The battery is disconnected, so Q is fixed. (i) Charge stored is unchanged. (ii) E=ε0σ=Aε0Q depends only on Q, so the field strength is unchanged on doubling the separation. (iii) C=dε0A halves, and U=2CQ2 with fixed Q therefore doubles (work is done against the plate attraction in separating them).
CBSE Class 12 Board · 2016 · All India · Subjective
