CBSE Class 12 · 2010 · CapacitanceShort Answer

A parallel-plate capacitor is charged to a potential difference

V

by a dc source. The capacitor is then disconnected from the source. If the distance between the plates is doubled, state with reason how the following change:
(i) electric field between the plates,
(ii) capacitance, and
(iii) energy stored in the capacitor.

Solution

The source is disconnected, so the charge

Q

(and hence

σ = Q / A

) is fixed. (i)

E = \frac{σ}{ε _{0}}

depends only on

σ

, so doubling

d

leaves

E

unchanged. (ii)

C = \frac{ε _{0} A}{d}

, so doubling

d

halves

C

. (iii)

U = \frac{Q ^{2}}{2 C}

; with

Q

fixed and

C

halved,

U

doubles (the extra energy comes from the work done in pulling the plates apart against their attraction).

CBSE Class 12 Board · 2010 · Subjective

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