A parallel plate capacitor is charged to a potential difference V by a d.c. source, The capacitor is then disconnected from the source. If the distance between the plates is doubled, state with reason how the following will change.
(i) Electric field between the plates,
(ii) Capacitance and
(iii) Energy stored in the capacitor

Solution

Given, a parallel plate capacitor is charged to a potential difference V. If the distance between the plates is doubled then,

i) Electric field in between the plates is given by $E = \frac{σ}{ε_{\circ}}$
Since, E does not depend upon distance between the capacitor plates, electric field remains unaffected.

ii) Capacitance is given by C= $\frac{ε_{\circ} A}{d}$
When distance between the plates is doubled C'= $\frac{ε_{\circ} A}{2 d} = \frac{1}{2} C$
That is, capacitance becomes half of it's initial value.

iii) Energy is given by U= $\frac{1}{2} \frac{Q^{2}}{C}$
There is no change in Q but, capacitance changes. Since Capacitance becomes half, stored energy of the capacitor becomes twice of it's initial value. [Capacitance and energy have inverse dependance].

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