A parallel plate capacitor, each with plate area A and separation d, is charged to a potential difference V. The battery used to charge it is then disconnected. A dielectric slab of thickness d and dielectric constant k is now placed between the plates. What change, if any, will take place in
(i) charge on the plates,
(ii) electric field intensity between the plates,
(iii) capacitance of the capacitor.
Justify your answer in each case.

Solution

Capacitance of the capacitor before a dielectric slab is placed in between is given by,
$C_{\circ} = \frac{ε_{\circ} A}{d}$

Potential difference = V
Initial charge on capacitor -q_{o= $C_{o} V$}

i) When, the battery is disconnected, charge on the capacitor remains unchanged.
i.e, q= q_o= $\frac{ε_{\circ} A}{d} V$ .

(ii) Initial electric field between the plates- $E_{o} = \frac{σ}{ε_{\circ}} = \frac{\frac{q}{A}}{ε_{\circ}} = \frac{q}{{Aε}_{\circ}}$

when, dielectric is introduced in between the plates, the permittivity of medium becomes $K ε_{\circ}$

Now, electric field in between the plates- $E = \frac{q}{{AKε}_{\circ}} = \frac{E_{o}}{K}$

Thus, electric field is reduced by a factor of $\frac{1}{K}$ .

(iii) The capacitance increases due to the decrease in potential difference and for any dielectric, K>1
$∴$ $C = {kC}_{0}$

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