A parallel plate capacitor of capacitance 12.5 pF is charged by a battery connected between its plates to potential difference of 12.0 V. The battery is now disconnected and a dielectric slab is inserted between the plates. The change in its potential en

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
A parallel plate capacitor of capacitance 12.5 pF is charged by a battery connected between its plates to potential difference of 12.0 V. The battery is now disconnected and a dielectric slab $(ε_{r} = 6)$ is inserted between the plates. The change in its potential energy after inserting the dielectric slab is _______ $\times 10^{- 12} J .$ [2024]

Ans.
(750) Before inserting dielectric, capacitance is given as $C_{0} = 12.5$ pF and charge on the capacitor is $Q = C_{0} V$ . After inserting dielectric capacitance will become $C_{f} = ε_{r} C_{0} .$

Change in potential energy of the capacitor

$∆ U = U_{0} - U_{f} = \frac{Q^{2}}{2 C_{0}} - \frac{Q^{2}}{2 C_{f}} = \frac{Q^{2}}{2 C_{0}} [1 - \frac{1}{ε_{r}}]$

$∆ U = \frac{{(C_{0} V)}^{2}}{2 C_{0}} [1 - \frac{1}{ε_{r}}] = \frac{1}{2} C_{0} V^{2} [1 - \frac{1}{ε_{r}}]$

Using $C_{0} = 12.5 p F,$ $V = 12$ $V,$ $ε_{r} = 6$

$∆ U = \frac{1}{2} (12.5) \times 12^{2} [1 - \frac{1}{6}] = \frac{1}{2} (12.5) \times 12^{2} \times \frac{5}{6}$

$= 750 p J = 750 \times 10^{- 12} J$

Want Us to Email you About Special Offers & Updates?