Question 282: Electrostats & Capacitance

Question

A parallel plate capacitor having capacitance 12 pF is charged by a battery to a potential difference of 10 V between its plates. The charging battery is now disconnected and a porcelain slab of dielectric constant 6.5 is slipped between the plates. The work done by the capacitor on the slab is :

Accepted Answer

Initial energy of capacitor
U
i
= $${1 \over 2}$$ $${{{v^2}} \over c}$$
=  $${1 \over 2}$$ $$ 	imes $$ $${{120 	imes 120} \over {12}}$$ = 600 J
Since battery is disconnected so charge remain same.
Final energy of capacitor
U
f
= $${1 \over 2}{{{v^2}} \over c}$$
=  $${1 \over 2} 	imes {{120 	imes 120} \over {12 	imes 6.5}}$$ = 92
W + U
f
= U
i
W = 508 J

Answer

508 pJ

Answer

692 pJ

Answer

560 pJ

Answer

600 pJ

Electrostats & Capacitance

A parallel plate capacitor having capacitance 12 pF is charged by a battery to a potential difference of 10 V between its plates. The charging battery is now disconnected and a porcelain slab of dielectric constant 6.5 is slipped between the plates. The work done by the capacitor on the slab is :

Choose the correct answer: