You are given an isolated capacitor of capacitance C (without dielectric) C charged to a potential difference V. Find how many times will each of the following becomes if a dielectric slab of dielectric constant 4 is inserted between its plates, completely filling the gap between the plates:

a. capacitance

b. potential difference

c. field between the plates

d. energy stored by the plates

Open in App

Solution

Recall the definition of dielectric
C'=kC as k=4 so C'=4C so capacitance increases by 4 times
now Q=C'V'=CV and C'=4C so V'=V/4 (capacitor is isolated so Q=constant) so V'=1/4 times V
now E=V'/d=V/4d=1/4 times E
and U=Q^2/2C'=Q^2/2(4C)=1/4 times Ui

Suggest Corrections

Similar questions

.A capacitor is charged to potential V and is disconnected . A dielectric of dielectric
constant 4 is inserted filling the whole space between the plates. How do the following
change? –Capacitance, potential difference, Field between the plates, energy stored by
the capacitors.

Q. A dielectric with a dielectric constant of

k

is inserted between the plates of the capacitor of capacitance

C

having a potential difference of

Δ V

between the two plates. Calculate the energy stored in the capacitor?

Q. An isolated capacitor with a capacitance C has a potential difference between the plates of V, which leaves on each plate a charge

Q

.
A dielectric is then inserted in-between the plates, what affect does this have on the electric field between the plates?

If the potential difference between the plates of a charged and isolated parallel-plate condenser of capacitance 1 μF changes from 12 V to 4 V when a dielectric slab of dielectric constant K is introduced to fill the entire space between the plates, then

Q. A dielectric slab is inserted between the plates of an isolated charged capacitor. Which of the following quantities will remain the same?
(a) The electric field in the capacitor
(b) The charge on the capacitor
(c) The potential difference between the plates
(d) The stored energy in the capacitor

Join BYJU'S Learning Program

Recall the definition of dielectric C'=kC as k=4 so C'=4C so capacitance increases by 4 times now Q=C'V'=CV and C'=4C so V'=V/4 (capacitor is isolated so Q=constant) so V'=1/4 times V now E=V'/d=V/4d=1/4 times E and U=Q^2/2C'=Q^2/2(4C)=1/4 times Ui

Recall the definition of dielectric
C'=kC as k=4 so C'=4C so capacitance increases by 4 times
now Q=C'V'=CV and C'=4C so V'=V/4 (capacitor is isolated so Q=constant) so V'=1/4 times V
now E=V'/d=V/4d=1/4 times E
and U=Q^2/2C'=Q^2/2(4C)=1/4 times Ui