A parallel plate capacitor is charged and then the battery is disconnected, When the plates of the capacitor are brought closer, then

energy stored in the capacitor decreases

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

the potential difference between the plates decreases

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

the capacitance increases

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

the electric field between the plates decreases

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct options are
A energy stored in the capacitor decreases
B the potential difference between the plates decreases
C the capacitance increases
As battery is disconnected so charge q is constant.
The parallel plate capacitance is $C = \frac{A ϵ_{0}}{d} o r C \propto \frac{1}{d}$
If the capacitor are brought closer, d will decrease so C will increase.
Energy stored in capacitor is $U = \frac{q^{2}}{2 C} o r U \propto \frac{1}{C}$
As C increases, so U will decrease.
The electric field between plates is $E = \frac{q}{A ϵ_{0}}$
As q is constant so E becomes unchanged.
Potential difference between the plates is $V = E d o r V \propto d$
As d decreases so V will decrease.

Suggest Corrections

Similar questions

Q. A parallel plate capacitor is charged by a battery, which is then disconnected. A dielectric slab is then inserted in the space between the plates. Explain what changes, if any, occur in the values of
(i) capacitance.
(ii) potential difference between the plates.
(iii) electric field between the plates.
(iv) the energy stored in the capacitor.

Statement 1 :

A parallel plate capacitor is charged by a battery of voltage

V

. The battery is then disconnected. If the space between the plates is filled with a dielectric, the energy stored in the capacitor will decrease.

Statement 2 :

The capacitance of a capacitor increases due to the introduction of a dielectric between the plates.

.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 capacitor

C

is charged to a potential difference

V

and the battery is disconnected. Now, if the capacitor plates are brought closer slowly by some distance

Q. A capacitor of capacitance

2 μ F

is charged with a

10 V

battery. Now battery is disconnected and a dielectric of dielectric constant

K = 4

is inserted between the plates. What is the potential difference between the plates of the capacitor?

Join BYJU'S Learning Program