A parallel plate capacitor of capacity C^° is charged to a potential V^° (I) .The energy stored in the capacitor when the battery is disconnected and the plate separation is doubled is E1(ii) the energy stored in the capacitor when the charging battery is kept connected and the separation between the capacitor plates is doubled is E2. then E1/E2 value is

Open in App

Solution

Suggest Corrections

Similar questions

Q. A parallel plate capacitor of capacity

C_{\circ}

is charged to a potential

V_{\circ}

.The energy stored in the capacitor when the battery is disconnected and the plate separation is doubled is

E_{1}

and the energy stored in the capacitor when the charging battery is kept connected and the separation between the capacitor plates is doubled is

E_{2}

. Then value of

\frac{E_{1}}{E_{2}}

is:

A capacitor of capacitance C is charged fully using a battery of e.m.f, E . It is then
disconnected form the battery . If the separation between the plates of the capacitor is
now doubled. What will happen to : Charge stored by the capacitor,P.D across it, field
strength between the plates and energy stored between the plates of the capacitor.. Answer the question again, if the battery is kept connected

Explain briefly the process of charging a parallel plate capacitor when it is connected across a d.c. battery.

A capacitor of capacitance ‘C’ is charged to ‘V’ volts by a battery. After some time the battery is disconnected and the distance between the plates is doubled. Now a slab of dielectric constant, 1 < k < 2, is introduced to fill the space between the plates. How will the following be affected?

(a) The electric field between the plates of the capacitor

(b) The energy stored in thecapacitor

Justify your answer by writing the necessary expressions

Q. A parallel plate capacitor is connected to an ideal battery which provides a fixed potential differences. Originally, the energy stored in the capacitor is