CapacitanceMedium
Question
A parallel plate capacitor has a dielectric slab in it. The slab just fills the space inside the capacitor. The capacitor is charged by a battery and the battery is disconnected. Now the slab is started to pull out uniformly at t = 0. If at time t, capacitance of the capacitor is C, potential difference across plate is V, and energy stored in it is U, then which of the following graphs are correct ?
Options
A.

B.

C.

D.

Solution
Capacitance of capacitor is = C0 = 

C =
C =
[x + l(L − x)] =
[kL − (k − 1)x] =
[kL − (k − 1) vt]
So, C decreases linearly with time
Charge on capacitor Q = C0V0 =
V0 = constant.
Potential difference across plate is V =
⇒ V ∞ 
V =
Potential energy U =
QV =
C0V0.V

C =
C =
So, C decreases linearly with time
Charge on capacitor Q = C0V0 =
Potential difference across plate is V =
V =
Potential energy U =
Create a free account to view solution
View Solution FreeMore Capacitance Questions
The resultant capacity between the points A and B in the adjoining circuit will be -...The potential difference between the points P and Q in the adjoining circuit will be-...If the plates of a parallel plate capacitor connected to a battery are moved close to each other, thenA. the charge stor...In the given fig., if the heat flowing through the 5ω resistance is 10 calorie then how much heat is flowing throug...A condenser of capacitor 50 μF is charged to 10 volt, the energy stored is :...