Question

A parallel plate capacitor is disconnected from the battery and a dielectric slab of dielectric constant (K > 1) is now inserted in it. Which of the following is a correct statement?
1. The capacitance of the capacitor decreases.
2. The potential difference between the plates increases.
3. The charge on the capacitor remains constant.
4. Electric field between the plates remains constant.

Answer 1

Correct Answer - Option 3 : The charge on the capacitor remains constant.

CONCEPT:

For the parallel plate capacitor, the capacitance is given by:

\(C= \frac{KA\varepsilon _{0}}{d}\)

where A is the area of the plate, d is the distance between plates, K is the dielectric constant of material and ϵ is constant.

• K = 1 for air or vacuum.
• When the configuration is not changed and the capacitor is not connected to any circuit, the charge always remains constant (by the law of conservation of charge).
  • Initial Charge = Final Charge

EXPLANATION:

• Option 1:

Let the initial capacitance is \(C= \frac{A\varepsilon _{0}}{d}\) and after inserting dielectric is \(C'= \frac{KA\varepsilon _{0}}{d}\)

C' = KC

• So the capacitance of the capacitor increases.
  • So option 1 is a wrong statement.
• Option 2:
• By the conservation of charge: Initial Charge = Final Charge

Q = Q'

CV = C'V'

\(V' = {V \over K}\).................(i)

• So the potential difference between the plates decreases K times.
  • ​So option 2 is a wrong statement.
• Option 3:
• When the configuration is not changed and the capacitor is not connected to any circuit, the charge always remains constant (by the law of conservation of charge).
• Initial Charge = Final Charge
  • ​So option 3 is a correct statement.​
• Option 4:

From eq (i)

\(V' = {V \over K}\)

divide by plate distance 'd'

\(V'/d = {V/d \over K}\)

\(E' = {E \over K}\)

• So the electric field between the plates decreases K times.
  • ​So option 4 is a wrong statement.
• Hence the correct statement is in option 3.