Question

An uncharged capacitor is connected to a 21-V battery until it is fully charged, after which it is disconnected from the battery. A slab of paraffin (K= 2.2) is then inserted between the plates.
What will now be the voltage between the plates?

Answer 1

Answer:

The voltage bewtween the plates will be 9.5V

Explanation:

Facts:

The capacitance of a parallel plate capacitor having plate area A and plate separation d is C=ϵ0A/d.  

Where ϵ0 is the permittivity of free space.  

A capacitor filled with dielectric slab of dielectric constant K, will have a new capacitance C1=ϵ0kA/d

        C1=K(ϵ0A/d)

        C1=KC   ----------- 1

Where C is the capacitance with no dielectric and C1 is the capacitance with dielectric.

The new capacitance is k times the capacitance of the capacitor without dielectric slab.  

This implies that the charge storing capacity of a capacitor increases k times that of the capacitor without dielectric slab.

Given points

• The terminal voltage of the battery to which the capacitor is connected to charge V=25V

• A dielectric slab of paraffin of dielectric constant K=2  is inserted in the space between the capacitor plates after the fully charged capacitor is disconnected

The charge stored in the original capacitor Q=CV

The charge stored in the original capacitor after inserting dielectric Q1=C1V1

The law of conservation of energy states that the energy stored is constant:

i.e Charge stored in the original capacitor is same as charge stored after the dielectric is inserted.

Q   =  Q1

CV = C1V1

  CV = C1V1  -------2

We derived C1=KC from equation 1. Inserting this into equation 2

   CV = KCV1

 V1 = (CV)/KC

         V/K

       = 21/2.2

      = 9.5