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
