Answer:
D) The potential difference between the plates increases.
Step-by-step explanation:
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 with increased distance, will have a new capacitance C1=ϵ0kA/d1
Where d1 = nd
since d1 > d
therefore n >1
n is a factor derived as a result of the increased distance
Therefore the new capacitance becomes:
C1=ϵ0A/d1
C1= ϵ0A/nd
C1= C/n -------1
Where C1 is the capacitance with increased distance.
This implies that the charge storing capacity of the capacitor with increased plate separation decreases by a factor of (1/n) compared to that of the capacitor with original distance.
Given points
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.
Charge before plate separation increase same as after plate separation increase
Q = Q1
CV = C1V1
CV = C1V1 -------2
We derived C1=C/n in equation 1. Inserting this into equation 2
CV = (CV1)/n
V1 = n(CV)/C
= n V
Since n > 1 as a result of the derived new distance, the new voltage will increase