91Ó°ÊÓ

The plates of the parallel capacitor are joined with the battery. Thus, the battery will establish a potential difference between the plates. The charge will be deposited on the plates.Thereafter, the battery will be disconnected from the parallel plates, and the charge will remain constant on the plates.

For the parallel plate’s capacitor, thecapacitance can be expressed as:

\(C = { \in _0}\frac{A}{d}\)

Here, A is the area of the plates, d is the separation distance, and\({ \in _0}\)is the permittivity of the free space.

The separation distance will be doubled without the loss of charge. Thus, the value of capacitance decreases.

The value of capacitance decreases when the plates are separated by a large distance. If the distance between the plates increases, the capacitance value will decrease.

The capacitance can be expressed as:

\(C = \frac{Q}{V}\)

Here, Q is the charge, and V is the potential difference. The above equation of capacitance gives the constant ratio value.

However, the charge should remain constant with a smaller value of capacitance. Moreover, there must be an increase in the voltage between the two parallel plates.

Thus, the capacitance decreases, and the voltage between the plates increases.