Consider a air-filled parallel-plate capacitor with plates of length 8 cm, width 5.52 cm, spaced a distance 1.99 cm apart. Now imagine that a dielectric slab with dielectric constant 2.6 is inserted a length 4.4 cm into the capacitor. The slab has the same width as the plates. The capacitor is completely filled with dielectric material down to a length of 4.4 cm. A battery is connected to the plates so that they are at a constant potential 0.8 V while the dielectric is inserted. Required:
What is the ratio of the new potential energy to the potential energy before the insertion of the dielectric?

The ratio of the new potential energy to the potential energy before the insertion of the dielectric is 0.58

Explanation:

Given that,

Length of plates = 8 cm

Width = 5.52 cm

Distance = 1.99 cm

Dielectric constant = 2.6

Length = 4.4 cm

Potential = 0.8 V

We need to calculate the initial capacitance

Using formula of capacitance

Put the value into the formula

We need to calculate the final capacitance

Using formula of capacitance

Put the value into the formula

We need to calculate the  ratio of the new potential energy to the potential energy before the insertion of the dielectric

Using formula of energy

Put the value into the formula

Hence, The ratio of the new potential energy to the potential energy before the insertion of the dielectric is 0.58

the ima (higher than 1m) will be greater than the ima with the height of 1m.