Question

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?

Answer 1

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

Step-by-step 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

`C=(\epsilon_(0)A)/(d)`

Put the value into the formula

`C=(8.85*10^(-12)*8*5.52*10^(-4))/(1.99*10^(-2))`

`C=1.96*10^(-12)`

We need to calculate the final capacitance

Using formula of capacitance

`C'=(\epsilon_(0)A_(1))/(d)+(k\epsilon_(0)A_(2))/(d)`

Put the value into the formula

`C'=((8.85*10^(-12))/(1.99*10^(-2)))((4.4*5.52)+(3.6*5.52)2.6)*10^(-4)`

`C'=3.37*10^(-12)`

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

`(E)/(E')=((1)/(2)CV^2)/((1)/(2)C'V^2)`

Put the value into the formula

`(E)/(E')=(1.96*10^(-12))/(3.37*10^(-12))`

`(E)/(E')=0.58`

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