A parallel plate capacitor of capacitance `100muF` is connected a power supply of `200 V`. A dielectric slab of dielectric constant 5 is now inserted into the gap between the plates. (a) Find the extra charge flown through the power supply and the work done by the supply. (b) Find the change in the electrostatic energy of the electric field in the capacitor.

Explanation

The original capacitance was `
`100 muF`
`. The charge on the capacitor before the insertion of the dielectric was, therefore, `
`Q_1 = 100 muF X 200 V = 20mC`
`. After the dielectric slab is introduced, the capacitance is increased to `
`500 muF`
`. The new charge on the capacitor is, therefore, `
`500muF X 200 V = 100mC`
`. The charge flown through the power supply is, therefore, `
`100mC - 20mC = 80mc`
`. the work done by the power supply is `
`200 V X 80 mC = 16J. (b) The electrostatic field energy of the capacitor without the dielectric slab is `
`U_1 = (1)/(2) CV^2`
` = (1)/(2) X(100 muF) X (200V)^2 = 2J`
` and that after the slab is inserted is `
`U_2= (1)/(2) X (500muF) x (200 V)^2 = 10 J`
`. Thus, the energy is increased by 8 J.