Surface area of each plate of a parallel plate capacitor is 50 `cm^2`. They are separated by 2mm in air. It is connected with a 100V power supply. Now a dielectric ( k =5) is inserted between its two paltes. What will happen
If the voltage source remains connected?

Explanation

We know, capacitance of parallel plate capacitor is `C = (k in_0alpha)/(d)`
For air, the capacitance,
`C_1= (in_0alpha)/(d) [in_0 = 8.854 xx 10^(-12)F*m^(-1), alpha = 50 cm^2 = 5xx10^(-3) m^2, d = 2mm = 2xx 10^(-3)m]`
`= (8.854xx10^(-12)xx5xx10^(-3))/(2xx10^(-3))`
` =2.21xx10^(-11)F`
For the dielectric (k = 5), the capacitance,
`C_2= (k in_0alpha)/(d) = 5xx2.21 xx10^(-11) = 1.11 xx10^(-10)F`
If a dielectric is inserted, capacitance of a parallel plate capacitance increases. Since the capacitor is still connceted to the power supply, its potential will remain constant.
When the intervening medium is air, charge on the capacitor,
`Q_0 = " capacitance " xx` potential
`= 2.21xx10^(-10)xx100 = 2.21 xx10^(-9)C`
When the intervening medium is the dielectric (k =5), charge on the capacitor,
`Q = 1.11 xx 10^(-10) xx 100= 1.11 xx 10^(-8)C`
Change in charge of the capacitor `= Q-Q_0`
`= 1.11 xx 10^(-8) - 2.21 xx 10^(-9)`
`= 8.854 xx 10^(-9) C`
Change in potential difference =0.