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 plates. What will happen
If the voltage source is absent during this insertion?

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 the battery is removed, charge stored remains the same. So change in charge of the capacitor =0.
According to the question, the potential difference between the plates of the capacitor before the insertion of the dielectric = 100V.
After the insertion of the dielectric, the potential difference between the plates is,
`V = (Q_0)/(C_2)= (2.21xx10^(-9))/(1.11xx10^(-10))= 19.91V`
So, the potential difference decreases.
Change in potential difference = 100-19.91 = 80.09 V`