Question

A capacitor having a capacitance of 100 µF is charged to a potential difference of 50 V. (a) What is the magnitude of the charge on each plate? (b) The charging battery is disconnected and a dielectric of dielectric constant 2⋅5 is inserted. Calculate the new potential difference between the plates. (c) What charge would have produced this potential difference in absence of the dielectric slab. (d) Find the charge induced at a surface of the dielectric slab.

Answer 1

a) Magnitude of the charge on each plate is

5 mC

b) New potential difference between the plates is 20 V

c) Potential difference in absence of the dielectric slab is 2 mC

d) The charge induced at a surface of the dielectric slab is 3 mC

Explanation:

a) Charge of the capacitor Q = CV

Q = 100 X  $10^{-6}$X 50 = 5 mC

b) We know,

Potential difference = initial potential / dielectric constant

         V = 50/2.5 = 20 V

c) Charge in the capacitance q = Capacitance (c) X Potential difference

       20 X 100 X $10^{-6}$ = 2 mC

d) The charge induced on surface q = q( 1 - 1/k)

     Where k = dielectric constant

                    = 5 (1 - 1/2.5) = 3 mC