Question

A parallel-plate capacitor made of circular plates of radius 25 cm separated by 0.20 cm is charged to a potential difference of 600 Volts by a battery. Then a sheet of polyethylene is pushed between the plates, completely filling the gap between them. How much additional charge flows from the battery to one of the plates when the polyethylene is inserted?

Answer 1

Charge on capacitor before inserting polyethylene sheet will be:

Q0 = C0*V0

Q0 = (e0*A/d)*V0

A = Area of plates of capacitor = pi*r^2 = pi*0.25^2 = 0.196 m^2

d = separation distance between plates = 0.20 cm = 0.002 m

V0 = 600 V

e0 = permittivity of free space = 8.854*10^-12

So,

Q0 = 8.854*10^-12*0.196*600/0.002

Q0 = 5.21*10^-7 C

Now when polyethylene sheet is inserted between the plates, then new charge on plates will be:

Q1 = C1*V1

V1 = V0, since battery is still connected

C1 = New capacitance = k*C0

k = dielectric constant of polyethylene = 2.6

Q1 = k*C0*V0

Q1 = k*Q0 = 2.6*5.21*10^-7 C

Now additional charge flow will be:

dQ = Q1 - Q0

dQ = 2.6*5.21*10^-7 - 5.21*10^-7

dQ = 1.6*5.21*10^-7

dQ = 8.34*10^-7 C = additional charge flow

Let me know if you've any query.