Question

A homemade capacitor is assembled by placing two 10-in. pie pans 5 cm apart and connecting them to the opposite terminals of a 9-V battery.

A) Estimate the capacitance.

B) Estimate the charge on each plate.

C) Estimate the electric field halfway between the plates.

D) Estimate the work done by the battery to charge the plates.

E) Which of the above values change if a dielectric is inserted?

Options:

Capacitance.

Charge.

Electric field.

Work done by the battery.

Answer 1

A) Diameter of plates of capacitor,

D = 10 inch

= 25.4 cm

Distancce between the plates,

d = 5 cm

= 0.05 m

Area of the plates,

A = 0.05067 m^{​​​​​​2}

Capacitance of the capacitor,

C = A₀ / d

= 8.973 × 10^-12 F

The capacitance would be 8.973 pF.

B) Charge stored in the capacitor due to the applied voltage,

q = C × V

Here,

V = 9 V

q = 8.07567 × 10^-11 C

C) In a capacitor a nearly uniform electric field is produced between the plates.

At the centre of the plates,

E = V / d

= 180 V / m

D) Energy stored in a capacitor,

U = (1 / 2)CV^{​​​​​2}

The energy stored in the capacitor would be equal to the work done by the battery to charge the capacitor.

Work done to charge the capacitor,

W = (1 / 2) × C × V^{​​​​​​2}

= 3.634 × 10^-10 J

E) When a dielectric is placed between the plates of the capacitor, the capacitance changes. Since the applied voltage is constant, the charge stored in the capacitor changes. Also, the work done by the battery to charge the capacitor also change

But the electric field between the plates,

E = V / d

does not change.

The capacitance, charge stored in the capacitor and the work done by the battery change when a dielectric is introduced between the plates of the capacitor.