A parallel-plate air capacitor of area A = 28.1 cm2 and plate separation of d = 3.80 mm is charged by a battery to a voltage of 69.0 V. What is the charge on the capacitor?

4.52*10⁻¹⁰ C

Explanation:

The charge stored in a capacitor is given as,

Q = CV ... Equation 1

Where Q = Charge stored in a capacitor, C = Capacitance of the capacitor, V = Voltage.

But,

C = e₀A/d ... Equation 2

Where e₀ = permitivity of free space, A = Area of the plates, d = distance of separation of the plates

Substitute equation 2 into equation 1

Q = e₀AV/d ... Equation 3

Given: A = 28.1 cm² = 0.00281 m², V = 69.0 V, d = 3.8 mm = 0.0038 m

Constant: e₀ = 8.85*10⁻¹² F/m.

Substitute into equation 3

Q = 8.85*10⁻¹²*0.00281*69/0.0038

Q = 4.52*10⁻¹⁰ C.

Hence the charge on the capacitor = 4.52*10⁻¹⁰ C