Question

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?

Answer 1

Answer:

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