The plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 77.0 nC. The plates are in vacuum. The electric field between the plates has a magnitude of 3.00*106 V/m.

Part A: What is the potential difference between the plates?

Part B:What is the area of each plate?

Part C:What is the capacitance?

Part A: 7500 V

Part B: 2.899*10⁻³ m²

Part C: 10.27 pF or 10.27*10⁻¹² F

Explanation:

Part A:

Applying,

E = V/d ... Equation 1

Where E = electric field intensity between the plates, V = potential difference between the plates, d = distance of separation between the plates

make V the subject of the equation above,

V = Ed ... Equation 2

Given: E = 3.0*10⁶ V/m, d = 2.5 mm = 2.5*10⁻³ m

Substitute into equation 2

V = 3.0*10⁶ (2.5*10⁻³)

V = 7.5*10³ V

V = 7500 V

Part B:

Using,

E = Q / (e₀A) ... Equation 3

Where Q = Charge on each plate of the capacitor, A = Area of each plate, e₀ = constant = dielectric = permitivity of free space

make A the subject of the equation,

A = Q / (e₀E) ... Equation 4

Given: Q = 77 nC = 77*10⁻⁹ C, E = 3.0*10⁶ V/m

Constant: e₀ = 8.854*10⁻¹² F/m

Substitute into equation 4

A = 77*10⁻⁹ / (8.854*10⁻¹² * 3.0*10⁶)

A = 77*10⁻⁹ / (26.562*10⁻⁶)

A = 2.899*10⁻³ m²

A = 2.899*10⁻³ m².

Part C:

Using,

Q = CV ... Equation 5

Where C = Capacitance of the capacitor

make C the subject of the equation

C = Q/V ... Equation 6

Given: Q = 77 nC = 77*10⁻⁹ C, V = 7500 V

Substitute into equation 6

C = 77*10⁻⁹/7500

C = 10.27*10⁻¹² F

C = 10.27 pF