A capacitor is constructed with two parallel metal plates each with an area of 0.52 m2 and separated byd = 0.80 cm. The two plates are connected to a 5.0-volt battery. The current continues until a charge of magnitude Q accumulates on each of the oppositely charged plates.

Find the electric field in the region between the two plates.
V/m

Find the charge Q.
C

Find the capacitance of the parallel plates.
? 10?6 F

(i) The electric field (E) between the plates of a parallel plate capacitor is related to the potential difference (V) between the plates and the distance (d) of separation between the plates as follows;

E = V / d ----------------(i)

From the question;

V = 5.0V

d = 0.80cm = 0.008m

Substitute these values into equation (i) as follows;

E = 5.0 / 0.008

Solve for E;

E = 625 V/m

Therefore, the electric field in the region between the two plates is 625 V/m.

(ii) To make things easier, let's calculate the capacitance of the parallel plates first.

The capacitance (C) of a parallel plate capacitor is given as;

C = A x ε₀ / d --------------------------(ii)

Where;

A = Area of either of the plates of the capacitor = 0.52m²

ε₀ = permittivity of free space = 8.85 x 10⁻¹² F/m

d = distance between the plates = 0.8cm = 0.008m

Substitute these values into equation (ii) as follows;

C = 0.52 x 8.85 x 10⁻¹² / 0.008

Solve for C;

C = 575.25 x 10⁻¹² F

The capacitance (C) is related to potential difference (V) and charge (Q) on the plates as follows;

Q = C x V -------------------------(iii)

Where;

C = 575.25 x 10⁻¹² F

V = 5.0V

Substitute these values into equation (iii)

Q = 575.25 x 10⁻¹² x 5

Q = 2876.25 x 10⁻¹² C

Therefore, the charge on the plates is 2876.25 x 10⁻¹² C

(iii) The capacitance (C) of the parallel plates has been calculated in (ii) above.

Its value is 575.25 x 10⁻¹² F = 0.000575.25 x 10⁻⁶F