An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by a distance of 1.80 mm. If a 17.0 V potential difference is applied to these plates, calculate the following: ]
(a) the electric field between the plates
kV/m
(b) the capacitance
pF
(c) the charge on each plate

Explanation:

A.

V = 17.0V

d = 1.80mm

= 0.0018m

Electric field, E between the plates = V/d

= 17/0.0018

= 9444. 44 V/m.

B.

Capacitance of the capacitor, C = A ε₀/d

Where,

A = area = 7.60cm²

= 0.00076m²

d = distance between the plates = 1.80mm

= 0.0018m

ε₀ = the permittivity of free space = 8.85 x 10⁻¹²

C = 0.00076 x 8.85 x 10⁻¹² / 0.0018

= 3.74 x 10⁻¹² F.

C.

The charge on each plate, Q= C x V

Where;

V = potential difference = 17 V

C = 3.73 x 10⁻¹² F

= 3.73 x 10⁻¹² x 17

Q = 63.52 x 10⁻¹² C.

d they tend to lose electrons and become cations.

explanation: