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
(b) the capacitance
(c) the charge on each plate

(a) 9444.4 V/m

(b) 3.72×10⁻¹² F

(c)  6.324×10⁻¹¹ C

Explanation:

(a)

The formula of electric field is given as

E = V/r Equation 1

Where E = Electric Field, V = Electric potential, r = distance between the plates of the capacitor.

Given: V = 17.0 V, r = 1.8 mm = 0.0018 m.

Substitute into equation 1

E = 17.0/0.0018

E = 9444.4 V/m.

(b)

The formula of the capacitance of a capacitor is given as,

C = ε₀A/d Equation 2

Where C = Capacitance of the capacitor, ε₀ = permitivity of free space, d = distance between the plates, A = Area of each plate.

Given: A = 7.6 cm² = 0.00076 m³, d = 1.8 mm = 0.0018 m.

Constant: 8.854×10⁻¹² F/m.

Substitute into equation 2

C = 8.854×10⁻¹²(0.00076)/0.0018

C = 3.72×10⁻¹² F.

(c)

The charge stored in a capacitor is given as,

q = CV Equation 3

Where q = charge, V = Voltage applied across the plate of the capacitor.

Given: V = 17.0 V, C = 3.72×10⁻¹² F

Substitute into equation 3

q = 17(3.72×10⁻¹²)

q = 63.24×10⁻¹²

q = 6.324×10⁻¹¹ C

(a) 9444.4 V/m

(b) 3.74 x 10⁻¹² F

(c) 63.58 x 10⁻¹² C

(a)The potential difference(V) applied across the plates of a parallel plate capacitor is given by;

V = E x d           (i)

Where;

E = electric field between the plates

d = distance of separation between the plates.

From the question,

V = 17.0V

d = 1.80mm = 0.0018m

Substitute these values into equation (i) as follows;

17.0 = E x 0.0018

Solve for E;

E = 17 / 0.0018

E = 9444.4V/m

Therefore, the electric field between the plates is 9444.4 V/m

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

C = A ε₀ / d          (ii)

Where;

A = area of each of the plates of the parallel plate

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

d = distance of separation of the plates

From the question;

A = 7.60cm² = 0.00076m²

d = 1.80mm = 0.0018m

Substitute these values into equation (ii) as follows;

C = 0.00076 x 8.85 x 10⁻¹² / 0.0018

C = 3.74 x 10⁻¹² F

Therefore, the capacitance of the capacitor is 3.74 x 10⁻¹² F

(c) The charge (Q) on each of the plates of a parallel plate, the capacitance (C) of the plate and the potential difference (V) across the plates are related as follows;

Q = C x V        (iii)

From the question;

V = 17.0V

C = 3.74 x 10⁻¹² F     [as calculated in (b) above]

Substitute these values into equation (iii) as follows;

Q = 3.74 x 10⁻¹² x 17.0

Q = 63.48 x 10⁻¹² C

Therefore, the charge on each plate is 63.58 x 10⁻¹² C

hi there!

out of the two, (b) precision would be more reasonable.   however, both a and b could be pretty reasonable.

-awesomerepublic   : )