Answer :
Answer:
(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
Answer:
(a) 9444.4 V/m
(b) 3.74 x 10⁻¹² F
(c) 63.58 x 10⁻¹² C
Explanation:
(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