Respuesta :
Answer:
(a) 9.444 kV/m
(b) 3.73 pF
(c) 63.41 x 10⁻¹² C or 63.41 pC
Explanation:
(a) The electric field (E) between the plates of a capacitor is related to the distance (d) and the potential difference between the plates as follows;
V = E x d; -----------------(i)
From the question, the following has been given;
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 / 0.0018
E = 9444V/m
Divide the result by 1000 to convert it to kV/m
=> E = 9.444 kV/m
Therefore, the electric field between the plates is 9.444 kV/m.
(b) The capacitance (C) of the capacitor with no dielectric material (i.e with air as separating medium) is directly proportional to the area (A) of either of the plates and inversely proportional to the distance (d) between the plates as follows;
C = A ε₀ / d --------------------(ii)
where;
ε₀ = the permittivity of free space = 8.85 x 10⁻¹² F/m
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
Solve for C;
C = 0.00076 x 8.85 x 10⁻¹² / 0.0018
C = 0.422 x 8.85 x 10⁻¹²
C = 3.73 x 10⁻¹² F
C = 3.73 pF
Therefore, the capacitance of the capacitor is 3.73pF
(c) The magnitude of the charge (Q) on each plate of a capacitor is the product of the capacitance (C) of the capacitor and the potential difference (V) between the plates. i.e
Q = C x V --------------------(iii)
Where;
C = 3.73pF = 3.73 x 10⁻¹² F (as calculated above)
V = 17.0V
Substitute these values into equation (iii) as follows;
Q = 3.73 x 10⁻¹² x 17.0
Q = 63.41 x 10⁻¹² C or 63.41 pC
Therefore, the charge on each plate of the capacitor is 63.41 x 10⁻¹² C or 63.41 pC
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.
