<h2>
Answer:</h2>
(i) 625 V/m
(ii) 2876.25 x 10⁻¹² C
(iii) 0.000575.25 x 10⁻⁶ F
<h2>
Explanation:</h2>
(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)
<em>From the question;</em>
V = 5.0V
d = 0.80cm = 0.008m
<em>Substitute these values into equation (i) as follows;</em>
E = 5.0 / 0.008
<em>Solve for E;</em>
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
<em>Substitute these values into equation (ii) as follows;</em>
C = 0.52 x 8.85 x 10⁻¹² / 0.008
<em>Solve for C;</em>
C = 575.25 x 10⁻¹² F
<em>The capacitance (C) is related to potential difference (V) and charge (Q) on the plates as follows;</em>
Q = C x V -------------------------(iii)
<em>Where;</em>
C = 575.25 x 10⁻¹² F
V = 5.0V
<em>Substitute these values into equation (iii)</em>
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
