Answer:
A = 0.188 m²
Explanation:
First we find the distance between the plates by using the formula of electric field intensity:
E = ΔV/d
d = ΔV/E
where,
d = distance between plates = ?
ΔV = Potential Difference = 4 KV = 4000 V
E = Electric Field = 2 x 10⁸ V/m
Therefore,
d = 4000 V/(2 x 10⁸ V/m)
d = 2 x 10⁻⁵ m
Now, we find the Area of Plates by using formula of capacitance:
C = A∈₀∈r/d
where,
C = Capacitance = 0.25 μF = 0.25 x 10⁻⁶ F
A = Area of Plates = ?
∈₀ = Permittivity of free space = 8.85 x 10⁻¹² C/N.m²
∈r = Dielectric Constant = 3
Therefore,
0.25 x 10⁻⁶ F = A(8.85 x 10⁻¹² C/N.m²)(3)/(2 x 10⁻⁵ m)
A = (0.25 x 10⁻⁶ F)(2 x 10⁻⁵ m)/(3)(8.85 x 10⁻¹² C/N.m²)
A = 0.188 m²