Question

A parallel-plate capacitor is constructed using adielectric material whose dielectric constant is 3.00 and whose dielectric strength is 2.00X108V/m. The desired capacitance is 0.250 μF, and the capacitor must withstand a maximum potential difference of 4.00 kV. Find the minimum area of the capacitor plates.

Answer 1

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