A capacitor is to be constructed to have a capacitance of 100uF. The area of the plates is 6.om by 0.030m and the relative permitivityof dielectric is 7.0 Find the necessary separation of the plates and the electric field strength if a potential difference of 12V is applied across the capacitor.

The answer is below

Explanation:

Given that:

The area of the plates is 6 m by 0.030 m, Therefore the area = 6 m × 0.03 m =  0.18 m²

the relative permittivity of dielectric (εr) is 7.0

Permittivity of free space (εo) = 8.854 × 10^(-12)

capacitance of 100uF

potential difference (V) of 12V

d = separation between plate

The capacitance (C) of a capacitor is given by:

The electric field between plates is given as:

E = V /d

the answer is ampere

the ampere is used to measure electricity. an ampere is a unit of electric current equal. a unit of measure for the flow of current in a circuit. one ampere is the amount of current flow provided when one volt of electrical pressure is applied against one ohm of resistance.