Question

The electric field in the region between the plates of a parallel plate capacitor has a magnitude of 6.3 105 V/m. If the plate separation is 0.56 mm, determine the potential difference between the plates.

Answer 1

Answer:

3.528×10² V.

Explanation:

potential difference: This is the work done when one coulomb of charge moves from one point to another in an electric field. The S.I unit of potential difference is volt. The formula of potential difference is given as,

V = kq/r..................... Equation 1

And

E = kq/r² .................. Equation 2

Comparing equation 1 and equation 2,

V = E×r............................. Equation 3

Where V = potential difference, E = Electric field between the plate of the capacitor, r = distance between the plate.

Given: E = 6.3×10⁵ V/m, r = 0.56 mm = 0.00056 m.

Substitute into equation 3,

V = 6.3×10⁵×0.00056

V = 3.528×10² V.

Hence the potential difference of the plate = 3.528×10² V.