Question

Two large, parallel, metal plates carry opposite charges of equal magnitude. They are separated by a distance of 40.0 mm, and the potential difference between them is 370 V
A. What is the magnitude of the electric field (assumed to be uniform) in the region between the plates?
B. What is the magnitude of the force this field exerts on a particle with a charge of 2.40 nC ?
C. Use the results of part (b) to compute the work done by the field on the particle as it moves from the higher-potential plate to the lower.
D. Compare the result of part (c) to the change of potential energy of the same charge, computed from the electric potential.

Answer 1

Answer:

Explanation:

A )

electric field E = V / d where V is potential difference between plates separated by distance d .

putting the given values

E = 370 / .040  V / m

= 9250 V / m

B )

Force on charged particle of charge q in electric field E

F = q E

F = 2.4 x 10⁻⁹ x 9250

= 22200 x 10⁻⁹

= 222 x 10⁻⁷ N .

C ) since field is uniform , force will be constant

work done by electric field putting up this force

= force x displacement

= 222 x 10⁻⁷  x 40 x 10⁻³

= 888 x 10⁻⁹ J

D )

change in potential energy

= q ( V₁ - V₂ )

= 2.40 X 10⁻⁹ x 370

= 888 x 10⁻⁹ J .