Question

Two large metal parallel plates carry opposite charges of equal magnitude. They are separated by 45.0 mm, and the potentialdifferencebetweenthemis360V.(a)Whatisthemagnitude 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 charge

Answer 1

a. 8000 N/C or 8000 V/m

b. Incomplete question

Step-by-step explanation:

a. The electric field, E, is the ratio of the potential difference,V, to the separation distance, d.

V = 360 V

d = 45 mm = 0.045 m

`E = (V)/(d) = (360)/(0.045) = 8000`

b. The force on a charge in an electric field is the product of the field and the charge.

`F = qE`

With the charge, q, known, the force can be calculated.