# Numerical problem based on electric field between charged parallel plates

Last updated on March 16th, 2022 at 11:36 am

Here, we will solve a set of Numerical Problems based on the formula of the electric field between two charged parallel plates.

## Numerical Problem (Physics) – based on the formula of the electric field between two charged parallel plates

1 )

A cathode-ray-tube (CRT) computer monitor accelerates electrons between charged parallel plates (Figure 1).
These electrons are then directed toward a screen to create an image.

If the plates are 1.2 x 10–2 m apart and have a potential difference of 2.5 x 104 V between them, determine the magnitude of the electric field between the plates.

Solution:

Data provided:

V =2.5 x 104 V

d =1.2 x 10–2 m

Required to find:
the magnitude of the electric field between the plates (E)

To calculate the magnitude of the electric field between the plates, we will use the equation

|E| = V/d = 2.5 x 104 / 1.2 x 10–2 = 2.1 x 106 V/m

Hence, The magnitude of the electric field between the plates is 2.1 x 106 V/m

## Worksheet

i ) Two charged parallel plates, separated by 5.0 x 10–4 m, have an electric field of 2.2 x 104 V/m between them. What is the potential difference between the plates?

ii ) Spark plugs in a car have electrodes whose faces can be considered to be parallel plates. These plates are separated by a gap of 5.00 x 10–3 m.
If the electric field between the electrodes is 3.00 x 106 V/m, calculate the potential difference between the electrode faces.

Hints: