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
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.
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
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.
i) 11 V
ii) 1.50 x 104 V