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A uniform electric field is created by two parallel plates separated by a distance of 0.04 m. What is the magnitude of the electric field established between the plates

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Complete question:

A uniform electric field is created by two parallel plates separated by a

distance of 0.04 m. What is the magnitude of the electric field established

between the plates if the potential of the first plate is +40V and the second

one is -40V?

Answer:

The magnitude of the electric field established between the plates is 2,000 V/m

Explanation:

Given;

distance between two parallel plates, d = 0.04 m

potential between first and second plate, = +40V and -40V respectively

The magnitude of the electric field established between the plates is calculated as;

E = ΔV / d

where;

ΔV is change in potential between two parallel plates;

d is the distance between the plates

ΔV = V₁ -V₂

ΔV = 40 - (-40)

ΔV = 40 + 40

ΔV = 80 V

E = ΔV / d

E = 80 / 0.04

E = 2,000 V/m

Therefore, the magnitude of the electric field established between the plates is 2,000 V/m

The magnitude of the electric field developed that lies between the plates should be considered as the 2,000 V/m.

Calculation of the Electric Field:

Since

The distance that lies between 2 parallel plates should be d = 0.04 m

The potential that lies between first and second plate should be like +40V and -40V

So, The magnitude of the electric field should be

E = ΔV / d

here,

ΔV represents the change in potential that lies between 2 parallel plates.

d represents the distance between the plates.

So,

ΔV = V₁ -V₂

ΔV = 40 - (-40)

ΔV = 40 + 40

ΔV = 80 V

And,

E = ΔV / d

E = 80 / 0.04

E = 2,000 V/m

Therefore, the magnitude of the electric field established between the plates is 2,000 V/m.

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