Two parallel conducting plates are separated by 12.0 cm, and one of them is taken to be at zero volts. (a) What is the magnitude of the electric field strength between them, if the potential 5.6 cm from the zero volt plate is 450 V

-8.036 kV/m

Explanation:

The electric field E = -ΔV/Δx where ΔV = change in electric potential = V - V' where V = electric potential at x = 5.6 cm =  450 V and V' = electric potential at x = 0 cm, = 0 V . So, ΔV = V - V' = 450 V - 0 V = 450 V.

Δx = distance between the 0 V plate and the 450 V point = 5.6 cm = 0.056 m

So, E = -ΔV/Δx

Substituting the values of the variables into the equation, we have

E = -ΔV/Δx

E = -450 V/0.056 m

E = -8035.7 V/m

E = -8.0357 kV/m

E ≅ -8.036 kV/m

Since the electric field between two parallel conducting plates is constant, the electric field between the plates is E = -8.036 kV/m

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