## A uniform electric field, with a magnitude of 449 N/C, is directed parallel to the positive x-axis. If the potential at x= 4.59 m is 966 V,

Question

A uniform electric field, with a magnitude of 449 N/C, is directed parallel to the positive x-axis. If the potential at x= 4.59 m is 966 V, what is the potential at x= 1.61 m?

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3 years 2021-08-17T16:28:28+00:00 1 Answers 7 views 0

2.334‬ kV

Explanation:

Since the electric field strength Ecosθ = -ΔV/Δx = -(V₂ – V₁)/(x₂ – x₁)

where V₁ = potential at x₁ = 966 V, x₁ = 4.59 m, V₂ = potential at x₂ = unknown, x₂ = 1.61 m and θ = angle between E and the x – axis.

Given that E = + 449 N/C and since it is directed parallel to the positive x – axis, θ = 0°

So,

Ecosθ = -ΔV/Δx

Ecos0° = -ΔV/Δx

E = -ΔV/Δx

E = -(V₂ – V₁)/(x₂ – x₁)

making V₂ subject of the formula, we have

-E(x₂ – x₁) = V₂ – V₁

-E(x₂ – x₁) + V₁ = V₂

V₂ = V₁ – E(x₂ – x₁)

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

V₂ = V₁ – E(x₂ – x₁)

V₂ = 966 V – 449 N/C(1.61 m – 4.59 m)

V₂ = 966 V – 449 N/C(-2.98 m)

V₂ = 966 V + 1,338.02‬ Nm/C

V₂ = 966 V + 1,338.02‬ V

V₂ = ‭2,334.02‬ V

V₂ ≅ ‭2,334‬ V

V₂ = ‭2.334‬ kV