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

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  1. Eirian
    0
    2021-08-17T16:30:12+00:00
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    Answer:

    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

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