ASK YOUR TEACHER The equipotential surfaces surrounding a point charge are concentric spheres with the charge at the center. If the electric

Question

ASK YOUR TEACHER The equipotential surfaces surrounding a point charge are concentric spheres with the charge at the center. If the electric potential of two such equipotential surfaces that surround a point charge +1.43 ✕ 10−8 C are 215 V and 71.5 V, what is the distance between these two surfaces?

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Delwyn 4 days 2021-07-22T09:28:22+00:00 1 Answers 3 views 0

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    2021-07-22T09:30:19+00:00

    Answer:

    1.203 m

    Explanation:

    The expression for electric potential is given as,

    V = kq/r

    Where V =  Electric potential, k = coulomb’s constant, q = charge, r = distance.

    From the question,

    For the first surface,

    V₁ = kq/r₁……………….. Equation 1

    Where V₁ = Electric potential of the first surface, r₁ = distance of the first surface from the charge.

    make r₁ the subject of the equation,

    r₁ = kq/V₁…………….. Equation 2

    Given: q = 1.43×10⁻⁸ C, k = 9×10⁹ Nm²/C², V₁ = 215 V

    Substitute into equation 2

    r₁ =  1.43×10⁻⁸(9×10⁹)/215

    r₁ = 128.7/215

    r₁ = 0.597 m

    For the second surface,

    Similarly,

    r₂ = kq/V₂………………. Equation 4

    Given: V₂ = 71.5 V

    Substitute into equation 4

    r₂ = 1.43×10⁻⁸(9×10⁹)/71.5

    r₂ = 128.7/71.5

    r₂ = 1.8 m.

    Hence the distance between the surface = r₂-r₁

    = 1.8-0.597 = 1.203 m

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