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7 October, 15:12

Charge q1 is distance s from the negative plate of a parallel-plate capacitor. Charge q2=q1/3 is distance 2s from the negative plate. What is the ratio U1/U2 of their potential energies?

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  1. 7 October, 18:39
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    The ratio (U₁/U₂) = 6

    Explanation:

    U, the potential energy is given as

    U = kqQ/r

    k = Coulomb's constant

    q = charge we're concerned about

    Q = charge of the negative plate of the capacitor

    r = distance of q from the negative plate of the capacitor.

    For charge q₁

    U₁ = kq₁Q/s

    U₂ = kq₂Q/2s

    But q₂ = q₁/3

    U₂ becomes U₂ = kq₁Q/6s

    U₁ = kq₁Q/s

    U₂ = kq₁Q/6s

    (U₁/U₂) = 6
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