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28 August, 14:23

As a particle moves 10.0 m along an electric field of strength 75 N/C, its electrical potential energy decreases by 4.8 * 10-16 J. What is the particle's charge?

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  1. 28 August, 15:22
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    6.4*10⁻¹⁹ C

    Explanation:

    From the question,

    V = E*d ... Equation 1

    Where V = electric potential, E = Electric Field, d = distance moved by the particle.

    Given: E = 75 N/C, d = 10 m.

    Substitute into equation 1

    V = 75*10 = 750 V.

    Also Using,

    W = qV ... Equation 2

    W = Work done by the particle when it moves, q = Particles charge

    make q the subject of the equation,

    q = W/V ... Equation 3

    Given: W = 4.8*10⁻¹⁶ J, V = 750 V

    Substitute into equation 3

    q = 4.8*10⁻¹⁶/750

    q = 6.4*10⁻¹⁹ C
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