An electron is moved from point A to point B in a uniform electric field and gains 4.66×10-15 J of electrostatic potential energy. Calculate

Question

An electron is moved from point A to point B in a uniform electric field and gains 4.66×10-15 J of electrostatic potential energy. Calculate the magnitude of the electrostatic potential difference between the two points. (in V)

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Latifah 5 years 2021-08-17T07:50:12+00:00 1 Answers 44 views 0

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  1. thaiduong
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    2021-08-17T07:51:44+00:00
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    Answer:

    2.9125 x 10⁴ V

    Explanation:

    The electrostatic potential difference (ΔV) between two points, say A and B, is denoted by V_{B} – V_{A} and is the quotient of the change in the electrostatic potential energy (ΔP_{E}) of a charge Q moved from A to B, and the charge. i.e

    ΔV = V_{B} – V_{A} = ΔP_{E} / Q               ————(i)

    From the question;

    ΔP_{E} = 4.66 x 10⁻¹⁵J

    Q = charge of an electron = -1.6 x 10⁻¹⁹C

    Substitute these values into equation (i) as follows;

    V_{B} – V_{A} = (4.66 x 10⁻¹⁵) / (-1.6 x 10⁻¹⁹)

    V_{B} – V_{A} = -2.9125 x 10⁴ V

    |V_{B} – V_{A}| = 2.9125 x 10⁴ V

    Therefore, the magnitude of the electrostatic potential difference between these two points is 2.9125 x 10⁴ V

       

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