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31 December, 09:43

A charge of 0.14 C is moved from a position where the electric potential is 20 V to a position where the electric potential is 50 V. What is the change in potential energy of the charge associated with the change in position?

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  1. 31 December, 10:58
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    The change in electric potential energy is given by:

    ΔU = ΔVq

    ΔU = change in PE, ΔV = potential difference, q = charge

    Given values:

    ΔV = 50V - 20V = 30V, q = 0.14C

    Plug in and solve for ΔU:

    ΔU = 30 (0.14)

    ΔU = 4.2J
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