A spherical drop of water carrying a charge of 30 pC has a potential of 500 V at its surface (with V 0 at infinity). (a) What is the radius

Question

A spherical drop of water carrying a charge of 30 pC has a potential of 500 V at its surface (with V 0 at infinity). (a) What is the radius of the drop? (b) If two such drops of the same charge and radius combine to form a single spherical drop, what is the potential at the surface of the new drop?

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Bơ 4 years 2021-08-31T13:12:54+00:00 1 Answers 31 views 0

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  1. ngockhue
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    2021-08-31T13:14:16+00:00
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    Answer:

    A. 5.4 * 10^(-4) m

    B. 500V

    Explanation:

    A. Electric potential, V is given as:

    V = kq/r

    This means that radius, r is

    r = kq/V

    r = (9 * 10^9 * 30 * 10^(-12))/500

    r = (270 * 10^(-3))/500

    r = 5.4 * 10^(-4) m

    B. Now the radius is doubled and the charge is doubled,

    V = (9 * 10^9 * 2 * 30 * 10^(-12))/(2 * 5.4 * 10^(-4) * 2)

    V = 500V

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