If the radius of the equipotential surface of the point charge is 14.3 m at a potential of 2.20 kV, what will be the magnitude of the point

Question

If the radius of the equipotential surface of the point charge is 14.3 m at a potential of 2.20 kV, what will be the magnitude of the point charge that generates the potential?

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Thanh Hà 5 years 2021-07-26T06:05:21+00:00 1 Answers 17 views 0

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  1. ngockhue
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    2021-07-26T06:06:35+00:00
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    Answer:

    The magnitude of the point charge is 3.496 x 10⁻⁶ C

    Explanation:

    Given;

    radius of the surface, r = 14.3 m

    magnitude of the potential, V = 2.2 kV = 2,200 V

    The magnitude of the point charge is calculated as follows;

    V = (\frac{1}{4\pi \epsilon _0} )(\frac{Q}{r} )\\\\V = \frac{KQ}{r} \\\\Q = \frac{Vr}{K} \\\\Q = \frac{2,200 \times 14.3}{9\times 10^9} \\\\Q = 3.496 \times 10^{-6} \ C\\\\Q = 3.496 \ \mu C

    Therefore, the magnitude of the point charge is 3.496 μC

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