35. A spherical conductor has a radius of 14.0 cm and a charge of 26.0 mC. Calculate the electric field and the electric poten- tial at (a)

Question

35. A spherical conductor has a radius of 14.0 cm and a charge of 26.0 mC. Calculate the electric field and the electric poten- tial at (a) r 5 10.0 cm, (b) r 5 20.0 cm, and (c) r 5 14.0 cm from the center.

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Neala 4 years 2021-08-19T03:48:19+00:00 1 Answers 270 views 0

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  1. thuthao
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    2021-08-19T03:49:49+00:00
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    Answer:

    Therefore,

    V_{10}=2.34\times 10^{6}\ V

    V_{20}=1.17\times 10^{6}\ V

    V_{14}=1.67\times 10^{6}\ V

    Explanation:

    Given:

    A spherical conductor has a radius of 14.0 cm

    Q = 26.0 μC ( Assume it to be microC in the question it is miliC )

    To Find:

    Electric potential at

    (a) r 5 10.0 cm, (b) r 5 20.0 cm, and (c) r 5 14.0 cm from the center.

    Solution:

    Electric potential due to point charge Q at any distance r from the charge is,

    V=\dfrac{kQ}{r}

    Where,

    V = Electric Potential

    k = Coulombs constant = 9 × 10⁹ Nm²/C².

    Q = Charge

    r = distance in meter

    Substituting the values we get

    At r = 10 cm = 0.1 m

    V=\dfrac{9\times 10^{9}\times 26\times 10^{-6}}{0.1}

    V=2.34\times 10^{6}\ V

    At r = 20 cm = 0.2 m

    V=\dfrac{9\times 10^{9}\times 26\times 10^{-6}}{0.2}

    V=1.17\times 10^{6}\ V

    At r = 14 cm = 0.14 m

    V=\dfrac{9\times 10^{9}\times 26\times 10^{-6}}{0.14}

    V=1.67\times 10^{6}\ V

    Therefore,

    V_{10}=2.34\times 10^{6}\ V

    V_{20}=1.17\times 10^{6}\ V

    V_{14}=1.67\times 10^{6}\ V

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