(a) What is the potential between two points situated 10 cm and 20 cm from a 3.0-μC point charge? (b) To what location should the point at 2

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(a) What is the potential between two points situated 10 cm and 20 cm from a 3.0-μC point charge? (b) To what location should the point at 20 cm be moved to increase this potential difference by a factor of two?

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Dâu 3 years 2021-08-15T19:51:08+00:00 1 Answers 73 views 0

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  1. thanhcong
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    2021-08-15T19:52:41+00:00
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    Answer:

    (a) 135 kV

    (b) The charge chould be moved to infinity

    Explanation:

    (a)

    The potential at a distance of r from a point charge, Q, is given by

    V = -\dfrac{kQ}{r}

    where k = 9\times 10^9 \text{ F/m}

    Difference in potential between the points is

    kQ\left[-\dfrac{1}{0.2\text{ m}} -\left( -\dfrac{1}{0.1\text{ m}}\right)\right] = \dfrac{kQ}{0.2\text{ m}} = \dfrac{9\times10^9\text{ F/m}\times3\times10^{-6}\text{ C}}{0.2\text{ m}}

    PD = 135\times 10^3\text{ V} = 135\text{ kV}

    (b)

    If this potential difference is increased by a factor of 2, then the new pd = 135 kV × 2 = 270 kV. Let the distance of the new location be x.

    270\times10^3 = kQ\left[-\dfrac{1}{x}-\left(-\dfrac{1}{0.1\text{ m}}\right)\right]

    10 - \dfrac{1}{x} = \dfrac{270000}{9\times10^9\times3\times10^{-6}} = 10

    \dfrac{1}{x} = 0

    x = \infty

    The charge chould be moved to infinity

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