A point charge q1 is at the center of a sphere of radius 20 cm. Another point charge q2 = 5 nC is located at a distance r = 30 cm from the c

Question

A point charge q1 is at the center of a sphere of radius 20 cm. Another point charge q2 = 5 nC is located at a distance r = 30 cm from the center of the sphere. If the net flux through the surface of the sphere is 800 N.m2 /C, find q

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Ngọc Hoa 3 years 2021-08-02T00:42:14+00:00 1 Answers 8 views 0

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  1. lethu
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    2021-08-02T00:43:50+00:00
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    Answer:

    q_1 =7.08*10^{-9}C.

    Explanation:

    Gauss’s Law says that the electric flux \Phi_E through a closed surface is directly proportional to the charge Q_{enc} inside it. More precisely,

    $\Phi_E=\oint_S E\cdot dA = \dfrac{Q_{enc}}{\epsilon_0}. $

    This means what is outside this closed surface S does not contribute to the flux through it because field lines that go in must come out, resulting a zero flux from an external charge.

    In our context, this means the charge q_2 which is outside the sphere will have zero flux through the surface; therefore, Gauss’s law will only be concerned with charge q_1 which is inside the sphere; Hence,

    $\Phi_E=\oint_S E\cdot dA = \dfrac{q_1}{\epsilon_0} = 800 N\cdot m^2/C. $

    Solving for q_1 gives

    $ q_1= (800 N\cdot m^2/C)\epsilon_0, $

    $ q_1= (800 N\cdot m^2/C)*(8.85*10^{-12}C^2/N\cdot m^2) $

    \boxed{q_1 =7.08*10^{-9}C. }

    which is the charge inside the sphere.

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