A solid nonconducting sphere of radius R has a charge Q uniformly distributed throughout its volume. A Gaussian surface of radius r with r &

Question

A solid nonconducting sphere of radius R has a charge Q uniformly distributed throughout its volume. A Gaussian surface of radius r with r < R is used to calculate the magnitude of the electric field E at a distance r from the
center of the sphere. Which of the following equations results
from a correct application of Gauss’s law for this situation? And Why?

1. E (4 p r^2) = (Q r^3)/( epsilon sub0 * R3)
2. E(4pr^2)= (Q 3r^3)/ (epsilon sub0 * 4 p R)
3. E (4 p r^2) = (Q) / (epsilon sub0)

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Dulcie 2 months 2021-08-22T11:07:45+00:00 1 Answers 3 views 0

Answers ( )

    0
    2021-08-22T11:09:42+00:00

    Answer:

    E(4 \pi r^{2})={\frac{ Q r^{3}}{R^{3}\epsilon _{0}}}

    or

    E(4 \pi)={\frac{ Q r}{R^{3}\epsilon _{0}}}

    Explanation:

    We know that Gauss’s law states that the Flux enclosed by a Gaussian surface is given by

    E.S=\frac{q}{\epsilon_{0}}

    Here , E is electric field and S is surface are and q is charge enclosed by the surface and e is electrical permeability of the medium.

    Here the Gaussian is of radius r<R so area of surface is

    S=4 \pi r^{2}

    Also, charge enclosed by the surface is

    Charge =\frac{Total \: Charge }{Total \:Volume} \times Volume \: of \: Gaussian \: surface

    therefore,

    q=\frac{Q}{\frac{4}{3} \pi R^{3} }\frac{4}{3} \pi r^{3} =\frac{ Q r^{3}}{R^{3}}

    Here Q is total charge,

    Insert values in Gauss’s law

    E(4 \pi r^{2})=\frac{\frac{ Q r^{3}}{R^{3}}}{\epsilon _{0}}

    Rearrange them

    E(4 \pi r^{2})={\frac{ Q r^{3}}{R^{3}\epsilon _{0}}}

    on further solving

    E(4 \pi)={\frac{ Q r}{R^{3}\epsilon _{0}}}

    This is the required form.

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