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(C) According to Gauss's Law,for a non-conducting sphere with uniform charge density $\rho$,the electric flux through a Gaussian surface of radius $r$

Vedclass · Accepted Answer

$\frac{\rho r}{3\varepsilon_0}$

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(C) According to Gauss's Law,for a non-conducting sphere with uniform charge density $\rho$,the electric flux through a Gaussian surface of radius $r$ (where $r Since the sphere is non-conducting and the charge is distributed throughout the volume,the enclosed charge $q_{enclosed} = \rho \cdot V = \rho \cdot (\frac{4}{3}\pi r^3)$.The surface area of the Gaussian sphere is $4\pi r^2$.Thus,$E \cdot (4\pi r^2) = \frac{\rho \cdot \frac{4}{3}\pi r^3}{\varepsilon_0}$.Solving for $E$,we get $E = \frac{\rho r}{3\varepsilon_0}$.

If an insulated non-conducting sphere of radius $R$ has a uniform charge density $\rho$,the electric field at a distance $r$ from the center of the sphere $(r < R)$ will be:

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