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Question

(D) The total charge $q$ on the hollow spherical shell is given by the product of surface charge density $\sigma$ and the surface area of the sphere $

Vedclass · Accepted Answer

$\frac{2 \pi r^2 \sigma}{3 \varepsilon_0}$

Vedclass · Answer

(D) The total charge $q$ on the hollow spherical shell is given by the product of surface charge density $\sigma$ and the surface area of the sphere $(4 \pi r^2)$: $q = \sigma 	imes 4 \pi r^2$. According to Gauss's law,the total electric flux $\phi_{total}$ through the closed surface of the cube is $\frac{q}{\varepsilon_0}$. Since the cube is a symmetric closed surface and the charge is at its center,the flux is distributed equally through all $6$ faces of the cube. Therefore,the electric flux $\phi'$ through one face of the cube is $\phi' = \frac{1}{6} 	imes \frac{q}{\varepsilon_0}$. Substituting the value of $q$: $\phi' = \frac{1}{6} 	imes \frac{\sigma 	imes 4 \pi r^2}{\varepsilon_0} = \frac{2 \pi r^2 \sigma}{3 \varepsilon_0}$.

$A$ hollow spherical shell of radius $r$ has a uniform charge density $\sigma$. It is kept in a cube of edge $3r$ such that the centres of the cube and the shell coincide. Then the electric flux coming out of one face of a cube is ($\varepsilon_0$ - permittivity of free space).

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