A thin spherical shell of radius R and surfac | vedclass

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Question

(A) The total charge $Q$ on the spherical shell is given by $Q = 	ext{Surface Area} 	imes 	ext{Surface Charge Density} = (4 \pi R^2) \sigma$.Accord

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The total charge $Q$ on the spherical shell is given by $Q = 	ext{Surface Area} 	imes 	ext{Surface Charge Density} = (4 \pi R^2) \sigma$.According to Gauss's Law,the total electric flux $\phi_{	ext{total}}$ through any closed surface enclosing a charge $Q$ is $\phi_{	ext{total}} = \frac{Q}{\varepsilon_0}$.Substituting the value of $Q$,we get $\phi_{	ext{total}} = \frac{4 \pi R^2 \sigma}{\varepsilon_0}$.Since the cube is a symmetric closed surface and the spherical shell is placed at its center,the total flux is distributed equally among the $6$ faces of the cube.Therefore,the electric flux through one face of the cube is $\phi_{	ext{face}} = \frac{\phi_{	ext{total}}}{6} = \frac{4 \pi R^2 \sigma}{6 \varepsilon_0} = \frac{2 \pi R^2 \sigma}{3 \varepsilon_0}$.

$A$ thin spherical shell of radius $R$ and surface charge density $\sigma$ is placed in a cube of side $5R$ with their centers coinciding. The electric flux through one face of the cube is $(\varepsilon_0 = \text{Permittivity of free space})$

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