You are given a dipole of charge +q and -q se | vedclass

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(D) According to Gauss's Law,the electric flux $\phi$ through any closed surface is given by $\phi = \frac{q_{	ext{enclosed}}}{\varepsilon_0}$,where

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$-q / \varepsilon_0$

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(D) According to Gauss's Law,the electric flux $\phi$ through any closed surface is given by $\phi = \frac{q_{	ext{enclosed}}}{\varepsilon_0}$,where $q_{	ext{enclosed}}$ is the total charge enclosed by the surface.In the given figure,sphere '$A$' has a radius ' $R$ ' and is centered at the midpoint of the dipole. The distance between the charges $+q$ and $-q$ is $2R$. Therefore,the distance of each charge from the center is $R$.Since sphere '$A$' has a radius ' $R$ ',it encloses only the charge $-q$ located at the center of the sphere.Thus,the charge enclosed by sphere '$A$' is $q_{	ext{enclosed}} = -q$.Therefore,the electric flux through sphere '$A$' is $\phi_A = \frac{-q}{\varepsilon_0}$.

You are given a dipole of charge $+q$ and $-q$ separated by a distance $2R$. $A$ sphere '$A$' of radius ' $R$ ' passes through the centre of the dipole as shown below and another sphere '$B$' of radius ' $2R$ ' passes through the charge $+q$. Then the electric flux through the sphere '$A$' is

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