Two point charges 2q and q are placed at vert | vedclass

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(D) According to Gauss's Law, the total electric flux $\phi$ through a closed surface is $\phi = \frac{Q_{	ext{in}}}{\varepsilon_0}$, where $Q_{	ext

Vedclass · Accepted Answer

$\frac{3q}{4\varepsilon_0}$

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(D) According to Gauss's Law, the total electric flux $\phi$ through a closed surface is $\phi = \frac{Q_{	ext{in}}}{\varepsilon_0}$, where $Q_{	ext{in}}$ is the net charge enclosed by the surface.$1$. The charge $2q$ is placed at vertex $A$. $A$ vertex of a cube is shared by $8$ adjacent cubes. Therefore, the fraction of charge $2q$ enclosed within the cube is $\frac{2q}{8} = \frac{q}{4}$.$2$. The charge $q$ is placed at the centre of the face $CDEF$. $A$ face of a cube is shared by $2$ adjacent cubes. Therefore, the fraction of charge $q$ enclosed within the cube is $\frac{q}{2}$.$3$. The total enclosed charge $Q_{	ext{in}}$ is the sum of these contributions: $Q_{	ext{in}} = \frac{q}{4} + \frac{q}{2} = \frac{q + 2q}{4} = \frac{3q}{4}$.$4$. Thus, the total electric flux passing through the cube is $\phi = \frac{Q_{	ext{in}}}{\varepsilon_0} = \frac{3q}{4\varepsilon_0}$.

Two point charges $2q$ and $q$ are placed at vertex $A$ and the centre of face $CDEF$ of the cube, respectively, as shown in the figure. The electric flux passing through the cube is:

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