Question

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

• A. 3q/ε₀
• B. 3q/(4ε₀)
• C. q/ε₀
• D. 3q/(2ε₀)

Hint:

To remember sharing fractions: a vertex is shared by 8 cubes, an edge by 4, and a face by 2.

Answer 1

The Correct Option is B

To solve the given problem, we need to determine the electric flux passing through the cube when two point charges are placed at specific positions.

Concept Involved: According to Gauss's Law, the electric flux \(\Phi\) through a closed surface is given by the formula:

\(\Phi = \frac{Q_{\text{enclosed}}}{\varepsilon_0}\)

where \(Q_{\text{enclosed}}\) is the total charge enclosed by the surface and \(\varepsilon_0\) is the permittivity of free space.

1. Charge \(2q\) is placed at vertex \(A\). A vertex is shared by 8 cubes.
2. Charge \(q\) is placed at the center of face \(CDEF\). A face center is shared by 2 cubes.

Calculation of Enclosed Charge:

1. \(2q\) at vertex \(A\): Contribution to this particular cube is \(\frac{2q}{8} = \frac{q}{4}\).
2. \(q\) at face center \(CDEF\): Contribution to this particular cube is \(\frac{q}{2}\).

Total charge \(Q_{\text{enclosed}}\) by the cube is:

\(Q_{\text{enclosed}} = \frac{q}{4} + \frac{q}{2} = \frac{q}{4} + \frac{2q}{4} = \frac{3q}{4}\)

Electric Flux:

According to Gauss's Law, the flux through the cube is:

\(\Phi = \frac{Q_{\text{enclosed}}}{\varepsilon_0} = \frac{\frac{3q}{4}}{\varepsilon_0} = \frac{3q}{4\varepsilon_0}\)

Thus, the electric flux passing through the cube is \(\frac{3q}{4\varepsilon_0}\).

The correct answer is:

3q/(4ε₀)