We are given a point charge \( q \) placed at the centre of a cube with side length \( L \). We are asked to find the electric flux linked with each face of the cube.
Step 1: We use Gauss's law to find the total electric flux through a closed surface. According to Gauss's law: \[ \Phi_{{total}} = \frac{q}{\epsilon_0} \] where \( \Phi_{{total}} \) is the total electric flux and \( \epsilon_0 \) is the permittivity of free space.
Step 2: The cube has 6 faces, and the point charge \( q \) is located at the centre of the cube. Since the electric flux is symmetric, the flux through each face of the cube is the same. Thus, the flux linked with each face of the cube is: \[ \Phi_{{face}} = \frac{\Phi_{{total}}}{6} = \frac{q}{6 \epsilon_0} \] Thus, the electric flux linked with each face of the cube is \( \frac{q}{6 \epsilon_0} \).
In the given circuit, the electric currents through $15\, \Omega$ and $6 \, \Omega$ respectively are

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