Question

The electric flux from a cube of side 1 m is \(\Phi\). When the side of the cube is made 3 m and the charge enclosed by the cube is made one-third of the original value, then the flux from the bigger cube will be:

• A. \(\Phi\)
• B. \(\Phi/3\)
• C. \(3 \Phi\)
• D. \(9 \Phi\)

Hint:

According to Gauss's law, the electric flux through a closed surface depends only on the charge enclosed within the surface and not on the size or shape of the surface.

Answer 1

The Correct Option is A

The electric flux \(\Phi\) through a closed surface is related to the charge \(Q\) enclosed within the surface by Gauss's law: \[ \Phi = \frac{Q}{\epsilon_0} \] Where: - \(\Phi\) is the electric flux, - \(Q\) is the charge enclosed by the surface, - \(\epsilon_0\) is the permittivity of free space. From Gauss's law, the electric flux depends only on the charge enclosed within the surface, not on the size or shape of the surface. In this case, the charge enclosed by the cube is reduced to one-third of its original value, while the size of the cube increases to 3 meters. However, since the flux depends only on the charge enclosed, the flux will remain the same, i.e., \(\Phi\). Thus, the electric flux from the bigger cube is still \(\Phi\).