Question

A cube is made from six thin insulating square faces, each square having side 'd' and carrying a uniformly distributed charge of Q. The magnitude of electric flux leaving the sixth face, due to the electric field of other five faces, is $\phi$. The magnitude of electrostatic force on each face of cube is F. Then,

• A. $\phi = \frac{Q}{\epsilon_0}$ and $F = \frac{Q^2}{2 \epsilon_0 d^2}$
• B. $\phi = \frac{Q}{\epsilon_0}$ and $F = \frac{Q^2}{\epsilon_0 d^2}$
• C. $\phi = \frac{6Q}{\epsilon_0}$ and $F = \frac{Q^2}{2 \epsilon_0 d^2}$
• D. $\phi = \frac{Q}{\epsilon_0}$ and $F = \frac{Q^2}{4 \epsilon_0 d^2}$

Answer 1

Answer: (A)

$\phi = \frac{Q}{\epsilon_0}$ and $F = \frac{Q^2}{2 \epsilon_0 d^2}$

Answer 2

The total charge on the cube is $6Q$. By Gauss's Law, the total flux leaving the cube is $\frac{6Q}{\epsilon_0}$. By symmetry, the flux leaving each face is $\frac{1}{6} \left(\frac{6Q}{\epsilon_0}\right) = \frac{Q}{\epsilon_0}$. The total flux through the sixth face ($\Phi_6$) is the sum of the flux due to the other five faces ($\phi$) and the flux due to the charge on the sixth face itself ($\Phi_{6,6}$). Thus, $\Phi_6 = \phi + \Phi_{6,6}$. The field lines originating from the charge on the sixth face must exit the cube through the other five faces. Therefore, the net flux of the field from the sixth face through the sixth face itself is zero, i.e., $\Phi_{6,6} = 0$. Hence, $\phi = \Phi_6 = \frac{Q}{\epsilon_0}$.

The electrostatic force on a charged surface is given by $F = \int \frac{1}{2} \sigma E_{\perp} dA$, where $\sigma$ is the surface charge density and $E_{\perp}$ is the perpendicular component of the electric field due to other charges. For a uniformly charged face, this simplifies to $F = \frac{1}{2} \sigma \phi$, where $\phi$ is the flux through the face due to the electric field of the other five faces. Given $\sigma = \frac{Q}{d^2}$ and $\phi = \frac{Q}{\epsilon_0}$, the force on each face is $F = \frac{1}{2} \left(\frac{Q}{d^2}\right) \left(\frac{Q}{\epsilon_0}\right) = \frac{Q^2}{2 \epsilon_0 d^2}$.