Question

A point charge q is moved from just inside a cube to just outside through one of its corner points. What is the change in electric flux through the highlighted (yellow) face that meets at that corner?

• A. 0
• B. q/(8*epsilon_0)
• C. -q/(8*epsilon_0)
• D. q/(4*epsilon_0)

Answer 1

Answer: (A)

0

Answer 2

Let the point charge be $q$. We are interested in the change in electric flux through a specific face of the cube as the charge is moved from just inside to just outside, passing through a corner. The electric flux $\Phi$ through a surface $S$ due to a point charge $q$ is given by $\Phi = \frac{q}{4\pi\epsilon_0} \Omega$, where $\Omega$ is the solid angle subtended by the surface $S$ at the location of the charge.

Consider a cube with side length $a$. Let the charge be moved through the corner at the origin $(0,0,0)$. The three faces meeting at this corner are in the $xy$-plane ($z=0$), $xz$-plane ($y=0$), and $yz$-plane ($x=0$). Let the yellow face be the one in the $xy$-plane ($z=0$, for $0 \le x \le a, 0 \le y \le a$).

When the charge is exactly at the corner $(0,0,0)$, the cube occupies one octant of the space around it. The total solid angle of a full sphere is $4\pi$. The solid angle subtended by the entire cube at its corner is $\frac{1}{8}$ of the total solid angle, which is $\frac{4\pi}{8} = \frac{\pi}{2}$ steradians.

The three faces meeting at the corner divide this solid angle. By symmetry, each of these three faces subtends an equal solid angle at the corner. Therefore, the solid angle subtended by each of these three faces at the corner is $\frac{\pi/2}{3} = \frac{\pi}{6}$. This reasoning is incorrect for the flux calculation through a single face when the charge is at the corner.

A more direct approach using symmetry for a charge at the corner: The total flux through the entire cube when the charge is at the corner is not $q/\epsilon_0$ (Gauss's Law applies when the charge is enclosed). However, when the charge is at the corner, the flux through the three faces meeting at that corner is equal, and the flux through the other three faces is zero (since the electric field is parallel to these faces at the corner). The total flux through the cube must be distributed.

Let's consider the flux through the yellow face when the charge is at the corner. The solid angle subtended by the yellow face at the corner is $\pi/2$. This is incorrect. The solid angle subtended by the entire cube at the corner is $\pi/2$.

A crucial insight comes from considering the flux through the three faces meeting at the corner. When the charge is exactly at the corner, the flux through each of these three faces is equal. Let this flux be $\Phi_{corner}$. Due to symmetry, the flux through the other three faces is 0. The total flux through the cube is not $q/\epsilon_0$ in this case.

However, if we consider the flux through the yellow face as the charge moves from just inside to just outside through the corner: Let the charge be at position $\mathbf{r}_{in}$ just inside the cube, near the corner. The flux through the yellow face is $\Phi_{in}$. Let the charge be at position $\mathbf{r}_{out}$ just outside the cube, near the corner. The flux through the yellow face is $\Phi_{out}$. We need to find $\Delta \Phi = \Phi_{out} - \Phi_{in}$.

It can be shown that when the charge is exactly at a corner, the flux through each of the three faces meeting at that corner is $\frac{q}{8\epsilon_0}$. When the charge is infinitesimally inside the cube (e.g., at $(-\delta, -\delta, -\delta)$ if the corner is at $(0,0,0)$ and the face is $z=0$), the flux through the yellow face is $\Phi_{in}$. When the charge is infinitesimally outside the cube (e.g., at $(\delta, \delta, \delta)$), the flux through the yellow face is $\Phi_{out}$.

Consider the three faces that meet at the corner. Let these be $F_1, F_2, F_3$, and let the yellow face be $F_1$. When the charge is just inside, the flux through $F_1$ is $\Phi_{1,in}$. When the charge is just outside, the flux through $F_1$ is $\Phi_{1,out}$. The change in flux is $\Delta \Phi_1 = \Phi_{1,out} - \Phi_{1,in}$.

Due to the symmetry of the situation and the nature of electric fields radiating from a point charge, as the charge passes through the corner, the flux through the specific face that meets at that corner changes negligibly. The electric field lines that would have passed through the face from the inside will now pass through it from the outside, and vice-versa, but the net change in flux through that single face is zero.

More formally, consider the flux through the three faces meeting at the corner. Let the charge be at $(-\delta, -\delta, -\delta)$ (inside) and $(\delta, \delta, \delta)$ (outside). The flux through the face $z=0$ (yellow face) is $\Phi_{in}$ and $\Phi_{out}$ respectively. The change is $\Delta \Phi = \Phi_{out} - \Phi_{in}$. It turns out that the flux through each of the three faces meeting at the corner is exactly the same whether the charge is infinitesimally inside or infinitesimally outside, as it passes through the corner. Therefore, the change in flux through the yellow face is 0.

The total flux through the cube changes from $q/\epsilon_0$ (when inside) to $0$ (when outside). This change of $-q/\epsilon_0$ is distributed among all faces. However, the change in flux through the specific face at the corner is zero. This is because the contribution to the flux from the charge inside and outside, as it passes through the corner, effectively cancels out for that particular face. The electric field lines that were directed into the face from inside are now directed out of the face from outside, and the geometry ensures these contributions balance to zero change for that face.