Animated Solution for Physics - Electrostatics: A cubical region of side $a$ has its centre at the origin. It encloses three fixed point charges, $-q$ at $(0,-a/4,0)$, $+3q$ at $(0,0,0)$ and $-q$ at $(0,+a/4,0)$. Choose the correct option(s).

Let's embark on a thrilling journey through symmetry and electric flux! We are given a cubical region of side $a$, perfectly centered at the origin. Inside this cube, three point charges are placed strictly along the y-axis.

Specifically, we have a $+3q$ charge sitting right at the origin $(0,0,0)$, and two $-q$ charges placed symmetrically at $(0,-a/4,0)$ and $(0,+a/4,0)$.

This specific linear arrangement of charges is the key to unlocking the entire problem without doing any heavy integration.

Gauss's Law

Before we dive into the individual faces, let's look at the big picture. What is the total electric flux crossing the entire cubical region?

According to Gauss's Law, the total electric flux through any closed surface depends exclusively on the net charge enclosed within it.

Let's sum up our charges: $q_{\text{in}}=-q+3q-q=+q$.

Therefore, the total flux crossing the entire region is simply $q/{\epsilon }_0$. This immediately confirms that Option (c) is absolutely correct.

Now, let's analyze the flux through the individual faces of the cube. We'll start with the faces perpendicular to the y-axis, located at $y=+a/2$ and $y=-a/2$.

Notice that the charge distribution is perfectly symmetric about the xz-plane ($y=0$). For every charge at a positive y-coordinate, there is an identical charge at the corresponding negative y-coordinate.

Because of this perfect mirror symmetry, the electric field pattern on the positive y-side is a flawless mirror image of the negative y-side. Consequently, the electric flux piercing the right face must be exactly equal to the flux piercing the left face.

Option (b) claims that the flux through one face is more than the other, which violates this symmetry. Thus, Option (b) is incorrect.

What about the faces perpendicular to the x-axis? These are the faces at $x=+a/2$ and $x=-a/2$.

Since all the charges lie entirely on the y-axis, the setup is perfectly symmetric to the left and right of the yz-plane ($x=0$). There is no charge shifted towards the positive x or negative x direction.

This mirror symmetry guarantees that the flux crossing the front face at $x=+a/2$ perfectly matches the flux crossing the back face at $x=-a/2$.

Therefore, Option (a) is correct.

Finally, let's evaluate Option (d), which compares the flux through the x-face ($x=+a/2$) and the z-face ($z=+a/2$).

Here is where the true elegance of physics shines. Because all charges lie exactly on the y-axis, the entire system possesses a $90^{\circ }$ rotational symmetry around the y-axis.

Imagine rotating the entire cube by $90^{\circ }$ around the y-axis. The x-face physically rotates to become the z-face. However, because the charges are on the axis of rotation, the charge distribution remains completely unchanged!

This mathematical invariance guarantees that the flux through the x-face must equal the flux through the z-face.

So, physically, Option (d) is also correct.