A disc of radius a/4 having a uniformly distributed charge 6C is placed in the x-y plane with its ce

Q: Why don't we consider the charge outside the cube for the flux calculation?

According to Gauss's Law, any electric field line originating from a charge outside a closed surface will enter the surface and then exit it, contributing zero net flux. Only charges enclosed within the surface contribute to a net outward or inward flux.

Q: How did we know exactly half the disc is inside the cube?

The disc is centered at $x = -a/2$, which is exactly on the left face of the cube. Since the disc lies in the $x-y$ plane and the cube extends from $y = -a/2$ to $a/2$ (which is larger than the disc's radius of $a/4$), the cube's face perfectly bisects the disc.

Q: What if the rod had a non-uniform charge density?

If the rod had a non-uniform charge density (e.g., $\lambda = kx$), we couldn't use a simple length ratio. We would have to integrate the charge density $\lambda(x)$ from the starting point inside the cube ($x = a/4$) to the boundary of the cube ($x = a/2$).

Q: Why is the $z$-coordinate of the point charges $0$ important?

The cube extends from $z = -a/2$ to $z = a/2$. Since the point charges have a $z$-coordinate of $0$, they lie perfectly on the $x-y$ plane, which is well within the $z$-boundaries of the cube. If their $z$-coordinate was, say, $a$, they would be outside the cube.

Q: Does the position of the enclosed charge matter for the total flux?

No! Gauss's Law ($\Phi = q_{\text{in}}/\varepsilon_0$) shows that the total flux depends only on the magnitude of the enclosed charge, not its location. Moving the $-7C$ charge around inside the cube would change the electric field at specific points on the surface, but the *total* integrated flux would remain exactly the same.

Question

A disc of radius $\frac{a}{4}$ having a uniformly distributed charge $6C$ is placed in the $x-y$ plane with its centre at $(-\frac{a}{2}, 0, 0)$. A rod of length $a$ carrying a uniformly distributed charge $8C$ is placed on the $x$-axis from $x = \frac{a}{4}$ to $x = \frac{5a}{4}$. Two point charges $-7C$ and $3C$ are placed at $(\frac{a}{4}, -\frac{a}{4}, 0)$ and $(-\frac{3a}{4}, \frac{3a}{4}, 0)$, respectively. Consider a cubical surface formed by six surfaces $x = \pm \frac{a}{2}$, $y = \pm \frac{a}{2}$, $z = \pm \frac{a}{2}$. The electric flux through this cubical surface is

Options:

(A) $\frac{-2C}{\varepsilon_0}$

(B) $\frac{2C}{\varepsilon_0}$

(C) $\frac{10C}{\varepsilon_0}$

(D) $\frac{12C}{\varepsilon_0}$

Sigma Percentile · Accepted Answer

### The Power of Gauss's Law

Imagine you are tasked with counting the number of water droplets flying out of a complex sprinkler system. Instead of tracking every single droplet, what if you could just measure the total water pressure at the source? This is the elegant power of **Gauss's Law** in electrostatics. It tells us that the net electric flux $\Phi$ through any closed surface depends *only* on the total charge enclosed within it, given by the master equation:

$$ \Phi = \frac{q_{	ext{in}}}{\varepsilon_0} $$

In this thrilling problem, we are given a cubical surface bounded by $x = \pm \frac{a}{2}$, $y = \pm \frac{a}{2}$, and $z = \pm \frac{a}{2}$. Our mission is simple yet requires surgical precision: we must dissect the space and find exactly how much charge is trapped inside this invisible box.

### Dissecting the Charge Distribution

We have a diverse cast of characters in this electrostatic play: a charged disc, a charged rod, and two point charges. To find the total enclosed charge, we must interrogate each one and determine what fraction of their charge lies within the cube's boundaries.

### The Disc: A Tale of Two Halves

Let's start with the disc. It carries a total charge of $6C$ and is perfectly centered at $(-\frac{a}{2}, 0, 0)$. Notice the $x$-coordinate of its center! The left face of our cubical surface lies exactly at $x = -\frac{a}{2}$.

Because the disc lies in the $x-y$ plane and the cube's face slices right through its center, exactly half of the disc is inside the cube, and the other half is outside. Therefore, the charge contributed by the disc is:

$$ q_{	ext{disc, in}} = \frac{6C}{2} = 3C $$

### The Rod: Crossing the Boundary

Next, we turn our attention to the rod. It has a total length of $a$ and carries a uniformly distributed charge of $8C$. It stretches along the $x$-axis from $x = \frac{a}{4}$ to $x = \frac{5a}{4}$.

Our cube's territory ends at $x = \frac{a}{2}$. So, how much of the rod manages to sneak inside? We calculate the length of the rod that lies between $x = \frac{a}{4}$ and $x = \frac{a}{2}$:

$$ 	ext{Length inside} = \frac{a}{2} - \frac{a}{4} = \frac{a}{4} $$

Since the rod's total length is $a$, exactly one-quarter of it is inside the cube. Using a simple unitary method, the enclosed charge from the rod is:

$$ q_{	ext{rod, in}} = \frac{8C}{a} 	imes \frac{a}{4} = 2C $$

### The Point Charges: Inside or Outside?

Now for the point charges. We must act as the bouncers of this cubical club and check their coordinates.

1. The first point charge is $-7C$, located at $(\frac{a}{4}, -\frac{a}{4}, 0)$. Since both $\frac{a}{4}$ and $-\frac{a}{4}$ fall comfortably within the range of $[-\frac{a}{2}, \frac{a}{2}]$, this charge is completely **inside** the cube.
2. The second point charge is $3C$, located at $(-\frac{3a}{4}, \frac{3a}{4}, 0)$. The $x$-coordinate $-\frac{3a}{4}$ is less than $-\frac{a}{2}$, meaning this charge is strictly **outside** the cube.

Thus, the total charge contributed by the point charges is simply:

$$ q_{	ext{points, in}} = -7C + 0 = -7C $$

### The Grand Summation and Final Flux

We have interrogated all the suspects. Now, we bring them together for the grand summation. The total enclosed charge $q_{	ext{in}}$ is the sum of the individual enclosed charges:

$$ q_{	ext{in}} = q_{	ext{disc, in}} + q_{	ext{rod, in}} + q_{	ext{points, in}} $$
$$ q_{	ext{in}} = 3C + 2C - 7C = -2C $$

With the total enclosed charge revealed, we return to our master equation, Gauss's Law. The net electric flux through the cubical surface is:

$$ \Phi = \frac{-2C}{\varepsilon_0} $$

This perfectly matches option (a). The beauty of this problem lies in its spatial reasoning. It teaches us a profound lesson: the universe might be filled with chaotic charge distributions, but when it comes to electric flux through a closed surface, only what's inside truly matters!

Select Answer:

Visualized Solution

$Φ = \frac{q _{in}}{ε _{0}}$

$q_{disc, in} = 3 C$

$q_{rod, in} = 2 C$

$q_{points, in} = - 7 C$

$q_{in} = - 2 C$

$Φ = \frac{- 2 C}{ε _{0}}$

\text{Final Answer}

The Sigma Insight: Electric Field Lines, Flux and Gauss's Law

The Power of Gauss's Law

Dissecting the Charge Distribution

The Disc

The Rod

The Point Charges

The Grand Summation and Final Flux

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