A circular disc of radius R carries surface charge density σ(r) = σ₀(1 - r/R), where σ₀ is a constan

Q: Why do we use a ring element instead of a disc element?

The charge density varies with the radial distance $r$. For a thin ring of radius $r$, the charge density is approximately constant over its entire area, allowing us to write $dq = \sigma(r) dA$.

Q: Does the shape of the enclosing surface matter?

No! According to Gauss's Law, the total electric flux through a closed surface depends *only* on the net charge enclosed, regardless of the surface's shape or size.

Q: Why is the area of the ring $2\pi r dr$?

If you cut the thin ring and unroll it, it forms a rectangle with length equal to its circumference ($2\pi r$) and width equal to its thickness ($dr$). Thus, the area is $2\pi r dr$.

Q: What happens if $\sigma_0$ is negative?

If $\sigma_0$ is negative, the total charge and the fluxes would be negative. However, since we are taking the ratio of the two fluxes, the negative signs would cancel out, leaving the final answer unchanged.

Q: Can we use the electric field to find the flux here?

While technically possible ($\phi = \oint \mathbf{E} \cdot d\mathbf{A}$), calculating the non-uniform electric field of this disc at all points on the spherical surfaces would be a mathematical nightmare. Gauss's Law provides a brilliant shortcut by relating flux directly to enclosed charge.

Question

A circular disc of radius $R$ carries surface charge density $\sigma(r) = \sigma_0 \left(1 - \frac{r}{R}\right)$, where $\sigma_0$ is a constant and $r$ is the distance from the center of the disc. Electric flux through a large spherical surface that encloses the charged disc completely is $\phi_0$. Electric flux through another spherical surface of radius $\frac{R}{4}$ and concentric with the disc is $\phi$. Then the ratio $\frac{\phi_0}{\phi}$ is_________.

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### The Setup: A Disc with a Gradient

Imagine a flat, circular disc of radius $R$ resting in space. If the charge were spread evenly across it, calculating the electric flux would be a walk in the park. But physics loves a good challenge! Here, the surface charge density $\sigma$ isn't constant; it fades as you move outward from the center.

The density is governed by the function $\sigma(r) = \sigma_0 \left(1 - \frac{r}{R}ight)$. At the very center ($r = 0$), the density is at its maximum, $\sigma_0$. As you travel to the edge ($r = R$), the density drops exactly to zero.

### The Master Tool: Gauss's Law

We are asked to find the ratio of electric fluxes through two different spherical surfaces. Whenever you hear "electric flux through a closed surface," your brain should immediately scream **Gauss's Law**.

Gauss's Law states that the total electric flux $\phi$ through any closed surface is directly proportional to the net charge enclosed within that surface:

$$\phi = \frac{q_{	ext{enclosed}}}{\epsilon_0}$$

This means we don't need to worry about the complex electric field vectors pointing in all directions. We just need to count the charge trapped inside our imaginary spheres!

### Slicing the Disc: The Elemental Ring

Because the charge density varies with the radial distance $r$, we cannot simply multiply the total area by a single density value. We must use the power of calculus.

Imagine slicing the disc into infinitely many thin, concentric rings. Let's pick one such elemental ring at a distance $r$ from the center, with an infinitesimally small thickness $dr$.

If we were to cut this ring and unroll it, it would look like a very long, thin rectangle. Its length is the circumference $2\pi r$, and its width is $dr$. Therefore, the area of this tiny ring is:

$$dA = 2\pi r \, dr$$

The amount of charge $dq$ sitting just on this ring is the area multiplied by the local charge density:

$$dq = \sigma(r) \, dA = \sigma_0 \left(1 - \frac{r}{R}ight) 2\pi r \, dr$$

### Calculating the Total Flux ($\phi_0$)

The problem defines $\phi_0$ as the flux through a *large* spherical surface that completely encloses the disc. This means the sphere traps $100\%$ of the disc's charge. To find this total charge, we integrate our $dq$ expression from the center ($r=0$) all the way to the edge ($r=R$):

$$q_{	ext{total}} = \int_0^R \sigma_0 \left(1 - \frac{r}{R}ight) 2\pi r \, dr$$

Let's pull out the constants and distribute the $r$:

$$q_{	ext{total}} = 2\pi\sigma_0 \int_0^R \left(r - \frac{r^2}{R}ight) dr$$

Integrating term by term gives:

$$q_{	ext{total}} = 2\pi\sigma_0 \left[ \frac{r^2}{2} - \frac{r^3}{3R} ight]_0^R$$

Substituting the upper limit $R$:

$$q_{	ext{total}} = 2\pi\sigma_0 \left( \frac{R^2}{2} - \frac{R^3}{3R} ight) = 2\pi\sigma_0 \left( \frac{R^2}{2} - \frac{R^2}{3} ight) = 2\pi\sigma_0 \left( \frac{R^2}{6} ight)$$

By Gauss's Law, the total flux is:

$$\phi_0 = \frac{2\pi\sigma_0}{\epsilon_0} \left( \frac{R^2}{6} ight)$$

### Calculating the Partial Flux ($\phi$)

Now, we look at the second spherical surface. This one is concentric with the disc but only has a radius of $\frac{R}{4}$. It acts like a smaller net, only catching the charge located in the central region of the disc up to $r = \frac{R}{4}$.

We use the exact same integral setup, but we change our upper limit to reflect the boundary of this new sphere:

$$q_{	ext{inner}} = 2\pi\sigma_0 \int_0^{R/4} \left(r - \frac{r^2}{R}ight) dr$$

$$q_{	ext{inner}} = 2\pi\sigma_0 \left[ \frac{r^2}{2} - \frac{r^3}{3R} ight]_0^{R/4}$$

Now, we must carefully substitute $r = \frac{R}{4}$:

$$q_{	ext{inner}} = 2\pi\sigma_0 \left( \frac{(R/4)^2}{2} - \frac{(R/4)^3}{3R} ight)$$

$$q_{	ext{inner}} = 2\pi\sigma_0 \left( \frac{R^2/16}{2} - \frac{R^3/64}{3R} ight) = 2\pi\sigma_0 \left( \frac{R^2}{32} - \frac{R^2}{192} ight)$$

To subtract these fractions, we find a common denominator of $192$:

$$q_{	ext{inner}} = 2\pi\sigma_0 \left( \frac{6R^2 - R^2}{192} ight) = 2\pi\sigma_0 \left( \frac{5R^2}{192} ight)$$

So, the flux through the smaller sphere is:

$$\phi = \frac{2\pi\sigma_0}{\epsilon_0} \left( \frac{5R^2}{192} ight)$$

### The Grand Finale: The Ratio

We have both fluxes! The final step is to find the ratio $\frac{\phi_0}{\phi}$. Watch how beautifully the physics constants melt away, leaving behind pure, elegant geometry:

$$\frac{\phi_0}{\phi} = \frac{\frac{2\pi\sigma_0}{\epsilon_0} \left( \frac{R^2}{6} ight)}{\frac{2\pi\sigma_0}{\epsilon_0} \left( \frac{5R^2}{192} ight)}$$

Cancel out $\frac{2\pi\sigma_0}{\epsilon_0}$ and $R^2$:

$$\frac{\phi_0}{\phi} = \frac{1/6}{5/192} = \frac{192}{6 	imes 5} = \frac{192}{30}$$

Dividing both the numerator and denominator by $6$ gives:

$$\frac{\phi_0}{\phi} = \frac{32}{5} = \mathbf{6.40}$$

And there we have it! A beautiful application of Gauss's Law and integral calculus.

Enter Numerical Value:

Visualized Solution

\text{Visualizing the Charged Disc}

\text{Gauss's Law}

\text{The Elemental Ring}

\text{Charge on the Elemental Ring}

\text{Total Flux } (\phi_0)

\text{Evaluating } \phi_0

\text{The Inner Sphere}

\text{Flux through Inner Sphere } (\phi)

\text{Evaluating } \phi

\text{The Final Ratio}

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

The Setup

The Master Tool

Slicing the Disc

Calculating the Total Flux ( $ϕ_{0}$ )

Calculating the Partial Flux ( $ϕ$ )

The Grand Finale

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