Demystifying I.E. Irodov Problem 3.25: Mastering Non-Uniform Charge Densities

If you are navigating the rigorous pathways of JEE Advanced preparation, training for Physics Olympiads, or simply possess a deep appreciation for challenging electrodynamics, you have undoubtedly crossed paths with I.E. Irodov's Problems in General Physics. For generations of physics students, this book has served as the ultimate testing ground. Its reputation for difficulty is well-earned, but the true brilliance of Irodov's problems lies in their ability to strip away formulaic thinking, forcing you to engage with core physical principles.

Today, we are taking a comprehensive, step-by-step journey through Problem 3.25 from the Electrodynamics section. This problem is a masterful exercise in applying Gauss's Law to a system featuring a radially varying, non-uniform volume charge density. It seamlessly bridges the gap between pure calculus and physical intuition.

Let's break it down.

The Physical Setup

Before diving into the math, it is crucial to understand the physical system we are dealing with. Here is the translated problem statement:

A ball of radius $R$ carries a positive charge whose volume density depends only on a separation $r$ from the ball's centre as $\rho = \rho_0 (1 - r/R)$, where $\rho_0$ is a constant. Assuming the permittivities of the ball and the environment to be equal to unity, find:

(a) the magnitude of the electric field strength as a function of the distance $r$ both inside and outside the ball;

(b) the maximum intensity $E_{max}$, and the corresponding distance $r_m$.

A Quick Note on Units: Irodov originally wrote his text using the CGS (Centimetre-Gram-Second) system, where the Coulomb constant is simply $1$. When he mentions "permittivities equal to unity," in our modern SI framework, this translates to utilizing the standard vacuum permittivity constant, $\varepsilon_0$. We will solve this using SI units for clarity.

Analyzing the Charge Distribution

The function $\rho(r) = \rho_0(1 - r/R)$ tells a fascinating physical story.

At the very center of the sphere ($r = 0$), the charge density is at its maximum, $\rho = \rho_0$.
As you move outward, the density linearly decreases.
At the very surface of the sphere ($r = R$), the charge density drops exactly to zero.

Because this distribution depends only on the radial distance $r$ from the center, the entire system possesses perfect spherical symmetry. This is our cue to deploy Gauss's Law.

Part A: Calculating the Electric Field

Gauss's Law states that the net electric flux through any closed surface is equal to the enclosed charge divided by the permittivity of free space:

$$\oint \mathbf{E} \cdot d\mathbf{A} = \frac{q_{in}}{\varepsilon_0}$$

Because of our spherical symmetry, we will use spherical Gaussian surfaces. On these surfaces, the electric field $\mathbf{E}$ points strictly radially outward and has a constant magnitude. This simplifies our flux integral tremendously to $E \cdot 4\pi r^2$.

1. The Field Inside the Sphere ($r < R$)

To find the field at a point inside the ball, we draw a Gaussian sphere of radius $r$. The immediate challenge is finding $q_{in}$, the total charge enclosed within this specific radius. Since the charge density is not uniform, we cannot simply multiply density by volume; we must integrate.

We consider an infinitesimally thin spherical shell of radius $x$ and thickness $dx$. The volume of this thin shell is $dV = 4\pi x^2 dx$. The charge within this shell is $dq = \rho(x) dV$.

To find the total enclosed charge, we integrate from the center ($0$) to our Gaussian radius ($r$):

$$q_{in} = \int_0^r \rho(x) dV = \int_0^r \rho_0 \left(1 - \frac{x}{R}\right) 4\pi x^2 dx$$

Let's expand and evaluate this integral:

$$q_{in} = 4\pi \rho_0 \int_0^r \left( x^2 - \frac{x^3}{R} \right) dx$$

$$q_{in} = 4\pi \rho_0 \left[ \frac{x^3}{3} - \frac{x^4}{4R} \right]_0^r$$

$$q_{in} = 4\pi \rho_0 \left( \frac{r^3}{3} - \frac{r^4}{4R} \right)$$

Now, we substitute this back into our simplified Gauss's Law equation:

$$E_{in} \cdot 4\pi r^2 = \frac{4\pi \rho_0}{\varepsilon_0} \left( \frac{r^3}{3} - \frac{r^4}{4R} \right)$$

Dividing both sides by $4\pi r^2$ isolates the electric field magnitude:

$$E_{in}(r) = \frac{\rho_0}{\varepsilon_0} \left( \frac{r}{3} - \frac{r^2}{4R} \right)$$

2. The Field Outside the Sphere ($r \ge R$)

For a point outside the ball, our Gaussian sphere encloses the entire charged ball. To find the total charge $Q$, we simply evaluate our $q_{in}$ expression at the boundary, where $r = R$:

$$Q = 4\pi \rho_0 \left( \frac{R^3}{3} - \frac{R^4}{4R} \right)$$

$$Q = 4\pi \rho_0 R^3 \left( \frac{1}{3} - \frac{1}{4} \right)$$

$$Q = 4\pi \rho_0 R^3 \left( \frac{1}{12} \right) = \frac{\pi \rho_0 R^3}{3}$$

Once we are outside a spherically symmetric charge distribution, the system behaves exactly as if all its charge were concentrated at a single point at the center. Applying Gauss's Law again:

$$E_{out} \cdot 4\pi r^2 = \frac{Q}{\varepsilon_0}$$

Substitute the total charge $Q$ to find the outer field:

$$E_{out}(r) = \frac{\pi \rho_0 R^3}{3 \varepsilon_0 (4\pi r^2)} = \frac{\rho_0 R^3}{12 \varepsilon_0 r^2}$$

Part B: Finding the Maximum Intensity ($E_{max}$)

Looking at our equation for the field outside the sphere ($E_{out}$), it follows an inverse-square law ($1/r^2$). This means the field strictly weakens as you move away from the surface. Consequently, the maximum electric field intensity must occur somewhere inside the ball.

To find the exact location of this peak intensity, we take the first derivative of our interior electric field function $E_{in}(r)$ with respect to $r$ and set it to zero.

$$\frac{dE_{in}}{dr} = \frac{d}{dr} \left[ \frac{\rho_0}{\varepsilon_0} \left( \frac{r}{3} - \frac{r^2}{4R} \right) \right] = 0$$

$$\frac{\rho_0}{\varepsilon_0} \left( \frac{1}{3} - \frac{2r}{4R} \right) = 0$$

Since the constants out front cannot equal zero, the terms inside the parenthesis must balance out:

$$\frac{1}{3} - \frac{r}{2R} = 0 \implies \frac{r}{2R} = \frac{1}{3}$$

Solving for $r$ gives us $r_m$, the distance at which the field is maximized:

$$r_m = \frac{2R}{3}$$

Finally, to discover the actual maximum field strength, $E_{max}$, we plug this optimal distance $r_m$ back into our original $E_{in}$ formula:

$$E_{max} = \frac{\rho_0}{\varepsilon_0} \left( \frac{1}{3}\left(\frac{2R}{3}\right) - \frac{1}{4R}\left(\frac{2R}{3}\right)^2 \right)$$

$$E_{max} = \frac{\rho_0}{\varepsilon_0} \left( \frac{2R}{9} - \frac{4R^2}{36R} \right)$$

$$E_{max} = \frac{\rho_0}{\varepsilon_0} \left( \frac{2R}{9} - \frac{R}{9} \right)$$

$$E_{max} = \frac{\rho_0 R}{9 \varepsilon_0}$$

Frequently Asked in Exams

Concepts from this problem are often used in:

JEE Advanced
NEET Physics
Olympiad exams

✔️ 3. Base for Advanced Topics

This concept is useful in:

Charged particle motion
Electrostatics applications
Electrodynamics problems

🚀 How to Solve Similar Problems

Follow this step-by-step approach:

Identify all forces acting on the particle
Write mathematical expressions for each force
Compare magnitudes of forces
Apply condition (greater than / equal / less than)
Solve for required variable

👉 This method works for almost all force comparison problems.

❌ Common Mistakes to Avoid

Ignoring direction of forces
Not applying inequality correctly
Confusing equilibrium with motion
Missing units or incorrect substitution

👉 Always double-check your force comparison.

💡 Pro Tips for JEE Advanced

Questions may include inclined planes or multiple forces
Electric field direction may be tricky
Sometimes acceleration is asked instead of condition

👉 Practice variations to master the concept.

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🔔 Conclusion

The I.E. Irodov Problem 3.25 solution is a perfect example of how basic physics concepts can solve complex problems. By comparing electric force and gravitational force, we can easily determine the condition for motion.

Mastering such problems will significantly improve your conceptual understanding and exam performance.