Charge is distributed within a sphere of radius R with a volume charge density rho(r) = (A/r^2) e^(-

Q: Why do we use $dV = 4\pi r^2 dr$ instead of $dx dy dz$?

Because the charge density $\rho(r)$ only depends on the radial distance $r$ from the center, the distribution is spherically symmetric. Using spherical coordinates makes the integration much simpler. The term $4\pi r^2 dr$ represents the volume of a thin spherical shell of radius $r$ and thickness $dr$.

Q: Why did the $r^2$ term disappear during the integration?

This is the mathematical elegance of the problem! The charge density was given as inversely proportional to $r^2$, while the volume element is directly proportional to $r^2$. When you multiply them together, they perfectly cancel out, leaving only the exponential term to integrate.

Q: What does the negative sign in the exponent mean physically?

The term $e^{-2r/a}$ indicates that the charge density decays exponentially as you move further away from the center. The constant $a$ acts as a 'decay length', determining how quickly the charge thins out.

Q: Why did we flip the terms inside the logarithm at the end?

We had a negative sign outside the logarithm: $-\frac{a}{2} \log(x)$. Using the property of logarithms, $- \log(x) = \log(x^{-1}) = \log(1/x)$. Flipping the terms inside the bracket allowed us to absorb the negative sign and match the format of the given options.

Q: Is $\log$ in the options base 10 or base $e$?

In higher mathematics and physics (like JEE), $\log$ without an explicit base almost always refers to the natural logarithm, $\ln$ (base $e$). This is especially true when integrating exponential functions $e^x$, as the inverse operation is the natural logarithm.

Question

Charge is distributed within a sphere of radius $R$ with a volume charge density $\rho(r) = \frac{A}{r^2} e^{-2r/a}$, where $A$ and $a$ are constants. If $Q$ is the total charge of this charge distribution, the radius $R$ is

Options:

(A) $a \log \left( \frac{1}{1 - \frac{Q}{2\pi a A}} \right)$

(B) $a \log \left( 1 - \frac{Q}{2\pi a A} \right)$

(C) $\frac{a}{2} \log \left( 1 - \frac{Q}{2\pi a A} \right)$

(D) $\frac{a}{2} \log \left( \frac{1}{1 - \frac{Q}{2\pi a A}} \right)$

Sigma Percentile · Accepted Answer

The problem of finding the total charge from a variable volume charge density is a classic application of Gauss's Law and integration in electrostatics. It tests your ability to set up a volume integral in spherical coordinates and execute it flawlessly. Let's dive into the elegant mathematics behind this charge distribution.

### The Anatomy of the Charge Distribution

Imagine a solid sphere of radius $R$. Unlike a simple conductor where charge resides only on the surface, here the charge is distributed throughout the entire volume. However, it's not spread evenly. The volume charge density is given by:

$$ \rho(r) = \frac{A}{r^2} e^{-2r/a} $$

This function tells us two things. First, the $1/r^2$ term means the charge is highly concentrated near the center and thins out as you move away. Second, the exponential term $e^{-2r/a}$ causes a rapid, exponential decay of the charge density.

### Slicing the Sphere: The Volume Element

To find the total charge $Q$, we cannot simply multiply density by volume. We must integrate. We slice the sphere into infinitely many thin, concentric spherical shells.

Consider one such shell at a distance $r$ from the center, with an infinitesimally small thickness $dr$. If we were to unroll this shell, it would resemble a flat sheet with a surface area of $4\pi r^2$ and a thickness of $dr$. Therefore, the volume of this tiny shell is:

$$ dV = 4\pi r^2 dr $$

### The Elegance of the Integral

The total charge $Q$ is the sum of the charges in all these tiny shells, from the center ($r = 0$) to the surface ($r = R$). We set up the integral:

$$ Q = \int_0^R \rho(r) dV $$

Substituting our expressions for $\rho(r)$ and $dV$:

$$ Q = \int_0^R \left( \frac{A}{r^2} e^{-2r/a} \right) (4\pi r^2 dr) $$

Here is where the mathematical elegance of the problem reveals itself. The $r^2$ in the denominator of the density perfectly cancels the $r^2$ in the volume element!

$$ Q = 4\pi A \int_0^R e^{-2r/a} dr $$

We are left with a straightforward exponential integral.

### Executing the Integration

Integrating $e^{-2r/a}$ with respect to $r$ yields:

$$ Q = 4\pi A \left[ \frac{e^{-2r/a}}{-2/a} \right]_0^R $$

Now, we apply the limits from $0$ to $R$:

$$ Q = 4\pi A \left( \frac{-a}{2} \right) \left[ e^{-2R/a} - e^0 \right] $$

Since $e^0 = 1$, we can simplify the expression:

$$ Q = -2\pi a A (e^{-2R/a} - 1) $$

To make the equation neater, we absorb the negative sign into the bracket:

$$ Q = 2\pi a A (1 - e^{-2R/a}) $$

### Rescuing the Radius

Our ultimate goal is to find the radius $R$. We need to isolate the exponential term. First, divide both sides by $2\pi a A$:

$$ 1 - e^{-2R/a} = \frac{Q}{2\pi a A} $$

Rearranging to isolate the exponential:

$$ e^{-2R/a} = 1 - \frac{Q}{2\pi a A} $$

To bring $R$ down from the exponent, we take the natural logarithm ($\ln$ or $\log_e$) of both sides:

$$ -\frac{2R}{a} = \log \left( 1 - \frac{Q}{2\pi a A} \right) $$

Finally, multiply by $-\frac{a}{2}$:

$$ R = -\frac{a}{2} \log \left( 1 - \frac{Q}{2\pi a A} \right) $$

Using the logarithmic property $-\log(x) = \log(x^{-1}) = \log(1/x)$, we can rewrite the final answer to match the given options:

$$ R = \frac{a}{2} \log \left( \frac{1}{1 - \frac{Q}{2\pi a A}} \right) $$

**This beautiful result shows exactly how the radius of the sphere depends on the total charge it contains and the parameters of its internal charge distribution.**

Animated Solution for Physics - Electrostatics: Charge is distributed within a sphere of radius $R$ with a volume charge density $ρ (r) = \frac{A}{r ^{2}} e^{- 2 r / a}$ , where $A$ and $a$ are constants. If $Q$ is the total charge of this charge distribution, the radius $R$ is

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Visualized Solution

Visualizing the Charge Distribution $ρ (r)$

The Volume Element $d V$

Setting up the Integral for $Q$

Simplifying the Integrand

Performing the Integration

Applying the Limits $0$ to $R$

Isolating the Exponential Term

Solving for $R$

The Way Forward: $Q \to 2 π a A$

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

The Anatomy of the Charge Distribution

Slicing the Sphere

The Elegance of the Integral

Executing the Integration

Rescuing the Radius

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