In the figure, the inner (shaded) region A represents a sphere of radius r_A = 1, within which the e

Q: Why do we use $4\pi r^2 dr$ as the volume element instead of just $dr$?

In spherical coordinates, the charge is distributed over a volume, not just a line. A thin spherical shell of radius $r$ and thickness $dr$ has a surface area of $4\pi r^2$, so its volume is $4\pi r^2 dr$.

Q: Why is the electric field outside the shell only dependent on the total enclosed charge?

This is the beauty of Gauss's Law! For any spherically symmetric charge distribution, the electric field at a point outside behaves exactly as if all the enclosed charge were concentrated at the center.

Q: How do we calculate the potential just outside the shell?

The potential $V$ at a distance $r$ outside a spherically symmetric charge distribution is given by $V = \frac{1}{4\pi\epsilon_0} \frac{q_{enclosed}}{r}$. We simply substitute our expression for $q_{net}$ and the given $r_B$.

Q: Why did we integrate the two regions separately?

The charge density function $\rho(r)$ changes its mathematical form at $r = 1$. To evaluate the integral correctly, we must split it at the boundary where the function changes.

Q: In Option A, why is $r_B = \sqrt{3/2}$ incorrect when $4r_B^2 - 3 = 0$ gives $r_B = \sqrt{3}/2$?

It's a classic trap! Solving $4r_B^2 - 3 = 0$ yields $r_B = \frac{\sqrt{3}}{2}$, which is approximately $0.866$. Option A suggests $r_B = \sqrt{\frac{3}{2}}$, which is approximately $1.22$. They are mathematically different values.

Question

In the figure, the inner (shaded) region A represents a sphere of radius $r_A = 1$, within which the electrostatic charge density varies with the radial distance $r$ from the center as $\rho_A = kr$, where $k$ is positive. In the spherical shell B of outer radius $r_B$, the electrostatic charge density varies as $\rho_B = \frac{2k}{r}$. Assume that dimensions are taken care of. All physical quantities are in their SI units. Which of the following statement(s) is(are) correct?

Options:

(A) If $r_B = \sqrt{\frac{3}{2}}$, then the electric field is zero everywhere outside B.

(B) If $r_B = \frac{3}{2}$, then the electric potential just outside B is $\frac{k}{\epsilon_0}$.

(C) If $r_B = 2$, then the total charge of the configuration is $15\pi k$.

(D) If $r_B = \frac{5}{2}$, then the magnitude of the electric field just outside B is $\frac{13\pi k}{\epsilon_0}$.

Sigma Percentile · Accepted Answer

The problem of finding the electric field and potential for a spherically symmetric charge distribution is a classic in electrostatics. But what happens when the charge density isn't uniform? What if it varies with the radial distance? This is where the true elegance of calculus and Gauss's Law comes into play.

### Analyzing the Setup
Imagine you are standing at the center of a solid sphere, Region A, which has a radius of $r_A = 1$. As you move outward, the charge density increases linearly according to the relation $ho_A = kr$.

Surrounding this inner sphere is a spherical shell, Region B, extending from $r = 1$ to an outer radius $r_B$. In this shell, the charge density behaves differently; it decreases with distance according to $ho_B = \frac{2k}{r}$.

Our mission is to determine the electric field and potential just outside this entire configuration. To do this, we must first find the total enclosed charge.

### The Master Equation: Integrating for Charge
Because the charge density varies, we cannot simply multiply density by volume. We must integrate the charge density over the volume of the spheres. The volume element for a thin spherical shell of radius $r$ and thickness $dr$ is $dV = 4\pi r^2 dr$.

The total charge is the sum of the charges in Region A and Region B:
$$q_{net} = \int_0^1 ho_A 4\pi r^2 dr + \int_1^{r_B} ho_B 4\pi r^2 dr$$

### Calculating Charge in Region A
Let's tackle the inner sphere first. Substituting $ho_A = kr$ into our integral:
$$q_A = \int_0^1 (kr) 4\pi r^2 dr = 4\pi k \int_0^1 r^3 dr$$

Integrating $r^3$ gives $\frac{r^4}{4}$. Evaluating this from $0$ to $1$:
$$q_A = 4\pi k \left[ \frac{1}{4} - 0 ight] = \pi k$$

### Calculating Charge in Region B
Now for the outer shell. Substituting $ho_B = \frac{2k}{r}$:
$$q_B = \int_1^{r_B} \left(\frac{2k}{r}ight) 4\pi r^2 dr = 8\pi k \int_1^{r_B} r dr$$

Notice how the $r$ in the denominator beautifully cancels one $r$ from the volume element! Integrating $r$ gives $\frac{r^2}{2}$. Evaluating from $1$ to $r_B$:
$$q_B = 8\pi k \left[ \frac{r_B^2}{2} - \frac{1}{2} ight] = 4\pi k (r_B^2 - 1)$$

### The Net Charge
Adding the charges from both regions gives us the total enclosed charge:
$$q_{net} = \pi k + 4\pi k (r_B^2 - 1) = \pi k (4r_B^2 - 3)$$

This is our master expression. With $q_{net}$ in hand, we can evaluate any scenario presented in the options.

### Evaluating the Options
**Option A:** For the electric field to be zero outside, the net charge must be zero. 
$$4r_B^2 - 3 = 0 \implies r_B = \frac{\sqrt{3}}{2}$$
However, $r_B$ must be greater than $1$ (since it's the outer radius of the shell). Moreover, Option A suggests $r_B = \sqrt{\frac{3}{2}}$, which is mathematically different. Thus, Option A is incorrect.

**Option B:** The electric potential just outside the shell is given by $V = \frac{1}{4\pi\epsilon_0} \frac{q_{net}}{r_B}$. Substituting our expression for $q_{net}$:
$$V = \frac{1}{4\pi\epsilon_0} \frac{\pi k (4r_B^2 - 3)}{r_B} = \frac{k}{4\epsilon_0} \left( 4r_B - \frac{3}{r_B} ight)$$
If $r_B = \frac{3}{2}$, we get:
$$V = \frac{k}{4\epsilon_0} \left( 4\left(\frac{3}{2}ight) - \frac{3}{3/2} ight) = \frac{k}{4\epsilon_0} (6 - 2) = \frac{k}{\epsilon_0}$$
This perfectly matches Option B!

**Options C and D:** For completeness, if $r_B = 2$, $q_{net} = 13\pi k 
eq 15\pi k$. If $r_B = \frac{5}{2}$, the electric field $E = \frac{1}{4\pi\epsilon_0} \frac{q_{net}}{r_B^2}$ evaluates to $\frac{22k}{25\epsilon_0}$, which does not match Option D.

### Conclusion
By systematically applying the principles of volume integration and Gauss's Law, we successfully navigated through a complex, non-uniform charge distribution. The key takeaway is to always trust the math—set up your integrals carefully, respect the boundaries of your regions, and let the equations guide you to the truth!

Select Answer:

Visualized Solution

\text{Analyzing the Charge Distribution}

\text{Total Charge Formula}

\text{Charge in Region A}

\text{Charge in Region B}

\text{Net Charge of the Configuration}

\text{Evaluating Option A}

\text{Evaluating Option B}

\text{Evaluating Options C and D}

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

Analyzing the Setup Imagine you are standing at the center of a solid sphere, Region A, which has a radius of $r_{A} = 1$

The Master Equation

Calculating Charge in Region A Let's tackle the inner sphere first

Calculating Charge in Region B Now for the outer shell

The Net Charge

Evaluating the Options Option A: For the electric field to be zero outside, the net charge must be zero.
$4 r_{B}^{2} - 3 = 0 ⟹ r_{B} = \frac{3}{2}$
However, $r_{B}$ must be greater than $1$ (since it's the outer radius of the shell)

Conclusion By systematically applying the principles of volume integration and Gauss's Law, we successfully navigated through a complex, non-uniform charge distribution

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