The region between two concentric spheres of radii a and b, respectively (see the figure), has volum

Q: Why do we only integrate the volume charge from $a$ to $r$?

Gauss's Law only cares about the charge *enclosed* within the Gaussian surface. The charge outside radius $r$ does not contribute to the net electric flux through that surface.

Q: Why is the area of the Gaussian surface $4\pi r^2$ and not the volume?

Gauss's Law relates the electric flux through a *closed surface* to the enclosed charge. Flux is the dot product of the electric field and the area vector, so we use the surface area of the sphere.

Q: Can the electric field really be constant everywhere between the spheres?

Yes! By carefully tailoring the volume charge density, the additional charge enclosed as you move outward perfectly compensates for the $1/r^2$ drop-off of the electric field, keeping it constant.

Q: What happens to the electric field for $r < a$?

For $r < a$, the only enclosed charge is the point charge $Q$. So, the electric field behaves exactly like a standard point charge field: $E = \frac{Q}{4\pi \varepsilon_0 r^2}$.

Q: Why did we set the numerator of the $1/r^2$ term to zero?

If the coefficient of $1/r^2$ is not zero, the electric field will change as $r$ changes. To make $E$ independent of $r$ (i.e., constant), that entire variable term must vanish.

Question

The region between two concentric spheres of radii $a$ and $b$, respectively (see the figure), has volume charge density $\rho = \frac{A}{r}$, where $A$ is a constant and $r$ is the distance from the centre. At the centre of the spheres is a point charge $Q$. The value of $A$ such that the electric field in the region between the spheres will be constant is

Options:

(A) $\frac{Q}{2\pi a^2}$

(B) $\frac{Q}{2\pi(b^2-a^2)}$

(C) $\frac{2Q}{\pi(a^2-b^2)}$

(D) $\frac{2Q}{\pi a^2}$

Sigma Percentile · Accepted Answer

### The Quest for a Constant Electric Field

Imagine a point charge sitting alone in space. Its electric field drops off rapidly, following the famous inverse-square law. But what if we wanted to create a region where the electric field doesn't fade away? What if we wanted it to remain perfectly constant as we move outward?

This problem challenges us to do exactly that. By surrounding a central point charge $Q$ with a carefully tailored cloud of charge (a volume charge density), we can perfectly compensate for the point charge's fading field. Let's dive into the mathematics of how this beautiful balance is achieved.

### Setting the Stage with Gauss's Law

To find the electric field in the region between the two concentric spheres (where $a < r < b$), we rely on our most powerful tool for symmetric charge distributions: **Gauss's Law**.

Gauss's Law states that the total electric flux through a closed surface is proportional to the total charge enclosed within it:

$$\oint \vec{E} \cdot d\vec{A} = \frac{q_{enc}}{\varepsilon_0}$$

We construct an imaginary spherical "Gaussian surface" of radius $r$ centered on the point charge. Because of the spherical symmetry, the electric field $\vec{E}$ is uniform in magnitude over this surface and points radially outward. Thus, the flux integral simplifies beautifully to $E(4\pi r^2)$.

### Calculating the Enclosed Charge

The real challenge lies in determining $q_{enc}$. The total enclosed charge consists of two parts: the central point charge $Q$, and the continuous volume charge trapped between the inner radius $a$ and our Gaussian surface at radius $r$.

To find the volume charge, we must integrate the given volume charge density $\rho = \frac{A}{x}$ over the spherical volume. We use a thin spherical shell of radius $x$ and thickness $dx$ as our differential volume element, $dV = 4\pi x^2 dx$.

$$q_{enc} = Q + \int_{a}^{r} \rho dV$$

Substituting our density and volume element:

$$q_{enc} = Q + \int_{a}^{r} \left(\frac{A}{x}\right) (4\pi x^2) dx$$

Notice how the $x$ in the denominator cancels one power of $x$ in the volume element. This leaves us with a very simple integral:

$$q_{enc} = Q + 4\pi A \int_{a}^{r} x dx$$

Evaluating the integral yields:

$$q_{enc} = Q + 4\pi A \left[ \frac{x^2}{2} \right]_{a}^{r} = Q + 2\pi A (r^2 - a^2)$$

### The Master Equation

Now, we bring our enclosed charge back to Gauss's Law to find the electric field $E$:

$$E (4\pi r^2) = \frac{Q + 2\pi A r^2 - 2\pi A a^2}{\varepsilon_0}$$

Dividing both sides by the surface area $4\pi r^2$, we can separate the expression into two distinct terms:

$$E = \frac{Q - 2\pi A a^2}{4\pi \varepsilon_0 r^2} + \frac{2\pi A r^2}{4\pi \varepsilon_0 r^2}$$

$$E = \frac{Q - 2\pi A a^2}{4\pi \varepsilon_0 r^2} + \frac{A}{2\varepsilon_0}$$

### The Grand Finale: Enforcing Constancy

Take a close look at our master equation for $E$. It has a constant term $\frac{A}{2\varepsilon_0}$ and a variable term that depends on $\frac{1}{r^2}$.

The problem demands that the electric field be **constant** throughout the region. For a function to be constant with respect to $r$, it must not contain $r$ at all! Therefore, the entire variable term must vanish.

We achieve this by setting the numerator of the variable term to zero:

$$Q - 2\pi A a^2 = 0$$

Solving for our unknown constant $A$, we get:

$$A = \frac{Q}{2\pi a^2}$$

And there we have it! By choosing this specific value for $A$, the volume charge density perfectly counteracts the inverse-square drop of the point charge, resulting in a beautifully uniform electric field.

Select Answer:

Visualized Solution

Visualizing the Setup

Gauss's Law

Enclosed Charge Setup

Evaluating the Integral

Electric Field Expression

Condition for Constant Field

The Way Forward

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

The Quest for a Constant Electric Field

Setting the Stage with Gauss's Law

Calculating the Enclosed Charge

The Master Equation

The Grand Finale

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