A thin spherical shell of radius R has charge Q spread uniformly over its surface. Which of the foll

Q: Why is the electric field exactly zero everywhere inside the shell? Shouldn't the charges on the surface exert some force?

They do exert forces! However, because the charge is distributed uniformly over a perfect sphere, the vector sum of the electric forces from all the tiny charge elements on the shell perfectly cancels out at any point inside. This is a direct consequence of the inverse-square nature of Coulomb's Law.

Q: Is the electric field continuous at the surface of the shell ($r = R$)?

No, it is discontinuous. Just inside the surface, the field is zero. Just outside, it abruptly jumps to its maximum value of $\frac{Q}{4\pi\varepsilon_0 R^2}$. This discontinuity is a characteristic feature whenever you cross a surface with a finite surface charge density.

Q: What if the sphere was solid and non-conducting, with charge distributed throughout its volume?

In that case, the field inside would not be zero. As you move outward from the center, a Gaussian surface encloses more and more charge. For a uniformly charged solid sphere, the internal electric field increases linearly with distance ($E \propto r$) until it reaches the surface.

Q: Why does the shell behave like a point charge when viewed from the outside?

This is known as the Shell Theorem, first proven by Isaac Newton for gravity. Because the electric force follows the same inverse-square law, the mathematical integration of forces from a uniform spherical shell to an outside point simplifies exactly to the formula of a point charge at the center.

Q: What is the exact value of the electric field exactly on the surface ($r = R$)?

Mathematically, the field is undefined exactly at the boundary due to the discontinuity. However, in physics, we often take the average of the field just inside and just outside, which gives $E = \frac{Q}{8\pi\varepsilon_0 R^2}$. But for practical purposes like this graph, we show the limit approaching from the outside.

Question

A thin spherical shell of radius $R$ has charge $Q$ spread uniformly over its surface. Which of the following graphs most closely represents the electric field $E(r)$ produced by the shell in the range $0 \le r < \infty$, where $r$ is the distance from the centre of the shell?

Sigma Percentile · Accepted Answer

### The Magic of Gauss's Law

Imagine you are standing inside a giant, perfectly spherical hollow room. The walls of this room are charged uniformly with electricity. What kind of electric forces would you feel floating inside? This classic problem is a beautiful demonstration of **Gauss's Law** and the elegance of spherical symmetry.

We are given a thin spherical shell of radius $R$ carrying a total charge $Q$. Our goal is to map out the electric field $E(r)$ at any distance $r$ from the center, both inside and outside the shell.

### Inside the Shell: The Zero-Gravity Zone

Let's start our journey inside the shell, where $r < R$. To find the electric field, we construct an imaginary spherical boundary—a **Gaussian surface**—with radius $r$ concentric with the shell.

Gauss's Law tells us that the total electric flux through this surface is proportional to the charge enclosed within it:

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

Because all the charge $Q$ is painted on the *outer* surface of the shell, our internal Gaussian sphere encloses absolutely nothing. Therefore, $q_{	ext{enc}} = 0$.

Since the flux is zero and the setup is perfectly symmetrical, the electric field must be zero everywhere inside:

$$E = 0 \quad 	ext{for} \quad r < R$$

This means if you place a test charge anywhere inside this hollow shell, it will experience no net electric force. The pulls and pushes from all the charges on the walls perfectly cancel each other out!

### Outside the Shell: The Point Charge Illusion

Now, let's step outside the shell, where $r \ge R$. We draw a new, larger Gaussian sphere that completely swallows the charged shell.

This time, the enclosed charge is the total charge of the shell, so $q_{	ext{enc}} = Q$. By symmetry, the electric field $\vec{E}$ points radially outward and has a constant magnitude over our Gaussian surface. The flux integral simplifies to the field multiplied by the surface area of the sphere:

$$E \cdot (4\pi r^2) = \frac{Q}{\varepsilon_0}$$

Solving for $E$, we get:

$$E = \frac{Q}{4\pi\varepsilon_0 r^2} \quad 	ext{for} \quad r \ge R$$

Notice something familiar? This is the exact same formula for the electric field of a **point charge**! From the outside, the shell's geometry doesn't matter. It acts as if all its charge is concentrated right at the center. The field drops off rapidly following the inverse-square law ($E \propto 1/r^2$).

### Constructing the Graph

To visualize this, we plot $E(r)$ against $r$:
1. From $r = 0$ to $r = R$, the field is dead zero. The graph is a flat line on the horizontal axis.
2. Exactly at $r = R$, the field suddenly jumps to its maximum value, $E_{	ext{max}} = \frac{Q}{4\pi\varepsilon_0 R^2}$.
3. For $r > R$, the field decays as a curve proportional to $1/r^2$.

Looking at the given options, **Option (a)** perfectly captures this behavior: a flat zero line inside, a sharp vertical jump at the surface, and an inverse-square decay outside.

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Visualized Solution

$Physical Setup$

$Gauss’s Law$

$Inside the Shell (r < R)$

$Electric Field Inside$

$Outside the Shell (r \geq R)$

$Electric Field Outside$

$Graphical Representation$

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

The Magic of Gauss's Law

Inside the Shell

Outside the Shell

Constructing the Graph

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Physical Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Gauss’s Law

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Inside the Shell (r<R)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Electric Field Inside

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Outside the Shell (r≥R)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Electric Field Outside

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Graphical Representation