A non-conducting solid sphere of radius R is uniformly charged. The magnitude of the electric field

Q: Why is the electric field zero at the center?

At the center, the enclosed charge is zero. Also, by symmetry, the pull from all directions cancels out perfectly.

Q: Is the electric field continuous at the surface?

Yes! If you calculate the limit from inside and outside as $r \to R$, both give the exact same value: $\frac{1}{4\pi\varepsilon_0}\frac{Q}{R^2}$.

Q: Why does the field increase inside but decrease outside?

Inside, as you move out, the amount of charge enclosed grows faster ($r^3$) than the surface area of your Gaussian sphere ($r^2$), leading to a net linear increase. Outside, the enclosed charge is constant, but the surface area keeps growing, diluting the field.

Q: What if the sphere was a conductor?

Great question! If it were a solid conducting sphere, all the charge would reside on the surface. The electric field inside would be exactly zero everywhere.

Q: Does this apply to gravity as well?

Absolutely! The gravitational field inside a uniform solid planet increases linearly with distance from the center, and decreases as $1/r^2$ outside.

Question

A non-conducting solid sphere of radius $R$ is uniformly charged. The magnitude of the electric field due to the sphere at a distance $r$ from its centre

Options:

(A) increases as $r$ increases for $r < R$

(B) decreases as $r$ increases for $0 < r < \infty$

(C) decreases as $r$ increases for $R < r < \infty$

(D) is discontinuous at $r = R$

Sigma Percentile · Accepted Answer

### The Magic of Gauss's Law

Imagine you are holding a solid, non-conducting sphere in your hands. Unlike a metal sphere where all the charge rushes to the surface, this sphere has charge distributed uniformly throughout its entire volume. It is like a perfectly baked chocolate chip cookie, where the chocolate chips (the charges) are spread evenly everywhere inside.

Our mission is to understand how the electric field behaves as we start from the very center of this sphere and travel outwards, eventually leaving the sphere and moving into the empty space beyond.

To embark on this journey, we need a powerful tool. That tool is **Gauss's Law**.

Gauss's Law is a beautiful mathematical statement that connects the electric flux passing through an imaginary closed surface (we call it a Gaussian surface) to the total charge enclosed within that surface.

Mathematically, it is written as:
$$ \oint \mathbf{E} \cdot d\mathbf{A} = \frac{q_{	ext{enclosed}}}{\varepsilon_0} $$

Let's break our journey into two distinct phases: traveling inside the sphere, and traveling outside the sphere.

### Phase 1: Journey Inside the Sphere ($r < R$)

We start at the center and move to a distance $r$ such that we are still inside the sphere. So, $r < R$.

To find the electric field at this point, we imagine a spherical Gaussian surface of radius $r$ concentric with our charged sphere.

The first question we must ask is: How much charge is trapped inside our imaginary Gaussian surface?

Since the charge is uniformly distributed, the charge density $ho$ (charge per unit volume) is constant. 
$$ ho = \frac{Q}{\frac{4}{3}\pi R^3} $$

The charge enclosed, $q_{	ext{enclosed}}$, is simply the volume of our Gaussian surface multiplied by this charge density.
$$ q_{	ext{enclosed}} = ho 	imes \left(\frac{4}{3}\pi r^3ight) $$

Substituting the value of $ho$, we get:
$$ q_{	ext{enclosed}} = \left( \frac{Q}{\frac{4}{3}\pi R^3} ight) 	imes \left(\frac{4}{3}\pi r^3ight) = Q \frac{r^3}{R^3} $$

Now, we apply Gauss's Law. By symmetry, the electric field $\mathbf{E}$ is radial and has the same magnitude everywhere on our Gaussian surface. The area of this surface is $4\pi r^2$.
$$ E 	imes (4\pi r^2) = \frac{Q r^3}{\varepsilon_0 R^3} $$

Solving for $E$, we find:
$$ E = \frac{1}{4\pi\varepsilon_0} \frac{Q}{R^3} r $$

Look closely at this beautiful result! All the terms in the fraction are constants. This means that inside the sphere, the electric field is **directly proportional** to the distance from the center. 
$$ E \propto r $$

At the exact center ($r = 0$), the electric field is zero. As you move outwards, the field increases linearly.

### Phase 2: Journey Outside the Sphere ($r \ge R$)

Now, let's step outside the sphere. We are at a distance $r$ such that $r \ge R$.

Again, we draw a spherical Gaussian surface of radius $r$.

How much charge is enclosed this time? Since our Gaussian surface is larger than the charged sphere, it encloses the **entire** charge $Q$. 
$$ q_{	ext{enclosed}} = Q $$

Applying Gauss's Law once more:
$$ E 	imes (4\pi r^2) = \frac{Q}{\varepsilon_0} $$

Solving for $E$, we get:
$$ E = \frac{1}{4\pi\varepsilon_0} \frac{Q}{r^2} $$

This is a very familiar equation! It is the exact same formula for the electric field of a point charge.

This tells us a profound secret of nature: for any point outside a uniformly charged spherical distribution, the sphere behaves exactly as if all its charge were concentrated at its very center.

Here, the electric field is **inversely proportional** to the square of the distance.
$$ E \propto \frac{1}{r^2} $$

As you move further away from the sphere, the electric field weakens rapidly.

### The Grand Finale: Connecting the Dots

Let's summarize our findings.

Inside the sphere ($r < R$), the electric field increases linearly with distance. 
Outside the sphere ($r > R$), the electric field decreases quadratically with distance.

But what happens exactly at the surface, where $r = R$?

If we plug $r = R$ into our "inside" formula, we get:
$$ E = \frac{1}{4\pi\varepsilon_0} \frac{Q}{R^2} $$

If we plug $r = R$ into our "outside" formula, we get the exact same thing:
$$ E = \frac{1}{4\pi\varepsilon_0} \frac{Q}{R^2} $$

This proves that the electric field is perfectly **continuous** at the surface. There are no sudden jumps or breaks. The field grows steadily from zero at the center, reaches its maximum value at the surface, and then fades away into the distance.

Returning to our original question, we can confidently evaluate the given options. 
Option (a) states that the field increases as $r$ increases for $r < R$. This is absolutely correct. 
Option (c) states that the field decreases as $r$ increases for $R < r < \infty$. This is also correct.

Therefore, the correct choices are (a) and (c).

Animated Solution for Physics - Electrostatics: A non-conducting solid sphere of radius $R$ is uniformly charged. The magnitude of the electric field due to the sphere at a distance $r$ from its centre

Select Answer:

Visualized Solution

Analyzing the Problem

Electric Field Inside the Sphere ( $r < R$ )

Charge Enclosed ( $r < R$ )

Calculating $E$ Inside

Electric Field Outside the Sphere ( $r \geq R$ )

Analyzing the Variation

Conclusion

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

The Magic of Gauss's Law

Phase 1

Phase 2

The Grand Finale

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