Animated Solution for Physics - Electrostatics: Charges $Q,2Q$ and $4Q$ are uniformly distributed in three dielectric solid spheres 1, 2 and 3 of radii $R/2,R$ and $2R$ respectively, as shown in figure. If magnitudes of the electric fields at point $P$ at a distance $R$ from the centre of spheres 1, 2 and 3 are $E_1,E_2$ and $E_3$ respectively, then

The journey to mastering electrostatics often brings us face-to-face with spherically symmetric charge distributions. This problem is a beautiful exercise in understanding how the electric field behaves both inside and outside a uniformly charged solid dielectric sphere.

Let's break down the physics and conquer this step-by-step!

Before we dive into the specific spheres, we need to arm ourselves with the right mathematical tools. For a solid dielectric sphere with a uniform charge distribution, the electric field $E$ at a distance $r$ from the center depends entirely on whether you are inside or outside the sphere.

If you are outside or on the surface ($r\ge R_{\text{sphere}}$), the sphere behaves exactly like a point charge concentrated at its center. The formula is the familiar inverse-square law:

However, if you are inside the sphere ($r<R_{\text{sphere}}$), the electric field grows linearly with the distance from the center. The formula becomes:

With these two weapons in our arsenal, let's analyze each sphere individually.

Our first sphere has a radius of $R/2$ and carries a total charge of $Q$. We are asked to find the electric field $E_1$ at a point $P$ located at a distance $r=R$ from the center.

Since the distance $R$ is greater than the sphere's radius $R/2$, point $P$ lies strictly outside the sphere. We can confidently use the external field formula:

This gives us our baseline value. Let's keep this in our back pocket.

Moving on to the second sphere, we see it has a radius of exactly $R$ and carries a charge of $2Q$. Point $P$ is again at a distance $r=R$.

This means point $P$ sits perfectly on the surface of the sphere. The external field formula still applies here. Let's substitute the values:

Notice how $E_2$ is exactly double the magnitude of $E_1$!

Now for the grand finale: the third sphere. This is a massive sphere with a radius of $2R$ and a hefty charge of $4Q$. But wait, point $P$ is still only at a distance $r=R$.

Because $R$ is less than $2R$, point $P$ is buried deep inside this third sphere. This is where many students fall into a trap. We cannot use the inverse-square law here; we must use the internal field formula!

Let's carefully substitute our raw values into the internal formula:

Don't forget to cube the entire radius in the denominator! Expanding the denominator gives us $8R^3$:

Simplifying the fraction, we get:

We have successfully calculated the electric field magnitudes for all three scenarios:

1. $E_1=1.0\frac{kQ}{R^2}$ 2. $E_2=2.0\frac{kQ}{R^2}$ 3. $E_3=0.5\frac{kQ}{R^2}$

Comparing these coefficients, it is crystal clear that $E_2$ is the strongest, followed by $E_1$, and $E_3$ is the weakest.

This perfectly matches option (c).

A Quick Thought Experiment: What if the problem had stated these were conducting spheres instead of dielectric? In a conductor, all excess charge resides on the surface, making the internal electric field exactly zero. In that case, $E_3$ would have been $0$! Always read the problem statement carefully. Keep visualizing, keep calculating, and you'll master electrostatics in no time!