1. Electric field due to a uniformly charged solid sphere

For a uniformly charged non-conducting solid sphere of radius $a$ and total charge $Q_{\text{tot}}$:

• Outside the sphere ($r \ge a$): $E=\frac{1}{4\pi\varepsilon_0}\frac{Q_{\text{tot}}}{r^2}$
• Inside the sphere ($r<a$): $E=\frac{1}{4\pi\varepsilon_0}\frac{Q_{\text{tot}}\,r}{a^3}$

We are to compare the fields at a point $P$ which is at distance $R$ from the center in each case.

2. Sphere 1

• Radius $a_1=\dfrac{R}{2}$
• Total charge $Q_1=Q$
• Observation point: $r=R$

Since $R>a_1$, point $P$ is outside sphere 1.

So, $E_1=\frac{1}{4\pi\varepsilon_0}\frac{Q}{R^2}$

3. Sphere 2

• Radius $a_2=R$
• Total charge $Q_2=2Q$
• Observation point: $r=R$

Here point $P$ is on the surface. Using either inside or outside formula gives same result: $E_2=\frac{1}{4\pi\varepsilon_0}\frac{2Q}{R^2}$

Thus, $E_2=2\left(\frac{1}{4\pi\varepsilon_0}\frac{Q}{R^2}\right)$

So clearly, $E_2>E_1$

4. Sphere 3

• Radius $a_3=2R$
• Total charge $Q_3=4Q$
• Observation point: $r=R$

Now $R<2R$, so point $P$ is inside sphere 3.

Using inside-field formula: $E_3=\frac{1}{4\pi\varepsilon_0}\frac{Q_3\,r}{a_3^3}$

Substitute values: $E_3=\frac{1}{4\pi\varepsilon_0}\frac{4Q\cdot R}{(2R)^3}$

$E_3=\frac{1}{4\pi\varepsilon_0}\frac{4Q\cdot R}{8R^3}$

$E_3=\frac{1}{4\pi\varepsilon_0}\frac{Q}{2R^2}$

So, $E_3=\frac{1}{2}\left(\frac{1}{4\pi\varepsilon_0}\frac{Q}{R^2}\right)=\frac{E_1}{2}$

Hence, $E_1>E_3$

5. Final comparison

We have: $E_1=\frac{1}{4\pi\varepsilon_0}\frac{Q}{R^2}$ $E_2=\frac{1}{4\pi\varepsilon_0}\frac{2Q}{R^2}$ $E_3=\frac{1}{4\pi\varepsilon_0}\frac{Q}{2R^2}$

Therefore, $E_2>E_1>E_3$

So the correct option is:

$\boxed{\text{C: } E_2>E_1>E_3}$

6. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

They match.