Comparison of Electric Field Magnitudes for Dielectric Spheres

To find the magnitudes of the electric fields $E_1$, $E_2$, and $E_3$ at a point $P$ located at a distance $r = R$ from the centers of the three uniformly charged dielectric solid spheres, we apply Gauss's Law to each sphere.

1. For Sphere 1:

• Total charge: $Q_1 = Q$
• Radius: $R_1 = \frac{R}{2}$
• Distance of point $P$ from the center: $r = R$

Since $r > R_1$, point $P$ lies outside the sphere. By Gauss's Law, the electric field outside a spherically symmetric charge distribution acts as if the entire charge were concentrated at the center: $E_1 = \frac{1}{4\pi\varepsilon_0} \frac{Q_1}{r^2} = \frac{1}{4\pi\varepsilon_0} \frac{Q}{R^2}$

2. For Sphere 2:

• Total charge: $Q_2 = 2Q$
• Radius: $R_2 = R$
• Distance of point $P$ from the center: $r = R$

Since $r = R_2$, point $P$ is on the surface of the sphere. The electric field is: $E_2 = \frac{1}{4\pi\varepsilon_0} \frac{Q_2}{R^2} = \frac{1}{4\pi\varepsilon_0} \frac{2Q}{R^2} = 2 \left(\frac{1}{4\pi\varepsilon_0} \frac{Q}{R^2}\right)$

3. For Sphere 3:

• Total charge: $Q_3 = 4Q$
• Radius: $R_3 = 2R$
• Distance of point $P$ from the center: $r = R$

Since $r < R_3$, point $P$ lies inside the sphere. For a uniformly charged solid sphere with charge density $\rho = \frac{Q_3}{\frac{4}{3}\pi R_3^3}$, the electric field at an interior distance $r$ is given by: $E_3 = \frac{1}{4\pi\varepsilon_0} \frac{Q_{\text{enclosed}}}{r^2} = \frac{1}{4\pi\varepsilon_0} \frac{Q_3 \, r}{R_3^3}$

Substituting the given values $Q_3 = 4Q$, $r = R$, and $R_3 = 2R$: $E_3 = \frac{1}{4\pi\varepsilon_0} \frac{(4Q)(R)}{(2R)^3} = \frac{1}{4\pi\varepsilon_0} \frac{4QR}{8R^3} = \frac{1}{2} \left(\frac{1}{4\pi\varepsilon_0} \frac{Q}{R^2}\right)$

Comparison: Let $E_0 = \frac{1}{4\pi\varepsilon_0}\frac{Q}{R^2}$. Then, $E_1 = E_0$ $E_2 = 2E_0$ $E_3 = 0.5E_0$

Thus, comparing the magnitudes, we have: $E_2 > E_1 > E_3$

Therefore, the correct option is (C).