A charge Q is distributed over two concentric hollow spheres of radii r and R (R > r) such that their surface densities are equal. The charge on the smaller and bigger shells is:
A. (Q * r²) / (r² + R²) and (Q * R²) / (r² + R²), respectively
B. Q (1 + (r² / R²)) and Q (1 + (R² / r²)), respectively
C. Q (1 - (r² / R²)) and Q (1 - (R² / r²)), respectively
D. (Q * R²) / (r² + R²) and (Q * r²) / (r² + R²), respectively

To solve this problem, we need to understand how the total charge QQ is distributed on two concentric spherical shells with surface charge densities equal.
Step-by-Step Approach:
1. Given:
o Two concentric spherical shells with radii rr and RR where R>rR > r.
o The total charge QQ is distributed over the shells such that the surface charge densities on both shells are equal.
o Let the charge on the smaller shell (radius rr) be Q1Q_1 and on the larger shell (radius RR) be Q2Q_2.
2. Surface charge density (σ\sigma): The surface charge density σ\sigma on a spherical shell is defined as the charge per unit area. For a spherical shell of radius rr, the surface area is 4πr24 \pi r^2, and the charge Q1Q_1 on it gives a surface charge density:
σ=Q14πr2\sigma = \frac{Q_1}{4\pi r^2}
Similarly, for the shell of radius RR, the surface area is 4πR24\pi R^2, and the charge Q2Q_2 on it gives:
σ=Q24πR2\sigma = \frac{Q_2}{4\pi R^2}
3. Equal surface charge densities: Since the surface charge densities are given to be equal, we equate the two expressions for σ\sigma:
Q14πr2=Q24πR2\frac{Q_1}{4\pi r^2} = \frac{Q_2}{4\pi R^2}
This simplifies to:
Q1r2=Q2R2\frac{Q_1}{r^2} = \frac{Q_2}{R^2}
So, we can write the relationship between Q1Q_1 and Q2Q_2 as:
Q1=Q2×r2R2Q_1 = Q_2 \times \frac{r^2}{R^2}
4. Total charge QQ: The total charge QQ is the sum of the charges on the two shells:
Q=Q1+Q2Q = Q_1 + Q_2
Using the relationship Q1=Q2×r2R2Q_1 = Q_2 \times \frac{r^2}{R^2}, substitute this into the total charge equation:
Q=Q2×r2R2+Q2Q = Q_2 \times \frac{r^2}{R^2} + Q_2
Factor out Q2Q_2:
Q=Q2(r2R2+1)Q = Q_2 \left( \frac{r^2}{R^2} + 1 \right)
Simplify:
Q=Q2×(r2+R2R2)Q = Q_2 \times \left( \frac{r^2 + R^2}{R^2} \right)
Solve for Q2Q_2:
Q2=QR2r2+R2Q_2 = \frac{Q R^2}{r^2 + R^2}
5. Charge on the smaller shell Q1Q_1: Using the relationship Q1=Q2×r2R2Q_1 = Q_2 \times \frac{r^2}{R^2}, substitute the value of Q2Q_2:
Q1=QR2r2+R2×r2R2Q_1 = \frac{Q R^2}{r^2 + R^2} \times \frac{r^2}{R^2}
Simplifying:
Q1=Qr2r2+R2Q_1 = \frac{Q r^2}{r^2 + R^2}
Final Answers:
• The charge on the smaller shell (Q1Q_1) is Qr2r2+R2\frac{Q r^2}{r^2 + R^2}.
• The charge on the larger shell (Q2Q_2) is QR2r2+R2\frac{Q R^2}{r^2 + R^2}.
Thus, the correct answer is:
A) Qr2r2+R2 and QR2r2+R2, respectively\boxed{A) \, \frac{Q r^2}{r^2 + R^2} \text{ and } \frac{Q R^2}{r^2 + R^2}, \text{ respectively}}