Two hollow conducting spheres of radii $R_1$ and $R_2$ ($R_1 \gg R_2$) are concentric and have equal charges.

Question

Two hollow conducting spheres of radii $R_1$ and $R_2$ ($R_1 \gg R_2$) are concentric and have equal charges. The potential would be:

Accepted Answer

1. **Setup:** Consider two concentric conducting spherical shells. Let the radius of the outer sphere be $R_1$ and the inner sphere be $R_2$ (given $R_1 \gg R_2$). Both have equal positive charge $Q$.
2. **Potential of Inner Sphere (Small):** The potential at the surface of the inner sphere ($V_{small}$) is the sum of the potential due to its own charge and the potential due to the outer sphere surrounding it. Inside a charged shell, potential is constant and equal to the surface value.
   $V_{small} = V_{self} + V_{due\_to\_outer} = \frac{kQ}{R_2} + \frac{kQ}{R_1}$.
3. **Potential of Outer Sphere (Big):** The potential at the surface of the outer sphere ($V_{big}$) is the sum of the potential due to its own charge and the potential due to the inner sphere (which acts as a point charge at the center for points outside).
   $V_{big} = V_{self} + V_{due\_to\_inner} = \frac{kQ}{R_1} + \frac{kQ}{R_1} = \frac{2kQ}{R_1}$.
4. **Comparison:** 
   Comparing terms: The term $\frac{kQ}{R_1}$ is common to both. The difference lies in the first term: $\frac{kQ}{R_2}$ vs $\frac{kQ}{R_1}$.
   Since $R_2 < R_1$, it follows that $\frac{1}{R_2} > \frac{1}{R_1}$.
   Therefore, $V_{small} > V_{big}$.
5. **Conclusion:** The potential is higher on the smaller sphere. (This concept is analogous to gravitational potential discussed in Class 11 Physics, Chapter 8, where potential inside a shell is constant).

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