Equal charges are given to two spheres of different radii. The potential will

Options

A. be more one the smaller sphere
B. be more on the bigger sphere
C. be equal on both the spheres
D. depend on the nature of the materials of the spheres

Answer

To solve the problem, we need to analyze the relationship between charge, potential, and radius for two spheres with equal charges but different radii.

### Step-by-Step Solution:

1. **Understand the Formula for Electric Potential**:
The electric potential \( V \) at the surface of a charged sphere is given by the formula:
\[ V = \frac{KQ}{R} \]
where \( K \) is Coulomb's constant, \( Q \) is the charge on the sphere, and \( R \) is the radius of the sphere.

2. **Identify the Given Information**:
We have two spheres with equal charges \( Q \) but different radii \( R_1 \) and \( R_2 \). Let's assume \( R_1 < R_2 \) (i.e., Sphere 1 is smaller than Sphere 2).

3. **Calculate the Potential for Each Sphere**:
- For Sphere 1 (smaller radius):
\[ V_1 = \frac{KQ}{R_1} \]
- For Sphere 2 (larger radius):
\[ V_2 = \frac{KQ}{R_2} \]

4. **Compare the Potentials**:
Since \( R_1 < R_2 \), we can see that:
\[ V_1 = \frac{KQ}{R_1} > V_2 = \frac{KQ}{R_2} \]
This indicates that the potential \( V_1 \) of the smaller sphere is greater than the potential \( V_2 \) of the larger sphere.

5. **Conclusion**:
Therefore, when equal charges are given to two spheres of different radii, the potential will be greater on the smaller sphere and less on the larger sphere.

### Final Answer:
The potential will be greater on the smaller sphere and less on the larger sphere.

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