There are two charged metallic spheres `S_(1) and S_(2)` of radii `R_(1) and R_(2)` respectively. The electric fields `E_(1)` and `E_(1)` on their surfaces are such that `E_(1)//E_(2) = R_(1)/R_(2)`. Then the ratio `V_(1)//V_(2)` of their electrostatic potentials on each sphere is

Options

A. `((R_(1))/(R_(2)))^(3)`
B. `(R_(1))/(R_(2))`
C. `((R_(2))/(R_(1)))`
D. `((R_(1))/(R_(2)))^(2)`

Explanation

To solve the problem, we need to find the ratio of the electrostatic potentials \( V_1 \) and \( V_2 \) of two charged metallic spheres \( S_1 \) and \( S_2 \) with radii \( R_1 \) and \( R_2 \) respectively. We are given that the ratio of the electric fields on their surfaces is:

\[ \frac{E_1}{E_2} = \frac{R_1}{R_2} \]

### Step 1: Understand the relationship between electric field and potential
The electric field \( E \) at the surface of a charged sphere is related to the potential \( V \) by the formula:

\[ E = \frac{V}{R} \]

where \( R \) is the radius of the sphere. Therefore, we can express the potentials \( V_1 \) and \( V_2 \) in terms of the electric fields and the radii:

\[ V_1 = E_1 \cdot R_1 \]
\[ V_2 = E_2 \cdot R_2 \]

### Step 2: Find the ratio of the potentials
Now, we can find the ratio of the potentials \( \frac{V_1}{V_2} \):

\[ \frac{V_1}{V_2} = \frac{E_1 \cdot R_1}{E_2 \cdot R_2} \]

### Step 3: Substitute the given ratio of electric fields
From the problem, we know that:

\[ \frac{E_1}{E_2} = \frac{R_1}{R_2} \]

Substituting this into our expression for the ratio of potentials gives:

\[ \frac{V_1}{V_2} = \frac{R_1}{R_2} \cdot \frac{R_1}{R_2} \]

### Step 4: Simplify the expression
This simplifies to:

\[ \frac{V_1}{V_2} = \left( \frac{R_1}{R_2} \right)^2 \]

### Conclusion
Thus, the ratio of the electrostatic potentials on each sphere is:

\[ \frac{V_1}{V_2} = \left( \frac{R_1}{R_2} \right)^2 \]

### Final Answer
The final answer is:

\[ \frac{V_1}{V_2} = \left( \frac{R_1}{R_2} \right)^2 \]

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