Consider two charged metallic spheres `S_(1) and S_(2)` of radii 3R and R, respectively. The electric potential `V_(1) (on S_(1) ) and (on S_(2))` on their surfaces are such that `(V_(1))/(V_(2)) = (4)/(1)`. Then the ratio `E_(1)(on S_(1)) / E_(2) (on S_(2))` of the electric fields on their surfaces is:

Options

A. (a)`((5)/(4))`
B. (b)`((4)/(3))`
C. (c)`((4)/(5))`
D. (d)`((3)/(4))`

Explanation

To solve the problem, we need to find the ratio of the electric fields on the surfaces of two charged metallic spheres \( S_1 \) and \( S_2 \) given their electric potentials.

### Step-by-Step Solution:

1. **Understanding the Electric Potential**:
The electric potential \( V \) on 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. **Applying the Formula for Both Spheres**:
For sphere \( S_1 \) with radius \( 3R \):
\[ V_1 = \frac{kQ_1}{3R} \]
For sphere \( S_2 \) with radius \( R \):
\[ V_2 = \frac{kQ_2}{R} \]

3. **Setting Up the Ratio of Potentials**:
We are given that:
\[ \frac{V_1}{V_2} = \frac{4}{1} \]
Substituting the expressions for \( V_1 \) and \( V_2 \):
\[ \frac{\frac{kQ_1}{3R}}{\frac{kQ_2}{R}} = \frac{4}{1} \]

4. **Simplifying the Ratio**:
The \( k \) and \( R \) terms cancel out:
\[ \frac{Q_1}{3Q_2} = 4 \]
This implies:
\[ Q_1 = 12Q_2 \]

5. **Finding the Electric Fields**:
The electric field \( E \) on the surface of a charged sphere is given by:
\[ E = \frac{kQ}{R^2} \]
For sphere \( S_1 \):
\[ E_1 = \frac{kQ_1}{(3R)^2} = \frac{kQ_1}{9R^2} \]
For sphere \( S_2 \):
\[ E_2 = \frac{kQ_2}{R^2} \]

6. **Setting Up the Ratio of Electric Fields**:
Now we find the ratio of the electric fields:
\[ \frac{E_1}{E_2} = \frac{\frac{kQ_1}{9R^2}}{\frac{kQ_2}{R^2}} = \frac{Q_1}{9Q_2} \]
The \( k \) and \( R^2 \) terms cancel out.

7. **Substituting the Charge Ratio**:
We already found that \( Q_1 = 12Q_2 \):
\[ \frac{E_1}{E_2} = \frac{12Q_2}{9Q_2} = \frac{12}{9} = \frac{4}{3} \]

### Final Answer:
The ratio of the electric fields on the surfaces of spheres \( S_1 \) and \( S_2 \) is:
\[ \frac{E_1}{E_2} = \frac{4}{3} \]