The answer to each of the following questions is a single-digit integer ranging from 0 to 9. Darken the correct digit.
There are two metallic spheres A and B of radii R and 3R respectively. Both the spheres are given charge so that they may attain the same potential. If `E_(A)` is the electric field on the surface of sphere A and `E_(B)` is the electric field intensity on the surface of sphere B then calculate `E_(A)//E_(B)`.

Explanation

To solve the problem, we need to calculate the ratio of the electric fields on the surfaces of two metallic spheres A and B, given that they are at the same electric potential. Here are the steps to arrive at the solution:

### Step 1: Understand the potential of metallic spheres
The electric potential \( V \) of a metallic sphere of radius \( R \) carrying charge \( Q \) is given by the formula:
\[ V = \frac{KQ}{R} \]
where \( K \) is the electrostatic constant, which can also be expressed as \( \frac{1}{4\pi\epsilon_0} \).

### Step 2: Write the potential equations for both spheres
For sphere A (radius \( R \) and charge \( Q_A \)):
\[ V_A = \frac{KQ_A}{R} \]

For sphere B (radius \( 3R \) and charge \( Q_B \)):
\[ V_B = \frac{KQ_B}{3R} \]

### Step 3: Set the potentials equal to each other
Since both spheres are at the same potential, we can set \( V_A \) equal to \( V_B \):
\[ \frac{KQ_A}{R} = \frac{KQ_B}{3R} \]

### Step 4: Simplify the equation
Cancel \( K \) and \( R \) from both sides:
\[ Q_A = \frac{Q_B}{3} \]

### Step 5: Write the electric field equations for both spheres
The electric field \( E \) at the surface of a metallic sphere is given by:
\[ E = \frac{KQ}{R^2} \]

For sphere A:
\[ E_A = \frac{KQ_A}{R^2} \]

For sphere B:
\[ E_B = \frac{KQ_B}{(3R)^2} = \frac{KQ_B}{9R^2} \]

### Step 6: Find the ratio \( \frac{E_A}{E_B} \)
Now, we can find the ratio of the electric fields:
\[ \frac{E_A}{E_B} = \frac{\frac{KQ_A}{R^2}}{\frac{KQ_B}{9R^2}} = \frac{Q_A}{Q_B} \cdot 9 \]

### Step 7: Substitute \( Q_A \) in terms of \( Q_B \)
From Step 4, we know \( Q_A = \frac{Q_B}{3} \):
\[ \frac{E_A}{E_B} = \frac{\frac{Q_B}{3}}{Q_B} \cdot 9 = \frac{1}{3} \cdot 9 = 3 \]

### Final Answer
Thus, the ratio of the electric fields \( \frac{E_A}{E_B} \) is:
\[ \frac{E_A}{E_B} = 3 \]