Two spherical conductors A and B of radii R and 2R respectively , are separated by a large distance . If some charge is given to both the spheres and later they are connected by a conducting wire , then in equilibrium condition , the ratio of the magnitude of the electric fields at the surface of spheres A and B is

Options

A. `1:4`
B. `4:1`
C. `1:2`
D. `2:1`

Answer

To solve the problem, we need to find the ratio of the electric fields at the surfaces of two spherical conductors A and B after they are connected by a conducting wire. The steps are as follows:

### Step 1: Understand the Setup
We have two spherical conductors:
- Sphere A with radius \( R \)
- Sphere B with radius \( 2R \)

They are initially charged with charges \( q_1 \) and \( q_2 \) respectively. When connected by a wire, they will reach the same electric potential.

### Step 2: Write the Expression for Electric Potential
The electric potential \( V \) at the surface of a charged sphere is given by:
\[ V = k \frac{q}{r} \]
where \( k \) is Coulomb's constant, \( q \) is the charge, and \( r \) is the radius of the sphere.

For Sphere A:
\[ V_A = k \frac{q_1}{R} \]
For Sphere B:
\[ V_B = k \frac{q_2}{2R} \]

### Step 3: Set the Potentials Equal
Since the spheres are connected by a wire, their potentials are equal:
\[ V_A = V_B \]
Substituting the expressions for \( V_A \) and \( V_B \):
\[ k \frac{q_1}{R} = k \frac{q_2}{2R} \]
We can cancel \( k \) and \( R \) (as long as \( R \neq 0 \)):
\[ q_1 = \frac{1}{2} q_2 \]

### Step 4: Write the Expression for Electric Field
The electric field \( E \) at the surface of a charged sphere is given by:
\[ E = k \frac{q}{r^2} \]
For Sphere A:
\[ E_A = k \frac{q_1}{R^2} \]
For Sphere B:
\[ E_B = k \frac{q_2}{(2R)^2} = k \frac{q_2}{4R^2} \]

### Step 5: Find the Ratio of Electric Fields
We want to find the ratio \( \frac{E_A}{E_B} \):
\[ \frac{E_A}{E_B} = \frac{k \frac{q_1}{R^2}}{k \frac{q_2}{4R^2}} = \frac{q_1}{q_2} \cdot 4 \]
The \( k \) and \( R^2 \) terms cancel out.

### Step 6: Substitute the Charge Ratio
From Step 3, we found that \( q_1 = \frac{1}{2} q_2 \). Substituting this into the ratio:
\[ \frac{E_A}{E_B} = \frac{\frac{1}{2} q_2}{q_2} \cdot 4 = \frac{1}{2} \cdot 4 = 2 \]

### Final Answer
Thus, the ratio of the magnitudes of the electric fields at the surfaces of spheres A and B is:
\[ \frac{E_A}{E_B} = 2:1 \]