Two spheres of radius `a` and `b` respectively are charged and joined by a wire. The ratio of electric field of the spheres is

Options

A. `a//b`
B. `b//a`
C. `a^(2)//b^(2)`
D. `b^(2)//a`

Explanation

To solve the problem of finding the ratio of the electric fields of two charged spheres connected by a wire, we can follow these steps:

### Step 1: Understand the Setup
We have two spheres with radii \( a \) and \( b \) that are charged and connected by a wire. When connected, the potential across both spheres becomes equal.

### Step 2: Electric Field Formula
The electric field \( E \) due to a charged sphere at its surface is given by the formula:
\[ E = \frac{V}{R} \]
where \( V \) is the electric potential and \( R \) is the radius of the sphere.

### Step 3: Electric Potential of a Sphere
The electric potential \( V \) of a charged sphere is given by:
\[ V = \frac{kQ}{R} \]
where \( k \) is Coulomb's constant, \( Q \) is the charge on the sphere, and \( R \) is the radius of the sphere.

### Step 4: Setting Up the Ratio of Electric Fields
Since both spheres are connected by a wire, their potentials \( V_1 \) and \( V_2 \) will be equal:
\[ V_1 = V_2 \]
Thus, we can express the electric fields \( E_1 \) and \( E_2 \) for the two spheres:
\[ E_1 = \frac{V_1}{a} \quad \text{and} \quad E_2 = \frac{V_2}{b} \]

### Step 5: Equating the Potentials
Since \( V_1 = V_2 \), we can write:
\[ E_1 = \frac{V}{a} \quad \text{and} \quad E_2 = \frac{V}{b} \]

### Step 6: Finding the Ratio of Electric Fields
Now, we can find the ratio of the electric fields:
\[ \frac{E_1}{E_2} = \frac{\frac{V}{a}}{\frac{V}{b}} = \frac{b}{a} \]

### Conclusion
Thus, the ratio of the electric fields of the two spheres is:
\[ \frac{E_1}{E_2} = \frac{b}{a} \]

### Final Answer
The ratio of the electric fields of the spheres is \( \frac{b}{a} \).

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