Two spheres of same metal have radii a and b . They have been connected to a conducting wire. Find the ratio of the electric field intensity upon them.

Options

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

Explanation

To solve the problem of finding the ratio of electric field intensity upon two spheres of the same metal with radii \( a \) and \( b \) connected by a conducting wire, we can follow these steps:

### Step 1: Understand the Setup
We have two spheres of the same metal with radii \( a \) and \( b \). When connected by a conducting wire, they will share charge until they reach the same electric potential.

### Step 2: Electric Field Due to a Sphere
The electric field \( E \) due to a charged sphere at its surface is given by the formula:
\[ E = \frac{kQ}{r^2} \]
where \( k \) is Coulomb's constant, \( Q \) is the charge on the sphere, and \( r \) is the radius of the sphere.

### Step 3: Charge Distribution
Let the charges on the spheres be \( Q_1 \) and \( Q_2 \). After connecting them with a wire, the charges will redistribute, and they will have the same potential. The total charge \( Q \) on the system is:
\[ Q = Q_1 + Q_2 \]
Since they are connected, the potential \( V \) on both spheres will be equal:
\[ V_a = V_b \]
This implies:
\[ \frac{Q_1}{a} = \frac{Q_2}{b} \]

### Step 4: Express Charges in Terms of Total Charge
From the potential equality, we can express \( Q_1 \) and \( Q_2 \) in terms of a common charge \( Q \):
\[ Q_1 = \frac{Q a}{(a + b)} \quad \text{and} \quad Q_2 = \frac{Q b}{(a + b)} \]

### Step 5: Electric Field Intensities
Now, we can find the electric field intensities at the surfaces of the spheres:
- For sphere with radius \( a \):
\[ E_a = \frac{kQ_1}{a^2} = \frac{k \left(\frac{Q a}{(a + b)}\right)}{a^2} = \frac{kQ}{a(a + b)} \]
- For sphere with radius \( b \):
\[ E_b = \frac{kQ_2}{b^2} = \frac{k \left(\frac{Q b}{(a + b)}\right)}{b^2} = \frac{kQ}{b(a + b)} \]

### Step 6: Ratio of Electric Field Intensities
Now, we can find the ratio of the electric field intensities:
\[ \frac{E_a}{E_b} = \frac{\frac{kQ}{a(a + b)}}{\frac{kQ}{b(a + b)}} = \frac{b}{a} \]

### Final Answer
Thus, the ratio of the electric field intensity upon the two spheres is:
\[ \frac{E_a}{E_b} = \frac{b}{a} \]