Two metallic s pheres of radii `R_(1)` and `R_(2)` are connected b y a thin wire. If `+q_(1)~ and `+q_(2)` are the charges on the two s pheres then

Options

A. `q_(1)// q_(2)=R_(1)^(2)//R_(2)^(2)`
B. `q_(1)//q_(2)=R_(1)//R_(2)`
C. `q_(1)//q _(2)=R_(1)^(3)//R_(2) ^(3)`
D. `q_(1)//q_(2)=(R_(1)^(2)-R_(2)^(2))//(R_(1)^(2)+R_(2)^(2))`

Explanation

To solve the problem, we need to analyze the situation of two metallic spheres connected by a thin wire. When two conductors are connected by a wire, they will have the same electric potential.

### Step-by-Step Solution:

1. **Understanding the System**:
We have two metallic spheres with radii \( R_1 \) and \( R_2 \), and they carry charges \( +q_1 \) and \( +q_2 \) respectively. Since they are connected by a wire, they must be at the same electric potential.

2. **Electric Potential of a Sphere**:
The electric potential \( V \) of a charged metallic sphere is given by the formula:
\[ V = \frac{Q}{4 \pi \epsilon_0 R} \]
where \( Q \) is the charge on the sphere, \( R \) is the radius of the sphere, and \( \epsilon_0 \) is the permittivity of free space.

3. **Setting Up the Equations**:
For sphere 1 (radius \( R_1 \) and charge \( q_1 \)):
\[ V_1 = \frac{q_1}{4 \pi \epsilon_0 R_1} \]
For sphere 2 (radius \( R_2 \) and charge \( q_2 \)):
\[ V_2 = \frac{q_2}{4 \pi \epsilon_0 R_2} \]

4. **Equating the Potentials**:
Since the spheres are connected by a wire, their potentials are equal:
\[ V_1 = V_2 \]
Substituting the expressions for \( V_1 \) and \( V_2 \):
\[ \frac{q_1}{4 \pi \epsilon_0 R_1} = \frac{q_2}{4 \pi \epsilon_0 R_2} \]

5. **Simplifying the Equation**:
The \( 4 \pi \epsilon_0 \) cancels out from both sides:
\[ \frac{q_1}{R_1} = \frac{q_2}{R_2} \]

6. **Finding the Relationship Between Charges**:
Rearranging the equation gives us:
\[ \frac{q_1}{q_2} = \frac{R_1}{R_2} \]
This means that the ratio of the charges on the two spheres is equal to the ratio of their radii.

### Final Result:
\[ \frac{q_1}{q_2} = \frac{R_1}{R_2} \]