Two metal spheres of radii 0.01 m and 0.02 m are given a charge of 15 mC and 45 mC, respectively. They are then connected by a wire. The final charge on the first is `…xx10^(-3) C`.

Options

A. 40
B. 30
C. 20
D. 10

Answer

To solve the problem step by step, we need to follow these logical steps:

### Step 1: Understand the Problem
We have two metal spheres with given radii and charges. When they are connected by a wire, the charge will redistribute between them until they reach the same electric potential.

### Step 2: Identify Given Values
- Radius of sphere 1, \( r_1 = 0.01 \, m \)
- Radius of sphere 2, \( r_2 = 0.02 \, m \)
- Charge on sphere 1, \( Q_1 = 15 \, mC = 15 \times 10^{-3} \, C \)
- Charge on sphere 2, \( Q_2 = 45 \, mC = 45 \times 10^{-3} \, C \)

### Step 3: Charge Conservation
The total charge before connecting the spheres is equal to the total charge after they are connected:
\[ Q_1 + Q_2 = Q_1' + Q_2' \]
where \( Q_1' \) and \( Q_2' \) are the final charges on spheres 1 and 2, respectively.

Calculating the total initial charge:
\[ Q_{total} = 15 \times 10^{-3} + 45 \times 10^{-3} = 60 \times 10^{-3} \, C \]

### Step 4: Equal Potential Condition
When the spheres are connected, they will have the same potential:
\[ V_1 = V_2 \]
Using the formula for electric potential \( V = \frac{KQ}{r} \), we can write:
\[ \frac{KQ_1'}{r_1} = \frac{KQ_2'}{r_2} \]
Since \( K \) is a constant and can be canceled out, we have:
\[ \frac{Q_1'}{r_1} = \frac{Q_2'}{r_2} \]

### Step 5: Relate the Charges
From the above equation, we can express \( Q_2' \) in terms of \( Q_1' \):
\[ Q_2' = \frac{r_2}{r_1} Q_1' \]
Given that \( r_1 = 0.01 \, m \) and \( r_2 = 0.02 \, m \):
\[ \frac{r_2}{r_1} = \frac{0.02}{0.01} = 2 \]
Thus, we have:
\[ Q_2' = 2Q_1' \]

### Step 6: Substitute into Charge Conservation Equation
Now substitute \( Q_2' \) into the charge conservation equation:
\[ Q_1' + Q_2' = 60 \times 10^{-3} \]
Substituting \( Q_2' = 2Q_1' \):
\[ Q_1' + 2Q_1' = 60 \times 10^{-3} \]
\[ 3Q_1' = 60 \times 10^{-3} \]
\[ Q_1' = \frac{60 \times 10^{-3}}{3} = 20 \times 10^{-3} \, C \]

### Step 7: Final Answer
The final charge on the first sphere is:
\[ Q_1' = 20 \times 10^{-3} \, C \]