The radii of two charged metal spheres are 5cm and 10cm both having the same charge 60mC. If they are connected by a wire

Options

A. A charge of 20mC flows through the wire from larger to smaller sphere
B. A charge of 20mC flows through the wire from smaller to larger sphere
C. A charge of 20mC flows through the wire from smaller to larger sphere
D. No charge flows through the wire because both spheres have same charge

Explanation

To solve the problem of two charged metal spheres connected by a wire, we will follow these steps:

### Step 1: Identify the given data
- Radius of sphere 1, \( r_1 = 5 \, \text{cm} = 0.05 \, \text{m} \)
- Radius of sphere 2, \( r_2 = 10 \, \text{cm} = 0.10 \, \text{m} \)
- Charge on sphere 1, \( Q_1 = 60 \, \text{mC} = 60 \times 10^{-3} \, \text{C} \)
- Charge on sphere 2, \( Q_2 = 60 \, \text{mC} = 60 \times 10^{-3} \, \text{C} \)

### Step 2: Calculate the initial potentials of the spheres
The potential \( V \) of a charged sphere is given by the formula:
\[ V = \frac{k \cdot Q}{r} \]
where \( k = \frac{1}{4 \pi \epsilon_0} \approx 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \).

**For sphere 1:**
\[ V_1 = \frac{k \cdot Q_1}{r_1} = \frac{9 \times 10^9 \cdot 60 \times 10^{-3}}{0.05} \]
Calculating:
\[ V_1 = \frac{9 \times 10^9 \cdot 60 \times 10^{-3}}{0.05} = 1.08 \times 10^6 \, \text{V} \]

**For sphere 2:**
\[ V_2 = \frac{k \cdot Q_2}{r_2} = \frac{9 \times 10^9 \cdot 60 \times 10^{-3}}{0.10} \]
Calculating:
\[ V_2 = \frac{9 \times 10^9 \cdot 60 \times 10^{-3}}{0.10} = 0.54 \times 10^6 \, \text{V} \]

### Step 3: Connect the spheres with a wire
When the spheres are connected by a wire, charge will flow until the potentials are equal. Let the final charge on sphere 1 be \( Q_1' \) and on sphere 2 be \( Q_2' \).

### Step 4: Set up the equations for charge conservation and equal potential
1. **Charge conservation:**
\[ Q_1' + Q_2' = 120 \, \text{mC} \]

2. **Equal potential condition:**
\[ V_1' = V_2' \]
This gives:
\[ \frac{k \cdot Q_1'}{r_1} = \frac{k \cdot Q_2'}{r_2} \]
Simplifying (since \( k \) cancels out):
\[ \frac{Q_1'}{0.05} = \frac{Q_2'}{0.10} \]
This implies:
\[ Q_2' = 2 \cdot Q_1' \]

### Step 5: Substitute \( Q_2' \) into the charge conservation equation
Substituting \( Q_2' = 2 \cdot Q_1' \) into the charge conservation equation:
\[ Q_1' + 2 \cdot Q_1' = 120 \, \text{mC} \]
This simplifies to:
\[ 3 \cdot Q_1' = 120 \, \text{mC} \]
Thus:
\[ Q_1' = 40 \, \text{mC} \]

### Step 6: Find \( Q_2' \)
Using \( Q_2' = 2 \cdot Q_1' \):
\[ Q_2' = 2 \cdot 40 \, \text{mC} = 80 \, \text{mC} \]

### Step 7: Determine the charge that flowed
The initial charge on sphere 1 was \( 60 \, \text{mC} \) and the final charge is \( 40 \, \text{mC} \). Therefore, the charge that flowed from sphere 1 to sphere 2 is:
\[ \Delta Q = 60 \, \text{mC} - 40 \, \text{mC} = 20 \, \text{mC} \]

### Final Answer
The charge that flowed from the smaller sphere to the larger sphere is \( 20 \, \text{mC} \).

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