Two bodies X and Y carry charges −6.6μC and −5μC. How many electrons should be transferred from X to Y so that they acquire equal charges?

Options

A. `2X10^12`
B. `5X10^13`
C. `5X10^12`
D. `2X10^13`

Answer

To solve the problem, we need to determine how many electrons should be transferred from body X to body Y so that they both have equal charges.

### Step-by-Step Solution:

1. **Identify Initial Charges:**
- Charge on body X, \( Q_x = -6.6 \, \mu C = -6.6 \times 10^{-6} \, C \)
- Charge on body Y, \( Q_y = -5 \, \mu C = -5 \times 10^{-6} \, C \)

2. **Define the Charge Transfer:**
- Let \( n \) be the number of electrons transferred from X to Y.
- The charge of one electron is \( e = -1.6 \times 10^{-19} \, C \).

3. **Calculate Final Charges:**
- After transferring \( n \) electrons, the new charge on X will be:
\[ Q_x' = Q_x + n \cdot e = -6.6 \times 10^{-6} + n \cdot (1.6 \times 10^{-19}) \]
- The new charge on Y will be:
\[ Q_y' = Q_y - n \cdot e = -5 \times 10^{-6} - n \cdot (1.6 \times 10^{-19}) \]

4. **Set the Final Charges Equal:**
- Since we want the final charges to be equal:
\[ Q_x' = Q_y' \]
\[ -6.6 \times 10^{-6} + n \cdot (1.6 \times 10^{-19}) = -5 \times 10^{-6} - n \cdot (1.6 \times 10^{-19}) \]

5. **Rearranging the Equation:**
- Rearranging gives:
\[ -6.6 \times 10^{-6} + 5 \times 10^{-6} = -2n \cdot (1.6 \times 10^{-19}) \]
\[ -1.6 \times 10^{-6} = -2n \cdot (1.6 \times 10^{-19}) \]

6. **Solve for \( n \):**
- Dividing both sides by \(-2 \cdot (1.6 \times 10^{-19})\):
\[ n = \frac{1.6 \times 10^{-6}}{2 \cdot 1.6 \times 10^{-19}} = \frac{1.6 \times 10^{-6}}{3.2 \times 10^{-19}} = 5 \times 10^{12} \]

### Final Answer:
The number of electrons that should be transferred from X to Y is \( n = 5 \times 10^{12} \).