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 of how many electrons should be transferred from body X to body Y so that they acquire equal charges, we can follow these steps:

### Step 1: Define the initial charges
Let the charge on body X be \( Q_X = -6.6 \, \mu C = -6.6 \times 10^{-6} \, C \) and the charge on body Y be \( Q_Y = -5 \, \mu C = -5 \times 10^{-6} \, C \).

### Step 2: Set up the equation for equal charges
When \( n \) electrons are transferred from X to Y, the new charge on X will be:
\[ Q_X' = Q_X + n \cdot e \]
And the new charge on Y will be:
\[ Q_Y' = Q_Y - n \cdot e \]
where \( e \) is the charge of one electron, \( e = 1.6 \times 10^{-19} \, C \).

### Step 3: Write the equation for equal charges
To find \( n \), we set the new charges equal to each other:
\[ Q_X' = Q_Y' \]
Substituting the expressions for \( Q_X' \) and \( Q_Y' \):
\[ -6.6 \times 10^{-6} + n \cdot e = -5 \times 10^{-6} - n \cdot e \]

### Step 4: Rearrange the equation
Rearranging gives:
\[ -6.6 \times 10^{-6} + 5 \times 10^{-6} = -2n \cdot e \]
\[ -1.6 \times 10^{-6} = -2n \cdot e \]
\[ 1.6 \times 10^{-6} = 2n \cdot e \]

### Step 5: Solve for \( n \)
Now, we can solve for \( n \):
\[ n = \frac{1.6 \times 10^{-6}}{2 \cdot e} \]
Substituting \( e = 1.6 \times 10^{-19} \, C \):
\[ n = \frac{1.6 \times 10^{-6}}{2 \cdot 1.6 \times 10^{-19}} \]
\[ n = \frac{1.6 \times 10^{-6}}{3.2 \times 10^{-19}} = 0.5 \times 10^{13} \]
\[ n = 5 \times 10^{12} \]

### Conclusion
Thus, the number of electrons that should be transferred from X to Y is \( 5 \times 10^{12} \).

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