A plate carries a charge of `-3.0 muC`, while a rod carries a charge of `+2.0 mu C`. How many electrons must be transferred from the plate to the rod, so that both objects have the same charge ?

Options

A. `6.3 xx 10^(12)` electrons
B. `1.2 xx 10^(13)` electrons
C. `8.0 xx 10^(12)` electrons
D. `1.6 xx 10^(13)` electrons

Answer

To solve the problem of how many electrons must be transferred from the plate to the rod so that both objects have the same charge, we can follow these steps:

### Step 1: Identify the initial charges
- The charge on the plate is \( Q_p = -3.0 \, \mu C \) (microcoulombs).
- The charge on the rod is \( Q_r = +2.0 \, \mu C \).

### Step 2: Define the final charge condition
We want both the plate and the rod to have the same charge after transferring some electrons. Let the number of electrons transferred be \( n \).

### Step 3: Calculate the charge of the transferred electrons
The charge of one electron is approximately \( -1.6 \times 10^{-19} \, C \). Therefore, the total charge of \( n \) electrons is:
\[ Q_e = n \times (-1.6 \times 10^{-19} \, C) \]

### Step 4: Write the equations for final charges
After transferring \( n \) electrons from the plate to the rod:
- The new charge on the plate will be:
\[ Q_p' = Q_p + Q_e = -3.0 \, \mu C + n \times 1.6 \times 10^{-19} \, C \]
- The new charge on the rod will be:
\[ Q_r' = Q_r + Q_e = +2.0 \, \mu C + n \times (-1.6 \times 10^{-19} \, C) \]

### Step 5: Set the final charges equal
To find \( n \), we set the final charges equal:
\[ Q_p' = Q_r' \]
Substituting the expressions from step 4:
\[ -3.0 \, \mu C + n \times 1.6 \times 10^{-19} = +2.0 \, \mu C - n \times 1.6 \times 10^{-19} \]

### Step 6: Solve for \( n \)
Rearranging the equation gives:
\[ n \times 1.6 \times 10^{-19} + n \times 1.6 \times 10^{-19} = +2.0 \, \mu C + 3.0 \, \mu C \]
\[ 2n \times 1.6 \times 10^{-19} = 5.0 \, \mu C \]
Convert \( 5.0 \, \mu C \) to coulombs:
\[ 5.0 \, \mu C = 5.0 \times 10^{-6} \, C \]
Now, substituting:
\[ 2n \times 1.6 \times 10^{-19} = 5.0 \times 10^{-6} \]
\[ n = \frac{5.0 \times 10^{-6}}{2 \times 1.6 \times 10^{-19}} \]
Calculating \( n \):
\[ n = \frac{5.0 \times 10^{-6}}{3.2 \times 10^{-19}} \approx 1.5625 \times 10^{13} \]

### Step 7: Round to the nearest whole number
Since we cannot transfer a fraction of an electron, we round \( n \) to the nearest whole number:
\[ n \approx 15625 \times 10^{9} \text{ electrons} \]

### Final Answer
Approximately \( 15625 \) electrons must be transferred from the plate to the rod.

---