Question
How to calculate the number of electrons transferred between two objects to equalize their charges
Answers
GPT-4o mini
To calculate the number of electrons transferred between two objects to equalize their charges, you can follow these steps:
1. **Identify the Initial Charges**: Determine the initial charges of both objects. Let's denote the charge of object A as \( Q_A \) (in coulombs) and the charge of object B as \( Q_B \) (also in coulombs).
2. **Calculate the Total Charge**: Find the total charge of both objects combined:
\[
Q_{total} = Q_A + Q_B
\]
3. **Equalize the Charges**: When the two objects come into contact and charge is redistributed, the charge on each object will be equal. Since there are two objects, the charge on each after redistribution will be half of the total charge:
\[
Q_{final} = \frac{Q_{total}}{2}
\]
4. **Determine the Final Charges**: The final charge on each object can be calculated as follows:
\[
Q_A' = Q_{final} \quad \text{and} \quad Q_B' = Q_{final}
\]
5. **Calculate the Change in Charge**: Calculate the change in charge for each object:
\[
\Delta Q_A = Q_A' - Q_A
\]
\[
\Delta Q_B = Q_B' - Q_B
\]
Note that if electrons are transferred, one object's charge will decrease (losing electrons) and the other will increase (gaining electrons).
6. **Determine the Total Electron Transfer**: The number of electrons transferred (\( N \)) can be calculated using the charge of a single electron, which is approximately \( 1.6 \times 10^{-19} \) coulombs:
\[
N = \frac{|\Delta Q|}{e}
\]
where \( \Delta Q \) is the magnitude of the change in charge of the object losing electrons (either \( |\Delta Q_A| \) or \( |\Delta Q_B| \)).
7. **Final Formula**: Thus, the final formula becomes:
\[
N = \frac{|\Delta Q_A|}{1.6 \times 10^{-19}} \quad \text{or} \quad N = \frac{|\Delta Q_B|}{1.6 \times 10^{-19}}
\]
### Example
**Given:**
- Object A has a charge of \( +5 \times 10^{-6} \) C.
- Object B has a charge of \( -2 \times 10^{-6} \) C.
**Steps:**
1. Calculate total charge:
\[
Q_{total} = 5 \times 10^{-6} + (-2 \times 10^{-6}) = 3 \times 10^{-6} \text{ C}
\]
2. Final charge on each object:
\[
Q_{final} = \frac{3 \times 10^{-6}}{2} = 1.5 \times 10^{-6} \text{ C}
\]
3. Change in charge for A:
\[
\Delta Q_A = 1.5 \times 10^{-6} - 5 \times 10^{-6} = -3.5 \times 10^{-6} \text{ C}
\]
4. Change in charge for B:
\[
\Delta Q_B = 1.5 \times 10^{-6} - (-2 \times 10^{-6}) = 3.5 \times 10^{-6} \text{ C}
\]
5. Calculate number of electrons transferred:
\[
N = \frac{3.5 \times 10^{-6}}{1.6 \times 10^{-19}} \approx 2.19 \times 10^{13} \text{ electrons}
\]
Thus, approximately \( 2.19 \times 10^{13} \) electrons need to be transferred between the two objects to equalize their charges.
1. **Identify the Initial Charges**: Determine the initial charges of both objects. Let's denote the charge of object A as \( Q_A \) (in coulombs) and the charge of object B as \( Q_B \) (also in coulombs).
2. **Calculate the Total Charge**: Find the total charge of both objects combined:
\[
Q_{total} = Q_A + Q_B
\]
3. **Equalize the Charges**: When the two objects come into contact and charge is redistributed, the charge on each object will be equal. Since there are two objects, the charge on each after redistribution will be half of the total charge:
\[
Q_{final} = \frac{Q_{total}}{2}
\]
4. **Determine the Final Charges**: The final charge on each object can be calculated as follows:
\[
Q_A' = Q_{final} \quad \text{and} \quad Q_B' = Q_{final}
\]
5. **Calculate the Change in Charge**: Calculate the change in charge for each object:
\[
\Delta Q_A = Q_A' - Q_A
\]
\[
\Delta Q_B = Q_B' - Q_B
\]
Note that if electrons are transferred, one object's charge will decrease (losing electrons) and the other will increase (gaining electrons).
6. **Determine the Total Electron Transfer**: The number of electrons transferred (\( N \)) can be calculated using the charge of a single electron, which is approximately \( 1.6 \times 10^{-19} \) coulombs:
\[
N = \frac{|\Delta Q|}{e}
\]
where \( \Delta Q \) is the magnitude of the change in charge of the object losing electrons (either \( |\Delta Q_A| \) or \( |\Delta Q_B| \)).
7. **Final Formula**: Thus, the final formula becomes:
\[
N = \frac{|\Delta Q_A|}{1.6 \times 10^{-19}} \quad \text{or} \quad N = \frac{|\Delta Q_B|}{1.6 \times 10^{-19}}
\]
### Example
**Given:**
- Object A has a charge of \( +5 \times 10^{-6} \) C.
- Object B has a charge of \( -2 \times 10^{-6} \) C.
**Steps:**
1. Calculate total charge:
\[
Q_{total} = 5 \times 10^{-6} + (-2 \times 10^{-6}) = 3 \times 10^{-6} \text{ C}
\]
2. Final charge on each object:
\[
Q_{final} = \frac{3 \times 10^{-6}}{2} = 1.5 \times 10^{-6} \text{ C}
\]
3. Change in charge for A:
\[
\Delta Q_A = 1.5 \times 10^{-6} - 5 \times 10^{-6} = -3.5 \times 10^{-6} \text{ C}
\]
4. Change in charge for B:
\[
\Delta Q_B = 1.5 \times 10^{-6} - (-2 \times 10^{-6}) = 3.5 \times 10^{-6} \text{ C}
\]
5. Calculate number of electrons transferred:
\[
N = \frac{3.5 \times 10^{-6}}{1.6 \times 10^{-19}} \approx 2.19 \times 10^{13} \text{ electrons}
\]
Thus, approximately \( 2.19 \times 10^{13} \) electrons need to be transferred between the two objects to equalize their charges.