Answer :
To solve the question of how many electrons the conductor has in excess or deficit, we need to understand that the charge on a conductor can result from either an excess or a deficit of electrons. Since the conductor has a positive charge, this implies a deficit of electrons.
Let's go through the steps to find the number of electrons the conductor is missing:
1. Given data:
- Charge on the conductor: [tex]\( 3.2 \times 10^{-19} \)[/tex] C (Coulombs)
- Charge of a single electron: [tex]\( 1.60 \times 10^{-19} \)[/tex] C (Coulombs)
2. Objective:
- Find the number of electrons that account for this charge deficit.
3. Theory:
- We can determine the number of electrons by dividing the total charge by the charge carried by one electron.
4. Calculation:
- Number of electrons, [tex]\(n\)[/tex], can be calculated as:
[tex]\[ n = \frac{\text{Charge on the conductor}}{\text{Charge of a single electron}} \][/tex]
5. Substitute the given values:
[tex]\[ n = \frac{3.2 \times 10^{-19} \, \text{C}}{1.60 \times 10^{-19} \, \text{C/electron}} \][/tex]
6. Perform the division:
[tex]\[ n = \frac{3.2}{1.60} \][/tex]
7. Result:
[tex]\[ n = 2.0 \][/tex]
Therefore, the conductor has a deficit of 2 electrons. This means it is missing 2 electrons, resulting in the positive charge observed.
Let's go through the steps to find the number of electrons the conductor is missing:
1. Given data:
- Charge on the conductor: [tex]\( 3.2 \times 10^{-19} \)[/tex] C (Coulombs)
- Charge of a single electron: [tex]\( 1.60 \times 10^{-19} \)[/tex] C (Coulombs)
2. Objective:
- Find the number of electrons that account for this charge deficit.
3. Theory:
- We can determine the number of electrons by dividing the total charge by the charge carried by one electron.
4. Calculation:
- Number of electrons, [tex]\(n\)[/tex], can be calculated as:
[tex]\[ n = \frac{\text{Charge on the conductor}}{\text{Charge of a single electron}} \][/tex]
5. Substitute the given values:
[tex]\[ n = \frac{3.2 \times 10^{-19} \, \text{C}}{1.60 \times 10^{-19} \, \text{C/electron}} \][/tex]
6. Perform the division:
[tex]\[ n = \frac{3.2}{1.60} \][/tex]
7. Result:
[tex]\[ n = 2.0 \][/tex]
Therefore, the conductor has a deficit of 2 electrons. This means it is missing 2 electrons, resulting in the positive charge observed.