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Sagot :
To solve the problem, we need to follow the given steps and find the time period speed of the electron with provided data:
### Step 1: Understand the given data
- Current (I): 1.1 milliamps (mA)
- Charge on an electron (e): [tex]\(1.6 \times 10^{-19}\)[/tex] Coulombs (C)
### Step 2: Convert current from milliamps to amps
Current in milliamps (mA) must be converted to amps (A), where 1 A = 1000 mA.
[tex]\[1.1 \text{ mA} = 1.1 \times 10^{-3} \text{ A}\][/tex]
### Step 3: Calculate the number of electrons passing per second
Current ([tex]\(I\)[/tex]) is defined as the rate of flow of charge ([tex]\(Q\)[/tex]) per unit time ([tex]\(t\)[/tex]):
[tex]\[ I = \frac{Q}{t} \][/tex]
Rearranging this formula to solve for the charge ([tex]\(Q\)[/tex]) over time:
[tex]\[ Q = I \times t \][/tex]
But, since we are focusing on the number of electrons per second, we next find out how many elementary charges ([tex]\(e\)[/tex]) pass in one second.
[tex]\[ \text{Number of electrons per second} = \frac{I}{e} \][/tex]
Given the current ([tex]\(I\)[/tex]) in amps (A) and the charge of an electron ([tex]\(e\)[/tex]) in Coulombs (C):
[tex]\[ \text{Number of electrons per second} = \frac{1.1 \times 10^{-3} \text{ A}}{1.6 \times 10^{-19} \text{ C}} \][/tex]
### Step 4: Calculate the numerical result
From the known values provided:
- Current ([tex]\(I\)[/tex]) = 0.0011 A
- Charge of an electron ([tex]\(e\)[/tex]) = [tex]\(1.6 \times 10^{-19}\)[/tex] C
[tex]\[ \text{Number of electrons per second} = 0.0011 A \times \left(\frac{1}{1.6 \times 10^{-19} \text{ C}}\right) \][/tex]
[tex]\[ \text{Number of electrons per second} = 6.875 \times 10^{15} \text{ electrons/second} \][/tex]
### Final Results
- Current in Amps: 0.0011 A
- Number of electrons passing per second: [tex]\(6.875 \times 10^{15}\)[/tex] electrons/second
This indicates that for a current of 1.1 mA, approximately [tex]\(6.875 \times 10^{15}\)[/tex] electrons pass through the conductor per second.
### Step 1: Understand the given data
- Current (I): 1.1 milliamps (mA)
- Charge on an electron (e): [tex]\(1.6 \times 10^{-19}\)[/tex] Coulombs (C)
### Step 2: Convert current from milliamps to amps
Current in milliamps (mA) must be converted to amps (A), where 1 A = 1000 mA.
[tex]\[1.1 \text{ mA} = 1.1 \times 10^{-3} \text{ A}\][/tex]
### Step 3: Calculate the number of electrons passing per second
Current ([tex]\(I\)[/tex]) is defined as the rate of flow of charge ([tex]\(Q\)[/tex]) per unit time ([tex]\(t\)[/tex]):
[tex]\[ I = \frac{Q}{t} \][/tex]
Rearranging this formula to solve for the charge ([tex]\(Q\)[/tex]) over time:
[tex]\[ Q = I \times t \][/tex]
But, since we are focusing on the number of electrons per second, we next find out how many elementary charges ([tex]\(e\)[/tex]) pass in one second.
[tex]\[ \text{Number of electrons per second} = \frac{I}{e} \][/tex]
Given the current ([tex]\(I\)[/tex]) in amps (A) and the charge of an electron ([tex]\(e\)[/tex]) in Coulombs (C):
[tex]\[ \text{Number of electrons per second} = \frac{1.1 \times 10^{-3} \text{ A}}{1.6 \times 10^{-19} \text{ C}} \][/tex]
### Step 4: Calculate the numerical result
From the known values provided:
- Current ([tex]\(I\)[/tex]) = 0.0011 A
- Charge of an electron ([tex]\(e\)[/tex]) = [tex]\(1.6 \times 10^{-19}\)[/tex] C
[tex]\[ \text{Number of electrons per second} = 0.0011 A \times \left(\frac{1}{1.6 \times 10^{-19} \text{ C}}\right) \][/tex]
[tex]\[ \text{Number of electrons per second} = 6.875 \times 10^{15} \text{ electrons/second} \][/tex]
### Final Results
- Current in Amps: 0.0011 A
- Number of electrons passing per second: [tex]\(6.875 \times 10^{15}\)[/tex] electrons/second
This indicates that for a current of 1.1 mA, approximately [tex]\(6.875 \times 10^{15}\)[/tex] electrons pass through the conductor per second.
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