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Sure! Let's walk through the solution step-by-step to find the distance a free proton must travel in an electric field to reach [tex]\(3.4\%\)[/tex] of the speed of light, starting from rest.
### Given:
- Electric field strength ([tex]\(E\)[/tex]): [tex]\(3.00 \times 10^6 \, \text{N/C}\)[/tex]
- Speed of light ([tex]\(c\)[/tex]) : [tex]\(3.00 \times 10^8 \, \text{m/s}\)[/tex]
- Percent speed of light for proton: [tex]\(3.4\% = \frac{3.4}{100} = 0.034\)[/tex]
- Mass of proton ([tex]\(m_p\)[/tex]): [tex]\(1.67 \times 10^{-27} \, \text{kg}\)[/tex]
- Charge of proton ([tex]\(q_p\)[/tex]): [tex]\(1.60 \times 10^{-19} \, \text{C}\)[/tex]
### Step-by-Step Solution:
1. Calculate the final speed of the proton:
[tex]\[ v = 0.034 \times c = 0.034 \times 3.00 \times 10^8 \, \text{m/s} = 1.02 \times 10^7 \, \text{m/s} \][/tex]
2. Determine the kinetic energy (KE) gained by the proton:
[tex]\[ \text{KE} = \frac{1}{2} m_p v^2 \][/tex]
Substituting the values:
[tex]\[ \text{KE} = \frac{1}{2} \times 1.67 \times 10^{-27} \, \text{kg} \times (1.02 \times 10^7 \, \text{m/s})^2 \][/tex]
[tex]\[ \text{KE} = \frac{1}{2} \times 1.67 \times 10^{-27} \times 1.0404 \times 10^{14} \, \text{m}^2/\text{s}^2 \][/tex]
[tex]\[ \text{KE} \approx 8.68734 \times 10^{-14} \, \text{J} \][/tex]
3. Calculate the work done (W) by the electric field to move the proton a certain distance (d):
[tex]\[ W = E \cdot q_p \cdot d \][/tex]
Rearrange to solve for [tex]\(d\)[/tex]:
[tex]\[ d = \frac{\text{KE}}{E \cdot q_p} \][/tex]
4. Find the work done per unit charge by the electric field:
[tex]\[ E \cdot q_p = 3.00 \times 10^6 \, \text{N/C} \times 1.60 \times 10^{-19} \, \text{C} \][/tex]
[tex]\[ E \cdot q_p = 4.80 \times 10^{-13} \, \text{J} \][/tex]
5. Calculate the distance (d):
[tex]\[ d = \frac{8.68734 \times 10^{-14} \, \text{J}}{4.80 \times 10^{-13} \, \text{J}} \][/tex]
[tex]\[ d \approx 0.18098625 \, \text{m} \][/tex]
Therefore, the distance a free proton must travel in the given electric field to reach [tex]\(3.4\%\)[/tex] of the speed of light is approximately:
[tex]\[ d \approx 0.181 \, \text{m} \][/tex]
### Given:
- Electric field strength ([tex]\(E\)[/tex]): [tex]\(3.00 \times 10^6 \, \text{N/C}\)[/tex]
- Speed of light ([tex]\(c\)[/tex]) : [tex]\(3.00 \times 10^8 \, \text{m/s}\)[/tex]
- Percent speed of light for proton: [tex]\(3.4\% = \frac{3.4}{100} = 0.034\)[/tex]
- Mass of proton ([tex]\(m_p\)[/tex]): [tex]\(1.67 \times 10^{-27} \, \text{kg}\)[/tex]
- Charge of proton ([tex]\(q_p\)[/tex]): [tex]\(1.60 \times 10^{-19} \, \text{C}\)[/tex]
### Step-by-Step Solution:
1. Calculate the final speed of the proton:
[tex]\[ v = 0.034 \times c = 0.034 \times 3.00 \times 10^8 \, \text{m/s} = 1.02 \times 10^7 \, \text{m/s} \][/tex]
2. Determine the kinetic energy (KE) gained by the proton:
[tex]\[ \text{KE} = \frac{1}{2} m_p v^2 \][/tex]
Substituting the values:
[tex]\[ \text{KE} = \frac{1}{2} \times 1.67 \times 10^{-27} \, \text{kg} \times (1.02 \times 10^7 \, \text{m/s})^2 \][/tex]
[tex]\[ \text{KE} = \frac{1}{2} \times 1.67 \times 10^{-27} \times 1.0404 \times 10^{14} \, \text{m}^2/\text{s}^2 \][/tex]
[tex]\[ \text{KE} \approx 8.68734 \times 10^{-14} \, \text{J} \][/tex]
3. Calculate the work done (W) by the electric field to move the proton a certain distance (d):
[tex]\[ W = E \cdot q_p \cdot d \][/tex]
Rearrange to solve for [tex]\(d\)[/tex]:
[tex]\[ d = \frac{\text{KE}}{E \cdot q_p} \][/tex]
4. Find the work done per unit charge by the electric field:
[tex]\[ E \cdot q_p = 3.00 \times 10^6 \, \text{N/C} \times 1.60 \times 10^{-19} \, \text{C} \][/tex]
[tex]\[ E \cdot q_p = 4.80 \times 10^{-13} \, \text{J} \][/tex]
5. Calculate the distance (d):
[tex]\[ d = \frac{8.68734 \times 10^{-14} \, \text{J}}{4.80 \times 10^{-13} \, \text{J}} \][/tex]
[tex]\[ d \approx 0.18098625 \, \text{m} \][/tex]
Therefore, the distance a free proton must travel in the given electric field to reach [tex]\(3.4\%\)[/tex] of the speed of light is approximately:
[tex]\[ d \approx 0.181 \, \text{m} \][/tex]
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