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To determine the change in potential difference ([tex]\(\Delta V\)[/tex]) experienced by an electron moving in an electric field when its velocity changes, we can use the given data and apply the principles of energy conservation and kinematics.
Given data:
- The charge-to-mass ratio ([tex]\(\frac{c}{m}\)[/tex]) of the electron is [tex]\(1.76 \times 10^{12} \, \frac{C}{kg}\)[/tex].
- Initial velocity ([tex]\(v_1\)[/tex]) is [tex]\(10^7 \, \text{m/s}\)[/tex].
- Final velocity ([tex]\(v_2\)[/tex]) is [tex]\(3 \times 10^7 \, \text{m/s}\)[/tex].
### Steps to Solve:
1. Kinetic Energy Change:
- The change in kinetic energy ([tex]\(\Delta KE\)[/tex]) of the electron as it moves would be given by the difference in the kinetic energy at the final and initial velocities.
- [tex]\[ \Delta KE = \frac{1}{2} m v_2^2 - \frac{1}{2} m v_1^2 \][/tex]
- This can be simplified to:
- [tex]\[ \Delta KE = \frac{1}{2} m (v_2^2 - v_1^2) \][/tex]
2. Relation with Potential Difference:
- The work done to change the kinetic energy of the electron is equal to the charge ([tex]\(q\)[/tex]) times the potential difference ([tex]\(\Delta V\)[/tex]).
- [tex]\[ \Delta KE = q \Delta V \][/tex]
- Hence,
- [tex]\[ q \Delta V = \frac{1}{2} m (v_2^2 - v_1^2) \][/tex]
3. Substitute the Electron's Charge to Mass Ratio:
- The charge-to-mass ratio ([tex]\(\frac{q}{m}\)[/tex]) is given as [tex]\(1.76 \times 10^{12} \, \frac{C}{kg}\)[/tex].
- Rearrange the above equation to express [tex]\(\Delta V\)[/tex]:
- [tex]\[ \Delta V = \frac{m}{q} \times \frac{1}{2} (v_2^2 - v_1^2) \][/tex]
- Since [tex]\(\frac{q}{m} = 1.76 \times 10^{12} \, \frac{C}{kg}\)[/tex], its reciprocal [tex]\(\frac{m}{q}\)[/tex] will be [tex]\(\frac{1}{1.76 \times 10^{12}}\)[/tex].
4. Calculate [tex]\(\Delta V\)[/tex]:
- Substitute the given values:
- [tex]\[ \Delta V = \frac{1}{1.76 \times 10^{12}} \times \frac{1}{2} [(3 \times 10^7)^2 - (10^7)^2] \][/tex]
- Calculate the difference in velocities squared:
- [tex]\[ (3 \times 10^7)^2 - (10^7)^2 = 9 \times 10^{14} - 1 \times 10^{14} = 8 \times 10^{14} \][/tex]
- Now, plug this back into the equation for [tex]\(\Delta V\)[/tex]:
- [tex]\[ \Delta V = \frac{1}{1.76 \times 10^{12}} \times \frac{1}{2} \times 8 \times 10^{14} \][/tex]
- Simplify further:
- [tex]\[ \Delta V = \frac{8 \times 10^{14}}{2 \times 1.76 \times 10^{12}} = \frac{8 \times 10^{14}}{3.52 \times 10^{12}} = 2.272727 \times 10^2 \][/tex]
- [tex]\[ \Delta V \approx 2.3 \times 10^3 \, \text{V} \][/tex]
### Conclusion:
The calculated potential difference is approximately [tex]\(2.3 \times 10^3 \, \text{V}\)[/tex]. Among the given options, the closest value is:
A. [tex]\(2.3 \times 10^3 \)[/tex]
Thus, the correct answer is:
A. [tex]\(2,3 \times 10^3\)[/tex]
Given data:
- The charge-to-mass ratio ([tex]\(\frac{c}{m}\)[/tex]) of the electron is [tex]\(1.76 \times 10^{12} \, \frac{C}{kg}\)[/tex].
- Initial velocity ([tex]\(v_1\)[/tex]) is [tex]\(10^7 \, \text{m/s}\)[/tex].
- Final velocity ([tex]\(v_2\)[/tex]) is [tex]\(3 \times 10^7 \, \text{m/s}\)[/tex].
### Steps to Solve:
1. Kinetic Energy Change:
- The change in kinetic energy ([tex]\(\Delta KE\)[/tex]) of the electron as it moves would be given by the difference in the kinetic energy at the final and initial velocities.
- [tex]\[ \Delta KE = \frac{1}{2} m v_2^2 - \frac{1}{2} m v_1^2 \][/tex]
- This can be simplified to:
- [tex]\[ \Delta KE = \frac{1}{2} m (v_2^2 - v_1^2) \][/tex]
2. Relation with Potential Difference:
- The work done to change the kinetic energy of the electron is equal to the charge ([tex]\(q\)[/tex]) times the potential difference ([tex]\(\Delta V\)[/tex]).
- [tex]\[ \Delta KE = q \Delta V \][/tex]
- Hence,
- [tex]\[ q \Delta V = \frac{1}{2} m (v_2^2 - v_1^2) \][/tex]
3. Substitute the Electron's Charge to Mass Ratio:
- The charge-to-mass ratio ([tex]\(\frac{q}{m}\)[/tex]) is given as [tex]\(1.76 \times 10^{12} \, \frac{C}{kg}\)[/tex].
- Rearrange the above equation to express [tex]\(\Delta V\)[/tex]:
- [tex]\[ \Delta V = \frac{m}{q} \times \frac{1}{2} (v_2^2 - v_1^2) \][/tex]
- Since [tex]\(\frac{q}{m} = 1.76 \times 10^{12} \, \frac{C}{kg}\)[/tex], its reciprocal [tex]\(\frac{m}{q}\)[/tex] will be [tex]\(\frac{1}{1.76 \times 10^{12}}\)[/tex].
4. Calculate [tex]\(\Delta V\)[/tex]:
- Substitute the given values:
- [tex]\[ \Delta V = \frac{1}{1.76 \times 10^{12}} \times \frac{1}{2} [(3 \times 10^7)^2 - (10^7)^2] \][/tex]
- Calculate the difference in velocities squared:
- [tex]\[ (3 \times 10^7)^2 - (10^7)^2 = 9 \times 10^{14} - 1 \times 10^{14} = 8 \times 10^{14} \][/tex]
- Now, plug this back into the equation for [tex]\(\Delta V\)[/tex]:
- [tex]\[ \Delta V = \frac{1}{1.76 \times 10^{12}} \times \frac{1}{2} \times 8 \times 10^{14} \][/tex]
- Simplify further:
- [tex]\[ \Delta V = \frac{8 \times 10^{14}}{2 \times 1.76 \times 10^{12}} = \frac{8 \times 10^{14}}{3.52 \times 10^{12}} = 2.272727 \times 10^2 \][/tex]
- [tex]\[ \Delta V \approx 2.3 \times 10^3 \, \text{V} \][/tex]
### Conclusion:
The calculated potential difference is approximately [tex]\(2.3 \times 10^3 \, \text{V}\)[/tex]. Among the given options, the closest value is:
A. [tex]\(2.3 \times 10^3 \)[/tex]
Thus, the correct answer is:
A. [tex]\(2,3 \times 10^3\)[/tex]
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