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In a region of uniform electric field as an electron travels from A to B it slows from Ua is equal to 6.1 ×10 ^-6 metres per second to Ub is equal to 4.6×10^-6 metres per second what is its potential change ∆V= Vb-Va in volts? Instead of the electron in above example suppose a Proton(m=1.67×10^-27 kg) was accelerated from rest by potential difference Vba = to - 5,000 volts what would be protons change in potential energy and final speed?
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In a region of uniform electric field as an electron travels from A to...
Calculation of Potential Change:
Given:
Initial velocity, Ua = 6.1 × 10^-6 m/s
Final velocity, Ub = 4.6 × 10^-6 m/s
The potential change (∆V) can be calculated using the formula:
∆V = Vb - Va
Since the electric field is uniform, the change in potential energy is equal to the change in kinetic energy.
Let's calculate the potential change:
∆V = (1/2) * m * (Ub^2 - Ua^2)
Where:
m is the mass of the electron (9.1 × 10^-31 kg)
∆V = (1/2) * (9.1 × 10^-31 kg) * ((4.6 × 10^-6 m/s)^2 - (6.1 × 10^-6 m/s)^2)
Simplifying the equation:
∆V = (1/2) * (9.1 × 10^-31 kg) * ((21.16 × 10^-12 m^2/s^2) - (37.21 × 10^-12 m^2/s^2))
∆V = (1/2) * (9.1 × 10^-31 kg) * (-16.05 × 10^-12 m^2/s^2)
∆V = -(7.325 × 10^-31 J)
To convert the potential change to volts, we can use the formula:
1 volt = 1 joule/coulomb
Let's calculate the potential change (∆V) in volts:
∆V = -(7.325 × 10^-31 J) / (1.6 × 10^-19 C)
∆V = -0.04565 volts
Therefore, the potential change ∆V = Vb - Va is approximately -0.04565 volts.
Calculation of Proton's Change in Potential Energy and Final Speed:
Given:
Mass of the proton, m = 1.67 × 10^-27 kg
Potential difference, Vba = -5,000 volts
The change in potential energy (∆PE) of the proton can be calculated using the formula:
∆PE = q * ∆V
Where:
q is the charge of the proton (1.6 × 10^-19 C)
Let's calculate the change in potential energy (∆PE):
∆PE = (1.6 × 10^-19 C) * (-5,000 volts)
∆PE = -8 × 10^-16 J
The final kinetic energy (KE) of the proton is equal to the change in potential energy (∆PE):
KE = ∆PE
Since the proton starts from rest, the initial kinetic energy (KEi) is zero.
Therefore, the final kinetic energy (KEf) can be calculated as:
KEf = KEi + ∆PE
KEf = 0 + (-8 × 10^-16 J)
KEf = -8 × 10^-16 J
The final speed (Vf) of the proton can be calculated using the formula:
KEf = (1/2) * m * (Vf^2)
Let's calculate the final speed (Vf):
Vf^2 = (2 * KEf
Given:
Initial velocity, Ua = 6.1 × 10^-6 m/s
Final velocity, Ub = 4.6 × 10^-6 m/s
The potential change (∆V) can be calculated using the formula:
∆V = Vb - Va
Since the electric field is uniform, the change in potential energy is equal to the change in kinetic energy.
Let's calculate the potential change:
∆V = (1/2) * m * (Ub^2 - Ua^2)
Where:
m is the mass of the electron (9.1 × 10^-31 kg)
∆V = (1/2) * (9.1 × 10^-31 kg) * ((4.6 × 10^-6 m/s)^2 - (6.1 × 10^-6 m/s)^2)
Simplifying the equation:
∆V = (1/2) * (9.1 × 10^-31 kg) * ((21.16 × 10^-12 m^2/s^2) - (37.21 × 10^-12 m^2/s^2))
∆V = (1/2) * (9.1 × 10^-31 kg) * (-16.05 × 10^-12 m^2/s^2)
∆V = -(7.325 × 10^-31 J)
To convert the potential change to volts, we can use the formula:
1 volt = 1 joule/coulomb
Let's calculate the potential change (∆V) in volts:
∆V = -(7.325 × 10^-31 J) / (1.6 × 10^-19 C)
∆V = -0.04565 volts
Therefore, the potential change ∆V = Vb - Va is approximately -0.04565 volts.
Calculation of Proton's Change in Potential Energy and Final Speed:
Given:
Mass of the proton, m = 1.67 × 10^-27 kg
Potential difference, Vba = -5,000 volts
The change in potential energy (∆PE) of the proton can be calculated using the formula:
∆PE = q * ∆V
Where:
q is the charge of the proton (1.6 × 10^-19 C)
Let's calculate the change in potential energy (∆PE):
∆PE = (1.6 × 10^-19 C) * (-5,000 volts)
∆PE = -8 × 10^-16 J
The final kinetic energy (KE) of the proton is equal to the change in potential energy (∆PE):
KE = ∆PE
Since the proton starts from rest, the initial kinetic energy (KEi) is zero.
Therefore, the final kinetic energy (KEf) can be calculated as:
KEf = KEi + ∆PE
KEf = 0 + (-8 × 10^-16 J)
KEf = -8 × 10^-16 J
The final speed (Vf) of the proton can be calculated using the formula:
KEf = (1/2) * m * (Vf^2)
Let's calculate the final speed (Vf):
Vf^2 = (2 * KEf
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In a region of uniform electric field as an electron travels from A to B it slows from Ua is equal to 6.1 ×10 ^-6 metres per second to Ub is equal to 4.6×10^-6 metres per second what is its potential change ∆V= Vb-Va in volts? Instead of the electron in above example suppose a Proton(m=1.67×10^-27 kg) was accelerated from rest by potential difference Vba = to - 5,000 volts what would be protons change in potential energy and final speed?
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In a region of uniform electric field as an electron travels from A to B it slows from Ua is equal to 6.1 ×10 ^-6 metres per second to Ub is equal to 4.6×10^-6 metres per second what is its potential change ∆V= Vb-Va in volts? Instead of the electron in above example suppose a Proton(m=1.67×10^-27 kg) was accelerated from rest by potential difference Vba = to - 5,000 volts what would be protons change in potential energy and final speed? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about In a region of uniform electric field as an electron travels from A to B it slows from Ua is equal to 6.1 ×10 ^-6 metres per second to Ub is equal to 4.6×10^-6 metres per second what is its potential change ∆V= Vb-Va in volts? Instead of the electron in above example suppose a Proton(m=1.67×10^-27 kg) was accelerated from rest by potential difference Vba = to - 5,000 volts what would be protons change in potential energy and final speed? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a region of uniform electric field as an electron travels from A to B it slows from Ua is equal to 6.1 ×10 ^-6 metres per second to Ub is equal to 4.6×10^-6 metres per second what is its potential change ∆V= Vb-Va in volts? Instead of the electron in above example suppose a Proton(m=1.67×10^-27 kg) was accelerated from rest by potential difference Vba = to - 5,000 volts what would be protons change in potential energy and final speed?.
In a region of uniform electric field as an electron travels from A to B it slows from Ua is equal to 6.1 ×10 ^-6 metres per second to Ub is equal to 4.6×10^-6 metres per second what is its potential change ∆V= Vb-Va in volts? Instead of the electron in above example suppose a Proton(m=1.67×10^-27 kg) was accelerated from rest by potential difference Vba = to - 5,000 volts what would be protons change in potential energy and final speed? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about In a region of uniform electric field as an electron travels from A to B it slows from Ua is equal to 6.1 ×10 ^-6 metres per second to Ub is equal to 4.6×10^-6 metres per second what is its potential change ∆V= Vb-Va in volts? Instead of the electron in above example suppose a Proton(m=1.67×10^-27 kg) was accelerated from rest by potential difference Vba = to - 5,000 volts what would be protons change in potential energy and final speed? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a region of uniform electric field as an electron travels from A to B it slows from Ua is equal to 6.1 ×10 ^-6 metres per second to Ub is equal to 4.6×10^-6 metres per second what is its potential change ∆V= Vb-Va in volts? Instead of the electron in above example suppose a Proton(m=1.67×10^-27 kg) was accelerated from rest by potential difference Vba = to - 5,000 volts what would be protons change in potential energy and final speed?.
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