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A uniform electric field e is equal to 91 ×10^-6 volt per metre is created between two parallel charged plates an electron enters the field symmetrically between the plates with the speed Vo is equal to 4 × 10 ^- 3 M per second the length of each plate is l is equal to 1 m find the angle of deviation of the path of the electron as it comes out of the field .(mass of electron is m=9.1×10^-31 and its charge is e=-1.6×10^-19C?
Most Upvoted Answer
A uniform electric field e is equal to 91 ×10^-6 volt per metre is cre...
**Given information:**
- Electric field strength (e) = 91 × 10^-6 V/m
- Initial velocity of the electron (Vo) = 4 × 10^-3 m/s
- Length of each plate (l) = 1 m
- Mass of electron (m) = 9.1 × 10^-31 kg
- Charge of electron (e) = -1.6 × 10^-19 C
**Calculating the force on the electron:**
The force experienced by a charged particle in an electric field is given by the equation F = qE, where F is the force, q is the charge, and E is the electric field strength.
In this case, the charge of the electron is e = -1.6 × 10^-19 C and the electric field strength is e = 91 × 10^-6 V/m.
Substituting these values into the equation, we can calculate the force on the electron:
F = (-1.6 × 10^-19 C) * (91 × 10^-6 V/m)
F = -1.456 × 10^-24 N
**Calculating the acceleration of the electron:**
The force acting on a particle is related to its acceleration through Newton's second law, F = ma, where F is the force, m is the mass, and a is the acceleration.
In this case, the force on the electron is F = -1.456 × 10^-24 N and the mass of the electron is m = 9.1 × 10^-31 kg.
Substituting these values into the equation, we can calculate the acceleration of the electron:
-1.456 × 10^-24 N = (9.1 × 10^-31 kg) * a
a = -1.596 × 10^6 m/s^2
**Calculating the time taken by the electron to travel between the plates:**
The initial velocity of the electron is Vo = 4 × 10^-3 m/s. The acceleration of the electron is a = -1.596 × 10^6 m/s^2. The distance traveled by the electron between the plates is l = 1 m.
We can use the equation of motion, s = ut + (1/2)at^2, where s is the distance, u is the initial velocity, a is the acceleration, and t is the time.
Substituting the given values into the equation, we can calculate the time taken by the electron to travel between the plates:
1 = (4 × 10^-3 m/s) * t + (1/2) * (-1.596 × 10^6 m/s^2) * t^2
0 = -0.798 × 10^6 m/s^2 * t^2 + 4 × 10^-3 m/s * t - 1
Using the quadratic formula, we can solve for t:
t = (-4 × 10^-3 m/s ± √((4 × 10^-3 m/s)^2 - 4 * (-0.798 × 10^6 m/s^2) * (-1))) / (2 * -0.798 × 10^6 m/s^2)
Simplifying the equation, we get:
t = 8.9 × 10^-
- Electric field strength (e) = 91 × 10^-6 V/m
- Initial velocity of the electron (Vo) = 4 × 10^-3 m/s
- Length of each plate (l) = 1 m
- Mass of electron (m) = 9.1 × 10^-31 kg
- Charge of electron (e) = -1.6 × 10^-19 C
**Calculating the force on the electron:**
The force experienced by a charged particle in an electric field is given by the equation F = qE, where F is the force, q is the charge, and E is the electric field strength.
In this case, the charge of the electron is e = -1.6 × 10^-19 C and the electric field strength is e = 91 × 10^-6 V/m.
Substituting these values into the equation, we can calculate the force on the electron:
F = (-1.6 × 10^-19 C) * (91 × 10^-6 V/m)
F = -1.456 × 10^-24 N
**Calculating the acceleration of the electron:**
The force acting on a particle is related to its acceleration through Newton's second law, F = ma, where F is the force, m is the mass, and a is the acceleration.
In this case, the force on the electron is F = -1.456 × 10^-24 N and the mass of the electron is m = 9.1 × 10^-31 kg.
Substituting these values into the equation, we can calculate the acceleration of the electron:
-1.456 × 10^-24 N = (9.1 × 10^-31 kg) * a
a = -1.596 × 10^6 m/s^2
**Calculating the time taken by the electron to travel between the plates:**
The initial velocity of the electron is Vo = 4 × 10^-3 m/s. The acceleration of the electron is a = -1.596 × 10^6 m/s^2. The distance traveled by the electron between the plates is l = 1 m.
We can use the equation of motion, s = ut + (1/2)at^2, where s is the distance, u is the initial velocity, a is the acceleration, and t is the time.
Substituting the given values into the equation, we can calculate the time taken by the electron to travel between the plates:
1 = (4 × 10^-3 m/s) * t + (1/2) * (-1.596 × 10^6 m/s^2) * t^2
0 = -0.798 × 10^6 m/s^2 * t^2 + 4 × 10^-3 m/s * t - 1
Using the quadratic formula, we can solve for t:
t = (-4 × 10^-3 m/s ± √((4 × 10^-3 m/s)^2 - 4 * (-0.798 × 10^6 m/s^2) * (-1))) / (2 * -0.798 × 10^6 m/s^2)
Simplifying the equation, we get:
t = 8.9 × 10^-
Community Answer
A uniform electric field e is equal to 91 ×10^-6 volt per metre is cre...
The acceleration of the electron is
in the upward direction. The horizontal velocity remains
as there is no acceleration in this direction. Thus, the time taken in crossing the field is:
The upward component of the velocity of the electron as it emerges from the field region is:
The horizontal component of the velocity remains
The angle θ made by the resultant velocity with the original direction is given by:
Thus, the electron deviates by an angle:
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A uniform electric field e is equal to 91 ×10^-6 volt per metre is created between two parallel charged plates an electron enters the field symmetrically between the plates with the speed Vo is equal to 4 × 10 ^- 3 M per second the length of each plate is l is equal to 1 m find the angle of deviation of the path of the electron as it comes out of the field .(mass of electron is m=9.1×10^-31 and its charge is e=-1.6×10^-19C?
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A uniform electric field e is equal to 91 ×10^-6 volt per metre is created between two parallel charged plates an electron enters the field symmetrically between the plates with the speed Vo is equal to 4 × 10 ^- 3 M per second the length of each plate is l is equal to 1 m find the angle of deviation of the path of the electron as it comes out of the field .(mass of electron is m=9.1×10^-31 and its charge is e=-1.6×10^-19C? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A uniform electric field e is equal to 91 ×10^-6 volt per metre is created between two parallel charged plates an electron enters the field symmetrically between the plates with the speed Vo is equal to 4 × 10 ^- 3 M per second the length of each plate is l is equal to 1 m find the angle of deviation of the path of the electron as it comes out of the field .(mass of electron is m=9.1×10^-31 and its charge is e=-1.6×10^-19C? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A uniform electric field e is equal to 91 ×10^-6 volt per metre is created between two parallel charged plates an electron enters the field symmetrically between the plates with the speed Vo is equal to 4 × 10 ^- 3 M per second the length of each plate is l is equal to 1 m find the angle of deviation of the path of the electron as it comes out of the field .(mass of electron is m=9.1×10^-31 and its charge is e=-1.6×10^-19C?.
A uniform electric field e is equal to 91 ×10^-6 volt per metre is created between two parallel charged plates an electron enters the field symmetrically between the plates with the speed Vo is equal to 4 × 10 ^- 3 M per second the length of each plate is l is equal to 1 m find the angle of deviation of the path of the electron as it comes out of the field .(mass of electron is m=9.1×10^-31 and its charge is e=-1.6×10^-19C? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A uniform electric field e is equal to 91 ×10^-6 volt per metre is created between two parallel charged plates an electron enters the field symmetrically between the plates with the speed Vo is equal to 4 × 10 ^- 3 M per second the length of each plate is l is equal to 1 m find the angle of deviation of the path of the electron as it comes out of the field .(mass of electron is m=9.1×10^-31 and its charge is e=-1.6×10^-19C? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A uniform electric field e is equal to 91 ×10^-6 volt per metre is created between two parallel charged plates an electron enters the field symmetrically between the plates with the speed Vo is equal to 4 × 10 ^- 3 M per second the length of each plate is l is equal to 1 m find the angle of deviation of the path of the electron as it comes out of the field .(mass of electron is m=9.1×10^-31 and its charge is e=-1.6×10^-19C?.
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