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A particle of mass m and charge q is placed atrest in a uniform electric field E and then released.The kinetic energy attained by the particle aftermoving a distance y is [1998]
- a)qEy2
- b)qE2 y
- c)qEy
- d)q2 Ey
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
A particle of mass m and charge q is placed atrest in a uniform electr...
K.E. = Force × distance = qE.y
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A particle of mass m and charge q is placed atrest in a uniform electr...
Explanation:
When a charged particle is placed in a uniform electric field, it experiences a force given by F=qE where q is the charge of the particle and E is the electric field strength.
If the particle is initially at rest, it will accelerate in the direction of the electric field and gain kinetic energy.
Let us consider the motion of the charged particle in the electric field.
Initial conditions:
- The particle is at rest, so its initial kinetic energy is zero.
- The electric field is uniform, so the force experienced by the particle is constant in magnitude and direction.
Using Newton's second law of motion, we can relate the force experienced by the particle to its acceleration as follows:
F=ma
where F is the force experienced by the particle, m is its mass, and a is its acceleration.
Since the acceleration is constant, we can use the following kinematic equation to relate the distance traveled by the particle to its final velocity:
y=1/2at2
where y is the distance traveled by the particle, t is the time taken, and a is the acceleration.
Rearranging this equation, we get:
t=√(2y/a)
Substituting this value of t in the equation F=ma, we get:
F=ma=qE
Substituting this value of F in the work-energy theorem, we get:
W=Fy=qEy
where W is the work done on the particle by the electric field.
The work done on the particle is equal to the kinetic energy gained by it, so we have:
K=W=qEy
where K is the kinetic energy gained by the particle.
Hence, the correct answer is option 'C' (qEy).
When a charged particle is placed in a uniform electric field, it experiences a force given by F=qE where q is the charge of the particle and E is the electric field strength.
If the particle is initially at rest, it will accelerate in the direction of the electric field and gain kinetic energy.
Let us consider the motion of the charged particle in the electric field.
Initial conditions:
- The particle is at rest, so its initial kinetic energy is zero.
- The electric field is uniform, so the force experienced by the particle is constant in magnitude and direction.
Using Newton's second law of motion, we can relate the force experienced by the particle to its acceleration as follows:
F=ma
where F is the force experienced by the particle, m is its mass, and a is its acceleration.
Since the acceleration is constant, we can use the following kinematic equation to relate the distance traveled by the particle to its final velocity:
y=1/2at2
where y is the distance traveled by the particle, t is the time taken, and a is the acceleration.
Rearranging this equation, we get:
t=√(2y/a)
Substituting this value of t in the equation F=ma, we get:
F=ma=qE
Substituting this value of F in the work-energy theorem, we get:
W=Fy=qEy
where W is the work done on the particle by the electric field.
The work done on the particle is equal to the kinetic energy gained by it, so we have:
K=W=qEy
where K is the kinetic energy gained by the particle.
Hence, the correct answer is option 'C' (qEy).
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