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A charged particle of mass m and charge q is released from rest in a uniform electric field E. Neglecting the effect of gravity, kinetic energy of the charged particle after 't' seconds is
- a)E q m/t
- b)E q2 m/2t2
- c)2 E2T2∕mq
- d)E2q2t2∕2m
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
A charged particle of mass m and charge q is released from rest in a u...
force=qE,impulse=qEt=momentum=p, k.e.=p2/2m
Or
1) qE + mg = ma
a = qE/m + g
2) v = 0 + at
v = (qE/m + g)t
3) K.E =1/2mv^2
= 1/2m [(qE/m + g)t]^2 Joule
Most Upvoted Answer
A charged particle of mass m and charge q is released from rest in a u...
Solution:
Given, mass of the charged particle = m
Charge on the particle = q
Electric field = E
Time elapsed = t
Initial velocity of the particle is zero.
The particle moves under the influence of electric field E.
Using Newton's second law of motion, force on the charged particle is given as:
F = ma = qE
where 'a' is the acceleration of the charged particle.
The work done by the electric field on the particle is given as:
W = F*d = qE*d
where 'd' is the distance travelled by the particle.
The kinetic energy of the particle after time t can be calculated using work-energy theorem as:
K.E. = W
The distance travelled by the particle can be calculated using the equation of motion:
d = (1/2)at^2
Substituting the value of 'a' in terms of 'E' from the above equation, we get:
d = (qE/2m)t^2
Substituting the value of 'd' in terms of 'q', 'E' and 't' in the equation for work done, we get:
W = qE(qE/2m)t^2
Substituting the value of 'W' in the equation for kinetic energy, we get:
K.E. = (1/2)qE(qE/2m)t^2
Simplifying the above equation, we get:
K.E. = (1/2)(q^2E^2t^2)/2m
Hence, the correct option is 'D': E^2q^2t^2/(2m).
Given, mass of the charged particle = m
Charge on the particle = q
Electric field = E
Time elapsed = t
Initial velocity of the particle is zero.
The particle moves under the influence of electric field E.
Using Newton's second law of motion, force on the charged particle is given as:
F = ma = qE
where 'a' is the acceleration of the charged particle.
The work done by the electric field on the particle is given as:
W = F*d = qE*d
where 'd' is the distance travelled by the particle.
The kinetic energy of the particle after time t can be calculated using work-energy theorem as:
K.E. = W
The distance travelled by the particle can be calculated using the equation of motion:
d = (1/2)at^2
Substituting the value of 'a' in terms of 'E' from the above equation, we get:
d = (qE/2m)t^2
Substituting the value of 'd' in terms of 'q', 'E' and 't' in the equation for work done, we get:
W = qE(qE/2m)t^2
Substituting the value of 'W' in the equation for kinetic energy, we get:
K.E. = (1/2)qE(qE/2m)t^2
Simplifying the above equation, we get:
K.E. = (1/2)(q^2E^2t^2)/2m
Hence, the correct option is 'D': E^2q^2t^2/(2m).
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Community Answer
A charged particle of mass m and charge q is released from rest in a u...
F= qE
a = qE/m -------(1)
start from rest so u= 0
v= u+ at
v= at
from 1
v= qEt/m
kinetic energy= 1/2 mv²
KE= q²E²t²/2m
a = qE/m -------(1)
start from rest so u= 0
v= u+ at
v= at
from 1
v= qEt/m
kinetic energy= 1/2 mv²
KE= q²E²t²/2m
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A charged particle of mass m and charge q is released from rest in a uniform electric field E. Neglecting the effect of gravity, kinetic energy of the charged particle after 't' seconds isa)E q m/tb)E q2 m/2t2c)2 E2T2∕mqd)E2q2t2∕2mCorrect answer is option 'D'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A charged particle of mass m and charge q is released from rest in a uniform electric field E. Neglecting the effect of gravity, kinetic energy of the charged particle after 't' seconds isa)E q m/tb)E q2 m/2t2c)2 E2T2∕mqd)E2q2t2∕2mCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A charged particle of mass m and charge q is released from rest in a uniform electric field E. Neglecting the effect of gravity, kinetic energy of the charged particle after 't' seconds isa)E q m/tb)E q2 m/2t2c)2 E2T2∕mqd)E2q2t2∕2mCorrect answer is option 'D'. Can you explain this answer?.
A charged particle of mass m and charge q is released from rest in a uniform electric field E. Neglecting the effect of gravity, kinetic energy of the charged particle after 't' seconds isa)E q m/tb)E q2 m/2t2c)2 E2T2∕mqd)E2q2t2∕2mCorrect answer is option 'D'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A charged particle of mass m and charge q is released from rest in a uniform electric field E. Neglecting the effect of gravity, kinetic energy of the charged particle after 't' seconds isa)E q m/tb)E q2 m/2t2c)2 E2T2∕mqd)E2q2t2∕2mCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A charged particle of mass m and charge q is released from rest in a uniform electric field E. Neglecting the effect of gravity, kinetic energy of the charged particle after 't' seconds isa)E q m/tb)E q2 m/2t2c)2 E2T2∕mqd)E2q2t2∕2mCorrect answer is option 'D'. Can you explain this answer?.
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A charged particle of mass m and charge q is released from rest in a uniform electric field E. Neglecting the effect of gravity, kinetic energy of the charged particle after 't' seconds isa)E q m/tb)E q2 m/2t2c)2 E2T2∕mqd)E2q2t2∕2mCorrect answer is option 'D'. Can you explain this answer?, a detailed solution for A charged particle of mass m and charge q is released from rest in a uniform electric field E. Neglecting the effect of gravity, kinetic energy of the charged particle after 't' seconds isa)E q m/tb)E q2 m/2t2c)2 E2T2∕mqd)E2q2t2∕2mCorrect answer is option 'D'. Can you explain this answer? has been provided alongside types of A charged particle of mass m and charge q is released from rest in a uniform electric field E. Neglecting the effect of gravity, kinetic energy of the charged particle after 't' seconds isa)E q m/tb)E q2 m/2t2c)2 E2T2∕mqd)E2q2t2∕2mCorrect answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
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