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A charged particle having some mass is resting in equilibrium at a height H above the centre of a uniformly charged non-conducting horizontal ring of radius R. The force of gravity acts downwards. The equilibrium of the particle will be stable -
- a)For all values of H
- b)Only if H >
- c)Only if H <
- d)Only if H =
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
A charged particle having some mass is resting in equilibrium at a hei...
electric field due to non-conducting ring along its axis is given by,
E=kqz/(Z2+r2)3/2
where z is the separation between point of observation to the centre of the ring, q is the charge on the ring and r is the radius of the ring.
We know, electric field will be maximum only when z = R/√2, meaning at this point the value of force must be maximum. and if we increase the value of z from R/√2 it will decrease and if we decrease the value of z from R/√2 , it will increase.
condition of stable equilibrium,
A small vertical displacement upwards should cause the resultant force on the particle to be downwards, to return it or a small vertical displacement downward should cause the resultant force on the particle to be downward.
hence, at z > R/√2 or, h > R/√2 system will be in stable equilibrium.
E=kqz/(Z2+r2)3/2
where z is the separation between point of observation to the centre of the ring, q is the charge on the ring and r is the radius of the ring.
We know, electric field will be maximum only when z = R/√2, meaning at this point the value of force must be maximum. and if we increase the value of z from R/√2 it will decrease and if we decrease the value of z from R/√2 , it will increase.
condition of stable equilibrium,
A small vertical displacement upwards should cause the resultant force on the particle to be downwards, to return it or a small vertical displacement downward should cause the resultant force on the particle to be downward.
hence, at z > R/√2 or, h > R/√2 system will be in stable equilibrium.
Most Upvoted Answer
A charged particle having some mass is resting in equilibrium at a hei...
electric field due to non-conducting ring along its axis is given by,
E=kqz/(Z2+r2)3/2
where z is the separation between point of observation to the centre of the ring, q is the charge on the ring and r is the radius of the ring.
We know, electric field will be maximum only when z = R/√2, meaning at this point the value of force must be maximum. and if we increase the value of z from R/√2 it will decrease and if we decrease the value of z from R/√2 , it will increase.
condition of stable equilibrium,
A small vertical displacement upwards should cause the resultant force on the particle to be downwards, to return it or a small vertical displacement downward should cause the resultant force on the particle to be downward.
hence, at z > R/√2 or, h > R/√2 system will be in stable equilibrium.
E=kqz/(Z2+r2)3/2
where z is the separation between point of observation to the centre of the ring, q is the charge on the ring and r is the radius of the ring.
We know, electric field will be maximum only when z = R/√2, meaning at this point the value of force must be maximum. and if we increase the value of z from R/√2 it will decrease and if we decrease the value of z from R/√2 , it will increase.
condition of stable equilibrium,
A small vertical displacement upwards should cause the resultant force on the particle to be downwards, to return it or a small vertical displacement downward should cause the resultant force on the particle to be downward.
hence, at z > R/√2 or, h > R/√2 system will be in stable equilibrium.
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A charged particle having some mass is resting in equilibrium at a hei...
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A charged particle having some mass is resting in equilibrium at a height H above the centre of a uniformly charged non-conducting horizontal ring of radius R. The force of gravity acts downwards. The equilibrium of the particle will be stable -a)For all values of Hb)Only if H >c)Only if H <d)Only if H =Correct answer is option 'B'. Can you explain this answer?
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A charged particle having some mass is resting in equilibrium at a height H above the centre of a uniformly charged non-conducting horizontal ring of radius R. The force of gravity acts downwards. The equilibrium of the particle will be stable -a)For all values of Hb)Only if H >c)Only if H <d)Only if H =Correct answer is option 'B'. Can you explain this answer? 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 A charged particle having some mass is resting in equilibrium at a height H above the centre of a uniformly charged non-conducting horizontal ring of radius R. The force of gravity acts downwards. The equilibrium of the particle will be stable -a)For all values of Hb)Only if H >c)Only if H <d)Only if H =Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A charged particle having some mass is resting in equilibrium at a height H above the centre of a uniformly charged non-conducting horizontal ring of radius R. The force of gravity acts downwards. The equilibrium of the particle will be stable -a)For all values of Hb)Only if H >c)Only if H <d)Only if H =Correct answer is option 'B'. Can you explain this answer?.
A charged particle having some mass is resting in equilibrium at a height H above the centre of a uniformly charged non-conducting horizontal ring of radius R. The force of gravity acts downwards. The equilibrium of the particle will be stable -a)For all values of Hb)Only if H >c)Only if H <d)Only if H =Correct answer is option 'B'. Can you explain this answer? 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 A charged particle having some mass is resting in equilibrium at a height H above the centre of a uniformly charged non-conducting horizontal ring of radius R. The force of gravity acts downwards. The equilibrium of the particle will be stable -a)For all values of Hb)Only if H >c)Only if H <d)Only if H =Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A charged particle having some mass is resting in equilibrium at a height H above the centre of a uniformly charged non-conducting horizontal ring of radius R. The force of gravity acts downwards. The equilibrium of the particle will be stable -a)For all values of Hb)Only if H >c)Only if H <d)Only if H =Correct answer is option 'B'. Can you explain this answer?.
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A charged particle having some mass is resting in equilibrium at a height H above the centre of a uniformly charged non-conducting horizontal ring of radius R. The force of gravity acts downwards. The equilibrium of the particle will be stable -a)For all values of Hb)Only if H >c)Only if H <d)Only if H =Correct answer is option 'B'. Can you explain this answer?, a detailed solution for A charged particle having some mass is resting in equilibrium at a height H above the centre of a uniformly charged non-conducting horizontal ring of radius R. The force of gravity acts downwards. The equilibrium of the particle will be stable -a)For all values of Hb)Only if H >c)Only if H <d)Only if H =Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of A charged particle having some mass is resting in equilibrium at a height H above the centre of a uniformly charged non-conducting horizontal ring of radius R. The force of gravity acts downwards. The equilibrium of the particle will be stable -a)For all values of Hb)Only if H >c)Only if H <d)Only if H =Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
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