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 Author Topic: charge movement in zero electric field  (Read 29729 times) 0 Members and 1 Guest are viewing this topic. Click to toggle author information(expand message area).
sridhar10chitta
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 « Embed this message on: September 15, 2005, 11:56:02 am »

Suppose the electric field is switched off when a charge is moving with a certain velocity (no acceleration) in it. What will be its velocity with time after the field is made zero ? There is no external magnetic field considered.
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Sridhar Chitta
Instrumentation and Control Engg
Vignan Institute of Technology and Science
Deshmukhi Nalgonda Dist
Fu-Kwun Hwang
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 « Embed this message Reply #1 on: October 22, 2005, 04:35:03 pm » posted from:Taipei,T'ai-pei,Taiwan

If the E field is always zero, there is no net field all the time, then the net force is zero.

But if the E field exists, at the time it is being turned off. We need to know the rate of change of the E field: (dE/dt , which will create a magnetic field. ) to calculate the effect.
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sridhar10chitta
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 « Embed this message Reply #2 on: October 23, 2005, 07:02:01 pm »

My question:
Suppose the electric field is switched off when a charge is moving with a certain velocity (no acceleration) in it. What will be its velocity with time after the field is made zero ? There is no external magnetic field considered.

If the E field is always zero, there is no net field all the time, then the net force is zero.

But if the E field exists, at the time it is being turned off. We need to know the rate of change of the E field: (dE/dt , which will create a magnetic field. ) to calculate the effect.

Assume that the Efield is a steady constant field.
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Sridhar Chitta
Instrumentation and Control Engg
Vignan Institute of Technology and Science
Deshmukhi Nalgonda Dist
Fu-Kwun Hwang
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 « Embed this message Reply #3 on: October 24, 2005, 07:42:54 am » posted from:Taipei,T'ai-pei,Taiwan

There is no way you can turn off an uniform electric field at an instance.
Just like you can not change velocity from some finite value to zero at the same time.

There is a transit effect due to the change in field. So you need to specify dE/dt .

May be you did not care about the transit effect (may be it is too small), then just assume the net force is zero.

You should know what is the answer if net force is zero.
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sridhar10chitta
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 « Embed this message Reply #4 on: November 07, 2005, 11:47:00 am »

Dear Professor Hwang

I am sorry for the long delay in replying to your mail as this was due to my preoccupation with the conduct of the laboratory examinations in my college.

1. Firstly I would like to correct my earlier assumption that the velocity of the particle initially is uniform, which should be ‘accelerating’ under the force of the electric field.

The evaluation of the motion will be governed by (Ref: Chapter on ‘Interaction of Fields and Matter’ in the book titled “ Electromagnetic Waves and Radiating Systems” by Ed. C. Jordan and Keith G. Balmain Publ: Prentice-Hall.) the following equations:

Work done by the field on the particle moving from point 1 to point 2

W = (integral) F.ds from Pt1 to Pt2 ----(1)
with F = m dv/dt
where v = velocity of the particle
and m = mass of the particle

and thus, (1) reduces after integration

W = 1/2 m (v2 squared - v1 squared) ------   (2)
where v2 = vel at point 2 and v1 = vel at point 1.
Eqn 2 states that the work done is equal to the increase in “kinetic energy”.

2. Now, if it considered that the electric field were suddenly brought to zero when the particle reached point 2, then the particle would continue to travel with velocity v2.

3. If this sequence of events were to be studied using a Cathode Ray Tube, the screen of the tube, if originally was at zero potential will acquire a (negative or positive) potential by the charge (negative or positive) on the particle after the field is switched OFF. The particle will lose its kinetic energy and produce some heat on the screen.

Please confirm whether my reasoning is correct.

Other books I referred are:

1.   Fundamentals of Television Engineering by Glen Glasford Publ: Tata McGraw Hill
2.   Physics by Resnick and Halliday
3.   Electronic and Radio Engineering by F.E. Terman et al . Publ: Mc Graw Hill
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Sridhar Chitta
Instrumentation and Control Engg
Vignan Institute of Technology and Science
Deshmukhi Nalgonda Dist
Fu-Kwun Hwang
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 « Embed this message Reply #5 on: November 08, 2005, 03:25:23 pm » posted from:Taipei,T'ai-pei,Taiwan

[quote:b8cf85debc]

the electric field were suddenly brought to zero when the particle reached point 2.
[/quote:b8cf85debc]

There is no way you can change the Electric field to zero at the same instance. You always need some time to do it(it could be very small), unless you do not care about the transit effect.
Then, there is no force on the charge particle and the particle will move with constant velocity.

Yes. the energy will become thermal energy when the charge particle hit the cathode ray tube.
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Mariam
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 « Embed this message Reply #6 on: March 30, 2007, 08:42:23 pm »

Discuss the motion of charged particle when it enters the electric field
1. parallel to the direction of field
2. perpendicular to the direction of field

Sir i need all the calculation regarding this but finding it diffiult to solve
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Fu-Kwun Hwang
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 « Embed this message Reply #7 on: March 30, 2007, 10:18:42 pm »

I think you should know, if the charge is q and the electric field is E, then the electric force is qE.
For you problem, think about other similar cases: projectile motion under gravitation force
1. initial velocity for the particle is in the horizontal direction only.
2. initial velocity for the particle is in the vertical direction only.

Did you find out the similarity!
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"In theory, theory and practice are the same. In practice, they are not." ..."Albert Einstein (1879~1955, Mathematical physicist, Nobel Prize 1921-Physics)"