The diagram is just a summo wrestler with the point on its streched hand. As you

Question

The diagram is just a summo wrestler with the point on its streched hand.

As you can see, he is pushing on charge Q=1 (coulomb).EA can clearly PUSH quite well but why does he has to push at all?

lets set the back ground.

A). Rules for pushing a charge from A ->B:

1. For the moment, EA does the pushing.

2. EA (remember, he is our External Agent) has to move the charge (and its mass) from A to B at constant velocity. (Why speed and not velocity?)

3. At first, the charge is at rest, but EA has to increase the speed to whatever speed he is to move the charge at.

4. Then EA does the work.

5. Then he stops the charge at the end of the trip.

A).How does the work in step 3 compare to the work needed in step 5? Briefly, why is this important? Hint: compare the Kinetic Energy to speed it up with the energy to slow it down at the end of the journey. Do we have to worry about this? Briefly explain.

B). i ). If a positive charge q travels a distance d from point A to point B, as in the diagram above, along a path parallel to a uniform electric field of magnitude E, is the work done by the field on the charge positive, negative or zero? Explain your reasoning. How does the form of this equation compare to the work done on a mass m traveling a distance d in the almost uniform gravitational field near the surface of the earth?

B). ii. (ii) If the charge were negative, would the work done by the field be positive, negative, or zero? Explain your reasoning.

B). (iii) If you were to calculate the work divided by the charge in parts (i) and (ii) above, would that quantity (work done by the electric field divided by the charge on which the work is done) be positive, negative, or zero?

Explanation / Answer

1. When external agent does the pushing

so in step 3, when the speed is brought to v from 0

a positive wwork is done by external agent, to increase KE of the charge from 0 to 0.5mv^2 ( where m is mass of particle)

instep 5, to stop the pushing, external agent has to do negative ework = -0.5mv^2

sum of these two works is 0, in absence of dissipative forces

2. i) when a positive charge travels along a path parallel to external electric field, work done by the field = F*d

F = qE

d = (Xb - Xa)

so W = q*E(Xb - Xa)

This work is positive as the particle is travelling in the same direction as the force is being applied on it by the electric field

ii) If the charge were negative, then the direction of the force on the particle would be -ve

for antiparallel force andd displacements, the work done by field would be -ve

iii) W = qE(Xb - Xa)

W/q = E(Xb - Xa)

this would be a positive quantiuty as E is directed towards the direction where Xb > Xa

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