How can two equal and opposite forces produce motion? [EM]

[Adjax]

I am having hard time understanding the definition of work done to move a charge in an electrostatic field book says that if charge A and B sitting at some position and now if I move the charge say A towards B from arbitrary direction that force required is exactly equal and opposite to the the Electrostatic field Force .
But since initially, the charges were at rest then it implies they also didn't have any v component
Therefore , the motion shouldn't be possible?

I referred to so many books and all of them defines it like this without giving a proper reason ?
Kindly Help!

 

[Happiness]

You may let the force be slightly smaller than the electrostatic force at the beginning to start the charge moving with a small velocity. Once it starts moving, make the two forces equal in magnitude.

 

[Adjax]

Kindly elaborate !.I mean explain the reasoning behind this decision and its implications

 

[Happiness]

Kindly elaborate !

Could you phrase your question more clearly?

I move the charge say A towards a from arbitrary direction

isn't clear.

Anyway, to answer the question in the title, it is possible for an object experiencing no net force to be moving, by Newton's 1st law. So all we need to move, with no net force, an object stationary at first, is to have the object experience no net force only later (and not at the start), that is, after or once the object acquires a (small) velocity.

 

[Adjax]

sorry for typo :') [POST Editedd]

 

[sophiecentaur]

I am not sure why you are considering two charges, A and B. Only one charge is needed in order to answer the question.
Work done on a charge doesn't include any Kinetic Energy that a charged body might gain by accelerating it. Hence, it's easiest to assume the movement is at virtually zero speed or that the speed at both ends is zero. This may be what has confused you because the 'equal and opposite' condition doesn't exactly apply.
It's the same as when you work out the work done in lifting a mass against g. Unless it arrives at the top of its path with zero speed, more work than mgh is done on it. I agree that this is not always made clear and you can feel daft asking about it. A good thing you did!

 

[Adjax]

Eh?
I still don't understand this, ,if charge is at rest then to move it one need acceleration at first places( which yields KE as byproduct)?

 

[Happiness]

Eh?
I still don't understand this, ,if charge is at rest then to move it one need acceleration at first places( which yields KE as byproduct)?

Suppose the total energy to accelerate the charge and to put it at the desired place is 100 J, and the KE gained as a byproduct is 10 J. Then the PE or the "would-have-been" work done "intended in the definition" in the hypothetical scenario of no KE gain is 100 J ##- ##10 J ##=## 90 J.

I agree that this should be made clear. Current physics textbooks can be improved!

 

[sophiecentaur]

Eh?
I still don't understand this, ,if charge is at rest then to move it one need acceleration at first places( which yields KE as byproduct)?

You can slow it down so it arrives with v=0. You have traveled in a lift (elevator) I presume. Did you have to jump on or off it?

 

[Dale]

I referred to so many books and all of them defines it like this without giving a proper reason ?

Do you have a specific online reference that you could link to as an example. I have seen similar things before, but usually they word it carefully to not say that they were initially at rest.

 

[sophiecentaur]

Eh?
I still don't understand this, if charge is at rest then to move it one need acceleration at first places( which yields KE as byproduct)?

If the electron ends up traveling at some speed then there has been MORE work done on it. The Force X Distance Integral will be greater. The "Byproduct" is included in the total calculation of work done.
There is a limit to how convoluted the explanations can be in textbooks and how many times you can include 'extras' in the basic definition. This is one of those things which you can only sort out for yourself or by asking the right question elsewhere and hopefully get a good answer. I hope you are getting somewhere with this.

 

[russ_watters]

@Adjax; forget charges. Your issue is not understanding Newton's 3rd law. Newton's 3rd law just says if you push on an object, it pushes back on you with an equal and opposite force. It does not say the forces *on* an object are always balanced: the forces in a Newton's 3rd law pair act on different objects!

 

[Adjax]

First of all Thankyou @everyone for taking out your precious time to help me
@Dale: I referred every book I can, indiscriminately : From school book to Purcell to Older Books like Smythe and U.Jean to some Russian texts for e.g. Irodov and finally Internet
@sophiecentaur @Happiness @russ_watters : Now I see due to the consequence N3 law although the force exerted on the field and particle will be a little different, the work done stays same,

since as I move a particle slowly from point 1 to another it exerts force on electric field(E(r) which depends on radial distance) which by N3 law consequently exert the reaction force
We don't know how much force is appied by the external agent but we do know the reactive force exerted by the electric field and therefore now we can calculate the word done by the agent ! (since displacement is against the direction of reactive force of field therefore we take -dr)

Am I right ?​

 

[Dale]

I referred every book I can, indiscriminately

Can you link to one then? I suspect that you are misreading, particularly if you think this mistake is in every book

 

[sophiecentaur]

: I referred every book I can, indiscriminately

If you could quote the sentence in one of those books that seems to be giving you the wrong impression, we may be able to help you. You will not find every book to have a mistake in it!

 

[Adjax]

@Dale @sophiecentaur
I have the e-books on my computer, I can certainly quote the pages : I have noticed that older books like Smythe, JEan and Irodov directly jumps to potential witout referencing work done.

In Purcell : Pg 12 Para 3: "s. The force that has to be applied to move one charge toward the other is equal and opposite to the Coulomb force."
ohhh..wait...
I think culprit is purcell here, I was going through Purcell and it states the above claim where as none of the texts say Smythe ,U . Jean , Ernst Weber and Irodov claimed as such
they only said that when you move a charge work is done against the electric field or requires work which is quite obvious and based on that they derive the work done to move a charge from one position to another unlike Purcell 's claim
And, since I learned the thing from Purcell first , I inherently applied the purcell's viewpoint in any derivation of the same concept by different author's

Full Text: Energy is a useful concept here because electrical forces are conservative. When you push charges around in electric fields, no energy is irrecoverably lost. Everything is perfectly reversible. Consider first the work that must be done on the system to bring some charged bodies into a particular arrangement. Let us start with two charged bodies or particles very far apart from one another, as indicated in Fig. 1.4(a), carrying charges q1 and q2. Whatever energy may have been needed to create these two concentrations of charge originally we shall leave entirely out of account. How much work does it take to bring the particles slowly together until the distance between them is r12? It makes no difference whether we bring q1 toward q2 or the other way around. In either case the work done is the integral of the product: force times displacement, where these are signed quantities. The force that has to be applied to move one charge toward the other is equal and opposite to the Coulomb force. Therefore, < Integral Equations>
...Note that because r is changing from ∞ to r12, the differential dr is negative. We know that the overall sign of the result is correct, because the work done on the system must be positive for charges of like sign; they have to be pushed together (consistent with the minus sign in the applied force). Both the displacement and the applied force are negative in this case, resulting in positive work being done on the system. With q1 and q2 in coulombs, and r12 in meters, Eq. (1.9) gives the work in joules. This work is the same whatever the path of approach"

 

[sophiecentaur]

I think culprit is purcell here,

Culprit? He says it exactly as it is. If you leave the charge to itself, it will be moved by the Coulomb force. To restrain it, you need to apply an equal and opposite force. If you want to move it away from the direction of the Coulomb force, you need an Equal force - plus an infinitesimal amount. But any force will cause acceleration which means gain of KE so you have to slow back down when you arrive at the destination. But we have said all this at least once already.

 

[Dale]

In Purcell : Pg 12 Para 3: "s. The force that has to be applied to move one charge toward the other is equal and opposite to the Coulomb force."

This is a correct statement (although maybe not complete and explicit). It does not state that the charge started at rest, so it can move without accelerating.

How much work does it take to bring the particles slowly together until the distance between them is r12?

This is a better statement (in my opinion). The word "slowly" used in this context implies a limiting procedure. Specifically, you move a charge from one location to another with some maximum speed and record the force, then you repeat the procedure with a smaller maximum speed and record the force, then you again reduce the maximum speed and record the force required. In the limit as the maximum speed goes to 0 (slowly) the force is equal and opposite to the Coulomb force.