Calculating Work Done by Electrical Force on a Moving Charge

[assaftolko]

Homework Statement

Two positive charges of q=3*10^-6 C are a distance 1 m apart.

What will be the work done by the electrical force in bringing one of the charges to a distance of 0.5 m from the other charge (which is fixed at its position through the whole process)?

Homework Equations

The Attempt at a Solution

Well from W=-ΔU I got w=-0.081 J which makes since because the electric force acts in contrast to the path of q as it moves closer to the other q - and so negative work is done. But when I tried to calculate the work directly from W1->2 = ∫[itex]^{2}_{1}[/itex]Fel [itex]\bullet[/itex]dr when 1=1m and 2=0.5m , I got the same result but positive - since the angle between Fel and dr is 180 deg... why is that? Isn't dr pointing to the direction of heading?

 

[rude man]

Positive work is done ON the charge being moved.
The work done "by the electrical force" is the negative of the work done ON the charge.
In any case, the formula is W = +ΔU for the work done ON the charge.

The sign of ∫F dx will always be positive, where F is the force exerted ON the moved charge. This force is always the negative of the force exerted ON the moved charge BY the fixed charge. If the two charges are placed on the x axis, one at the origin and one at +1m, and we move the latter charge to the left, then the integral is ∫-F* -dx = +. If you move the charge at the origin instead the integral is ∫F* dx = + also. The integral is the work done ON the charge in either case.

The important point is to distinguish between the force exerted on the moved charge vs. the force the electric field exerts on the moved charge. One is the negative of the other, all the time.

 

[assaftolko]

Im sorry but i don't see how the work is always positive when u calculate it through integration: in mechanics i got a lot of the times negative work through integration, in the case of friction for example and also for other general forces that we were asked to calculate their work

 

[rude man]

If the force is + and the displacement is +, then ∫F*ds is +. This is the work done BY the force.

In the case of kinetic friction, the force is always COUNTER to the direction of motion so the work done BY the friction force is always negative. The work done by the force PUSHING THE OBJECT is positive.

It all depends on which force you're talking about to determine whether the integral is positive or negative.

 

[assaftolko]

Ok but in this case the electrical force is outwards (repulsion) and the direction of motion is inwards (from 1m separation to 0.5 seperation) so why the integration comes out positive??

 

[rude man]

Ok but in this case the electrical force is outwards (repulsion) and the direction of motion is inwards (from 1m separation to 0.5 seperation) so why the integration comes out positive??

Which integration?

OK, you have q1 at x = 0 and q2 at x = 1m, and you are moving q2 from 1m to 0.5m.

If you integrate the force exerted by the field on the moved charge the integral is negative: ∫F*(-dx) < 0. If you're integrating the force needed to push the two charges together the integral is positive: ∫(-F)*(-dx) > 0.

 

[assaftolko]

Which integration?

OK, you have q1 at x = 0 and q2 at x = 1m, and you are moving q2 from 1m to 0.5m.

If you integrate the force exerted by the field on the moved charge the integral is negative: ∫F*(-dx) < 0. If you're integrating the force needed to push the two charges together the integral is positive: ∫(-F)*(-dx) > 0.

ok so i think that the problem is with my interpetion: they say in the question that at first they are 1m apart and then q2 is moved (let's say to the left towards q1) to 0.5m apart from q1. As I see it - the force of repulsion is positive with size kq1q2/r^2 and this is multiplied by dr and by cos180=-1: the cosine of the angle between F (repulsion) and dr (in the direction of movement, that is at the opposite direction of F). This integral is, as you pointed out, ∫-kq1q2/r^2 * dr - but if the integration limits are 1 for bottom and 0.5 for upper - you will get negative numerical result - you can check for yourself...

 

[rude man]

ok so i think that the problem is with my interpetion: they say in the question that at first they are 1m apart and then q2 is moved (let's say to the left towards q1) to 0.5m apart from q1. As I see it - the force of repulsion is positive with size kq1q2/r^2 and this is multiplied by dr and by cos180=-1: the cosine of the angle between F (repulsion) and dr (in the direction of movement, that is at the opposite direction of F). This integral is, as you pointed out, ∫-kq1q2/r^2 * dr - but if the integration limits are 1 for bottom and 0.5 for upper - you will get negative numerical result - you can check for yourself...

Yes, you have chosen to compute the work done BY the electric field, which is why F is positive. And the work done by the field is negative as it must be. Otherwise the E field would produce positive work AND increase the potential energy of the system. That would violate the conservation of energy. So instead, change in potential energy + work done by the field = 0. ΔP.E. > 0, W < 0 where W = work done by the field.

The integral is positive only if F is negative and dr is negative, then this integral is the work done BY the force pushing q2 towards q1, threby increasing the P.E. of the system.

 

[assaftolko]

Yes, you have chosen to compute the work done BY the electric field, which is why= F is positive. And the work done by the field is negative as it must be. Otherwise the E field would produce positive work AND increase the potential energy of the system. That would violate the conservation of energy. So instead, change in potential energy + work done by the field = 0. ΔP.E. > 0, W < 0 where W = work done by the field.

The integral is positive only if F is negative and dr is negative, then this integral is the work done BY the force pushing q2 towards q1, threby increasing the P.E. of the system.

No no no I'm sorry I got confused in my last post - I meant to say in the last sentence that you will get positive numerical result, and not negetive like you should:

∫-kq1q2/r^2 * dr = -kq1q2∫dr/r^2 = -kq1q2*-1/r = -kq1q2*(-1/0.5-(-1/1))=+0.081 J

 

[assaftolko]

Which integration?

OK, you have q1 at x = 0 and q2 at x = 1m, and you are moving q2 from 1m to 0.5m.

If you integrate the force exerted by the field on the moved charge the integral is negative: ∫F*(-dx) < 0.

It comes out positive: [itex]\vec{F}[/itex][itex]\bullet[/itex][itex]\vec{dr}[/itex] = Fdrcos180 = -Fdr. If you integrate this expression between r1=1m and r2=0.5m you will get positive numerical result!

F is the coloumb force, and this force acts to the right. dr is a displacement element in the direction of the displacement - so it's to the left: The size of both vectors is positive, and the angle between them at all times is 180 deg

 

[rude man]

It comes out positive: [itex]\vec{F}[/itex][itex]\bullet[/itex][itex]\vec{dr}[/itex] = Fdrcos180 = -Fdr. If you integrate this expression between r1=1m and r2=0.5m you will get positive numerical result!

F is the coloumb force, and this force acts to the right. dr is a displacement element in the direction of the displacement - so it's to the left: The size of both vectors is positive, and the angle between them at all times is 180 deg

OK, let's say you're moving q1 instead of q2. q1 is at x = -1 and q2 is at x = 0. Then dx > 0 for sure.
F on q1 due to q2 = -kq1*q2/x². So ∫Fdx = kq1*q2/x which evaluated from -1 to -0.5 is kq1*q2(1/-0.5 - 1/-1) = kq1*q2[-2 - (-1)] = -kq1*q2 < 0.

This is no different than putting q1 at 0 and q2 at x = 1, then moving q2 to the left:
F on q2 due to q1 = kq1*q2/x², ∫Fdx = -kq1*q2/x which evaluated from 1 to 0.5 is -kq1*q2(1/0.5 - 1/1) = -kq1*q2 < 0 also. Notice that the limits of integration go + in the first case and - in the second case. That takes care of the direction of motion being + in the 1st case and - in the second.

 

[assaftolko]

My friend it seems to me that you integrated 1/x^2 as 1/x and not -1/x...

 

[rude man]

My friend it seems to me that you integrated 1/x^2 as 1/x and not -1/x...

Where?

 

[assaftolko]

In your last paragraph: you wrote ...(1/0.5 - 1/1), but it supoose to be (-1/0.5-(-1/1))

 

[rude man]

In your last paragraph: you wrote ...(1/0.5 - 1/1), but it supoose to be (-1/0.5-(-1/1))

No. Read the last paragraph again. q1 is at x=0 and q2 is at x = +1. q2 is moved from x=1 to x=0.5. So the limits are 1 and 0.5.

 

[assaftolko]

I agree on the limits, but i don't get why you don't have a minus sign before 1/0.5 for example, since you should put the limit 0.5 into the expression -1/x and not 1/x...

bottom limit: 1m, upper limit: 0.5m. 1/x^2 integrates as -1/x. When you integrate you calculate the upper limit of integration before subtracting from it the bottom limit of integration, so you get: -1/0.5-(-1/1)= -1. This -1 is multiplied by the -1 that comes from cos180, and so we get 1. The other numerical values are of q1, q2 and k and are all positive also - so at the end we get a positive number:
[itex]\int^{0.5}_{1}[/itex][itex]\vec{F}[/itex][itex]\bullet[/itex][itex]\vec{dx}[/itex] = [itex]\int[/itex][itex]^{0.5}_{1}[/itex]Fdxcos180 = [itex]\int[/itex][itex]^{0.5}_{1}[/itex]kq1q2/x^2*dx*-1 = -kq1q2[itex]\int^{0.5}_{1}[/itex]1/x^2*dx = -kq1q2*(-1/x)[itex]\left|[/itex][itex]^{0.5}_{1}[/itex] = -kq1q2*(-1/0.5--1/1) = -kq1q2*-1 = kq1q2 = +


i don't get why this isn't true...

 

[rude man]

I agree on the limits, but i don't get why you don't have a minus sign before 1/0.5 for example, since you should put the limit 0.5 into the expression -1/x and not 1/x...

bottom limit: 1m, upper limit: 0.5m. 1/x^2 integrates as -1/x. When you integrate you calculate the upper limit of integration before subtracting from it the bottom limit of integration, so you get: -1/0.5-(-1/1)= -1. i don't get why this isn't true...

You are contradicting yourself. You're putting in negative limits when they are clearly positive.

You get 1/.5 - 1/1= 1, not -1/.5 - 1/1 = -1.

 

[assaftolko]

No I'm not! I'm putting positive limits, the minus comes from -1/x!

 

[vela]

Well from W=-ΔU I got w=-0.081 J which makes since because the electric force acts in contrast to the path of q as it moves closer to the other q - and so negative work is done. But when I tried to calculate the work directly from W1->2 = ∫[itex]^{2}_{1}[/itex]Fel [itex]\bullet[/itex]dr when 1=1m and 2=0.5m , I got the same result but positive - since the angle between Fel and dr is 180 deg... why is that? Isn't dr pointing to the direction of heading?

You just have to be a little more careful with the signs. You have
$$dW = \vec{F}\cdot d\vec{r} = \|\vec{F}\| \|d\vec{r}\| \cos(180^\circ) = -\|\vec{F}\| \|d\vec{r}\|.$$ Clearly, that quantity is negative as you'd expect. The problem arises when you change ##\|d\vec{r}\|## to ##dr##. Because the lower limit is greater than the upper limit, dr<0, so you have to have ##\|d\vec{r}\| = -dr##.

 

[assaftolko]

What? But why? Isnt dr pointing in the direction of the displacement? I thought that when i write the expression with cos180 then the vector dr just becomes to be dr which is its size - a positive number

 

[vela]

dr isn't positive. |dr| is positive. ##\|d\vec{r}\| = |dr| = -dr##.

 

[rude man]

No I'm not! I'm putting positive limits, the minus comes from -1/x!

Right, and so the result of the integration is negative!

(Thanks for joining in, vela. We desperately need your help!)

 

[assaftolko]

What do each of these symbols represent exactly?

 

[assaftolko]

My friend it's not.. See my calculaions i did several posts ago

 

[vela]

##\vec{r}## is the displacement vector from the fixed charge to the one being moved. ##r## is the distance between the charges. They are different quantities, which is why it's important to be consistent with the use of the arrow to distinguish which one you're talking about.

Consider a finite change. ##\|\Delta\vec{r}\|## is the magnitude of the vector ##\Delta\vec{r} = \vec{r}_f - \vec{r}_i##. Being a magnitude, it's non-negative by definition. The change in the distance ##r## is ##\Delta r = r_f - r_i = \|\vec{r}_f\| - \|\vec{r}_i\|##. This can be positive or negative depending on the relative positions of ##\vec{r}_i## and ##\vec{r}_f##. The same relationships hold with the infinitesimal quantities.

The limits of your final integral, 1.0 m and 0.5 m, are distances, so you're integrating with respect to ##r## in the end.

 

[assaftolko]

dr isn't positive. |dr| is positive. ##\|d\vec{r}\| = |dr| = -dr##.

When i write dr without the vector sign i write the size of the vector as my understanding of vector writing goes.. And so dr is suppose to be written as -dr only if r^ is defined for some reason in the direction oppossite to the direction of the displacement - meaning radially outwards.. Thats what I am familiar with...

 

[assaftolko]

##\vec{r}## is the displacement vector from the fixed charge to the one being moved. ##r## is the distance between the charges. They are different quantities, which is why it's important to be consistent with the use of the arrow to distinguish which one you're talking about.

Consider a finite change. ##\|\Delta\vec{r}\|## is the magnitude of the vector ##\Delta\vec{r} = \vec{r}_f - \vec{r}_i##. Being a magnitude, it's non-negative by definition. The change in the distance ##r## is ##\Delta r = r_f - r_i = \|\vec{r}_f\| - \|\vec{r}_i\|##. This can be positive or negative depending on the relative positions of ##\vec{r}_i## and ##\vec{r}_f##. The same relationships hold with the infinitesimal quantities.

The limits of your final integral, 1.0 m and 0.5 m, are distances, so you're integrating with respect to ##r## in the end.

Why is r pointing from the fixed charge to the ont being moved? Isnt r defined to point from the start point to the end point of the moved charge??

 

[vela]

I was just assuming the fixed charge is at the origin because the math is simpler that way, but you're right, you could choose any origin.

 

[assaftolko]

No but what I am saying that the way i see it: r points in the direction of movement: from the moved charge towards the stationary charge: r the displacement vector is suppose to point in the direction of movement from what i know

 

[vela]

##\vec{r}## is simply the position of the charge. The direction of this vector has nothing to do with the way the charge is moving at any instant.

The direction the charge is moving is given by its velocity, which is proportional to ##d\vec{r}##, not ##\vec{r}##.

 

[assaftolko]

Ok i always thought that when you reffer to Fdr then dr is a small displacement vector that always points in the direction of movement. Of course i know that the general position vector r is taken from some origin to where the moved object is at, i just thought that for this line integral dr has somewhat different role.. Thanks a lot to both of you!

 

[vela]

When you wrote Fdr, what did you mean? Did you mean ##F\,dr## or ##\vec{F}\cdot d\vec{r}##? If it's the former, dr isn't a vector. If it's the latter, it is the small displacement vector that points in the direction of movement.

 

[assaftolko]

I meant both of them in some way.. I don't quite get the transition dr (vector) does from being an entity that points in the direction of motion to: -dr (scalar)... Its not so clear to me what is the role of this minus, which is present in addition to the minus sign generated from these 2 vectors (force and dr) being 180 deg oriented

 

[vela]

I explained why the minus sign appears back in post 19.

 

[assaftolko]

I explained why the minus sign appears back in post 19.

From what I know: dr, when not written as a vector, represents the size of the vector dr, and size is a positive quantity. When you write -dr, as I see it, you are saying that the vector dr is pointing in the negative direction of the r^ axis, my analogue is that of forces in simple Newtonian problems: mg is a vector quantity and if I write it as: -mg in my y-axis force equations then that's because I decided that the positive y-axis is going up.

That's the kind of analogy I'm trying to do here, because I basiclly don't understand something you said earlier, that: |dr|= -dr... this equation is hard for me to grasp... and by the way - why did you write ||[itex]\vec{dr}[/itex]||? What does this 2 bars symbol represent?