Why is work negative when we push two opposite particles tog

[x86]

The below is what has me confused. It's essentially stating that if we apply a force on a positive charge and push it close to a negative charge, our applied work is negative.

Why is this? I am confused about it because W = Fd and both F and d have the same direction.

The potential energy U of the system is equal to the work we must do to assemble the system, bringing in each charge from an infinite distance.
Calculation:
Let's mentally build the system of Fig. 24-19, starting with one of the charges, say q1, in place and the others at infinity. Then we bring another one, say q2, in from infinity and put it in place. From Eq. 24-46 with d substituted for r, the potential energy U12 associated with the pair of charges q1 and q2 is

U12 = kq1q2/d

We then bring the last charge q3 in from infinity and put it in place. The work that we must do in this last step is equal to the sum of the work we must do to bring q3 near q1 and the work we must do to bring it near q2. From Eq. 24-46, with d substituted for r, that sum is

W13 + W23 = U13 + U23 = kq1q3/d + kq2q3/d

The total potential energy U of the three-charge system is the sum of the potential energies associated with the three pairs of charges. This sum (which is actually independent of the order in which the charges are brought together) is

 

[Bystander]

force on a positive charge and push it close to a negative charge, our applied work is negative.

Why is this?

Compare this to the work for pushing two charges of the same sign together.

 

[x86]

Compare this to the work for pushing two charges of the same sign together.

The work required to push two charges of the same sign together is positive. But if you push two charges with opposite signs, they are also affected by their respective electric fields. So doesn't this still mean the work is positive? The charge is just being "pushed more" essentially (two forces in the same direction)

 

[Bystander]

they are also affected by their respective electric fields.

And, what is that effect?

 

[x86]

And, what is that effect?

If the charges are of opposite sign, then they are being pulled towards each other (potential energy is decreasing). So this is positive work. But if someone is pushing them together even faster, this is also positive work

 

[Bystander]

work required to push two charges of the same sign together is positive

being pulled towards each other (potential energy is decreasing). So this is positive

Forget what "coach" told you about doing "work" in the weight room (that lifting and lowering are both work for your muscles). This is physics, not physical fitness. You have to push two charges of the same sign to reduce the distance between them. You have to restrain two charges of opposite signs to prevent them from reducing the distance between them. You can get work from the second system by letting the charges approach each other. You have to put work into the first to make the two same sign charges approach each other.

 

[x86]

Forget what "coach" told you about doing "work" in the weight room (that lifting and lowering are both work for your muscles). This is physics, not physical fitness. You have to push two charges of the same sign to reduce the distance between them. You have to restrain two charges of opposite signs to prevent them from reducing the distance between them. You can get work from the second system by letting the charges approach each other. You have to put work into the first to make the two same sign charges approach each other.

I am still slightly a bit confused. The way I'm picturing the above scenario is like putting charged marbles together. Say you have three marbles and want to form a triangle. I pick a spot where one of the marbles exists, then bring a second marble to it. Pretend they have the same sign. Now displacement has a 0 degree angle with the force direction, W = F.d > 0, so I do positive work to bring them together. Now the third marble has an opposite sign compared to the first two marbles. Meaning the electric field is pulling it in (and doing positive work on it) by W=F.d. So the electric field is doing positive work on that third marble to bring it in (but we don't care about this, since we only care about the work I'm doing).

Thus the work I'm doing has to be work to restrain this marble. But now displacement = 0, so W = F*0 = 0?

 

[Bystander]

I am still slightly a bit confused. The way I'm picturing the above scenario is like putting charged marbles together. Say you have three marbles and want to form a triangle.

Solve/understand ONE thing at a time --- it'll save a lot of confusion.
You have two charges. They are some known distance apart at some initial time. At some later, final time they are a distance apart that is less than the initial distance. That means there has been a displacement. These are electrical charges. That means there is a force between them. The sign of the force depends on the product of the signs of the two charges. The sign of the displacement, maintaining the convention for your initial post, is positive for a decrease in separation for the two charges. You now have two possibilities for sign for the work done "bringing" the two charges closer together; positive, or negative. Pick a convention and STAY with it. Life will be much easier.