0
$\begingroup$

I'm studying electrostatic and I'm getting pretty frustrated because with the definition of work I'm getting that it's always positive and it doesn't make any sense.

So here I have 2 positive particles. $q_1$ it's fixed in it's position, and I'm gonna move $q_2$ from $A$ to $B$.

enter image description here

To calculate the work done by $q_2$ from $A$ to $B$ I use the following equation:

$$ W=\int_a^b{\vec{F}.\vec{ds}} $$

enter image description here

Where $\vec{ds}$ is the direction of travel, and $\vec{F}$ it's the force that $q_1$ does to $q_2$. And turns into this:

$$ W=\int_a^b{F.ds.cos(\theta)} $$

Since $\vec{F}$ and $\vec{ds}$ are going in the same direction, the angle between them $\theta$ is $0$, and $cos(0)=1$ so the equation looks like this:

$$ W=\int_a^b{F.dr} $$

Ok so far so good. If I resolve that integral, since both particles are positive, the work should be positive, because I'm moving in the direction of the force.

And TBH I don't need to continue to get to the part that confuses me.

Say that now I want to go from B to A:

enter image description here

$$ W=\int_b^a{F.ds.cos(\theta)} $$

But now, since I'm moving against the force, $\theta=180^\circ$ so $cos(180^\circ)=-1$

So the equation ends up like this:

$$ W=-\int_b^a{F.dr} $$

Since $\int_a^b = -\int_b^a$, I get the following:

$$ W=\int_a^b{F.dr} $$

And it doesn't make any sense, because it's the same integral I was doing while pulling the particles apart! So there is definitively something I'm doing wrong, but I don't know what.

In a book I have, it just skips the fact that $\theta$ it's gonna be $180^\circ$, so I'm guessing it has to do with that.

And I'm really hoping it's not too dumb, so I don't feel like an idiot, but it probably is :P

Improve this question
$\endgroup$

2 Answers 2

Reset to default
3
$\begingroup$

You are using the repulsive force as the force acting to move the charge from B to A(which is not actually moving the charge). We need an external force to move the charge from B to A, which will be taken into consideration(to calculate workdone).

Improve this answer
$\endgroup$
5
  • 1
    $\begingroup$ Ok forget about the work. Replace everything with electric potential difference from $A$ to $B$, $\Delta V_{AB} = \int_a^b\vec{E}.\vec{ds}= \int_a^b E.dr.cos(0º)= \int_a^b E.dr$ and then measure the difference from $B$ to $A$, $\Delta V_{BA} = \int_b^a\vec{E}.\vec{ds}=\int_b^a E.dr.cos(180º)= -\int_b^a E.dr = \int_a^b E.dr$, so you get that $\Delta V_{AB} = \Delta V_{BA}$, which is obviously wrong. And of course, if you go by the clearer definition, $\Delta V_{AB}=V_B-V_A$ and $\Delta V_{BA}=V_A-V_B$ it's pretty clear that $\Delta V_{AB} = -\Delta V_{BA}$. $\endgroup$
    – Zequez
    Commented Apr 3, 2015 at 4:48
  • $\begingroup$ Yeah, thats a way for that too. $\endgroup$
    – Kishan Kumar
    Commented Apr 3, 2015 at 5:00
  • $\begingroup$ Yes, but the question it's still there, what's that I'm doing wrong with the integrals that I get that $\Delta V_{AB}= \Delta V_{BA}$? $\endgroup$
    – Zequez
    Commented Apr 3, 2015 at 5:14
  • $\begingroup$ Edit: Sorry the $V$ integrals on the first comment should be negative. But the conclusion doesn't really change. $\endgroup$
    – Zequez
    Commented Apr 3, 2015 at 5:27
  • 1
    $\begingroup$ Same thing here to. To move the charge from B to A we need an external electric field, which will move the charge. Not this electric field will cause this movement. @Zequez $\endgroup$
    – Kishan Kumar
    Commented Apr 3, 2015 at 8:01
0
$\begingroup$

You can change the sign of your angle, or you can swap a and b, but you can't do both. When you swap a and b the angle gets increased by 180 degrees (which is the same as changing the sign of your angle), if you do this and change the sign of the angle (again) the two cancel each other out.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.