# Why am I getting that work it's always the same in both directions?

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 $B$.

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

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

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:

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

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

• 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 $\Delta V_{AB} = -\Delta V_{BA}$. Commented Apr 3, 2015 at 4:48
• Yes, but the question it's still there, what's that I'm doing wrong with the integrals that I get that $\Delta V_{AB}= \Delta V_{BA}$? Commented Apr 3, 2015 at 5:14
• Edit: Sorry the $V$ integrals on the first comment should be negative. But the conclusion doesn't really change. Commented Apr 3, 2015 at 5:27