Potential energy of a system of two punctual charges along the X axis

[AndersF]

Homework Statement

Two punctual charges $+Q$ are placed along the $X$-axis at the points $x=−a$ and $x=a$. Find the work to move a third charge $−Q$ from the infinity to the origin.

Relevant Equations

$E_{p}=-k \frac{Q^2}{r}$

I have not clear how to solve this problem. Here it is my attempt at a solution:

Let the charge at $-a$ be the number one and the one at $+a$ the number two. the potential energy of the punctual charge $-Q$ due to each charge +Q will be then $E_{pi}=-k \frac{Q^2}{r_i}$, whit $r_i$ the distance between then. So, if the charge $-Q$ is at a distance $x$ from the origin, the total potential energy will be the sum of those ones due to each charge $+Q$:

$E_{p_1}=-k \frac{Q^2}{x+a}$
$E_{p_2}=-k \frac{Q^2}{x-a}$

$E_{p}=E_{p_1}+E_{p_2}=-kQ^2 (\frac{1}{x+a}+\frac{1}{x-a})$

As the electric field is conservative, the work from $\infty$ to $0$ will be the difference of potential energy between those points, so:

$W_{-\infty,0}=\Delta E_p=E(0)-E(-\infty)=0$

But this is not the solution at my textbook... Am I missing something?

By the way: as the energy needed to cross the $x=-a$ point is infinite, how would it be possible to arrive at the $O$ point?

 

[SammyS]

I don't recall having heard this term before: "punctual charge".
As near as I can tell it refers to a point charge.

 

[etotheipi]

By the way: as the energy needed to cross the $x=-a$ point is infinite, how would it be possible to arrive at the $O$ point?

You don't have to assemble the charges in this way. The electric force is conservative, which means that the work it does only depends on the start and end points. You could start off with the third charge at infinity, bring it toward the others, do a few laps around the other two, use it to play a quick game of table tennis and then finally slot it into position. The work done by the electric field on this charge will be the same!

 

[SammyS]

Homework Statement:: Two punctual charges $+Q$ are placed along the $X$-axis at the points $x=−a$ and $x=a$. Find the work to move a third charge $−Q$ from the infinity to the origin.
Relevant Equations:: $E_{p}=-k \frac{Q^2}{r}$

I have not clear how to solve this problem. Here it is my attempt at a solution:

Let the charge at $-a$ be the number one and the one at $+a$ the number two. the potential energy of the punctual charge $-Q$ due to each charge +Q will be then $E_{pi}=-k \frac{Q^2}{r_i}$, whit $r_i$ the distance between then. So, if the charge $-Q$ is at a distance $x$ from the origin, the total potential energy will be the sum of those ones due to each charge $+Q$:

$E_{p_1}=-k \frac{Q^2}{x+a}$
$E_{p_2}=-k \frac{Q^2}{x-a}$

$E_{p}=E_{p_1}+E_{p_2}=-kQ^2 (\frac{1}{x+a}+\frac{1}{x-a})$

As the electric field is conservative, the work from $\infty$ to $0$ will be the difference of potential energy between those points, so:

$W_{-\infty,0}=\Delta E_p=E(0)-E(-\infty)=0$

But this is not the solution at my textbook... Am I missing something?

By the way: as the energy needed to cross the $x=-a$ point is infinite, how would it be possible to arrive at the $O$ point?

When the location of the third charge is such that $-a < x < a$, your distances are incorrect. Distance is a non-negative quantity.

Another idea is to approach origin along the y-axis.

 

[AndersF]

I don't recall having heard this term before: "punctual charge".
As near as I can tell it refers to a point charge.

It's true, I was referring to point charges xD

 

[AndersF]

When the location of the third charge is such that $-a < x < a$, your distances are incorrect. Distance is a non-negative quantity.

Another idea is to approach origin along the y-axis.

Ok, so the expression of the potential energy should then be written in a different manner to avoid negative distances? What would be the equation?

I hadn't written it, but the correct result, according to my textbook, should be $W=\Delta E_{p}=-2 k \frac{Q^{2}}{a}$

 

[AndersF]

When the location of the third charge is such that $-a < x < a$, your distances are incorrect. Distance is a non-negative quantity.

Another idea is to approach origin along the y-axis.

Ok, thanks, here it is the correct expression:

$E_{p}=E_{p_1}+E_{p_2}=-kQ^2 (\frac{1}{|x+a|}+\frac{1}{|x-a|})$

 

[SammyS]

Ok, thanks, here it is the correct expression:

$E_{p}=E_{p_1}+E_{p_2}=-kQ^2 (\frac{1}{|x+a|}+\frac{1}{|x-a|})$

Correct. Using this expression should give you the correct result as given by the textbook.

By the way: