Calculating the work done using a line integral

[Burhan Uddin]

Homework Statement

1. a) A point charge + q is placed at the origin. By explicitly calculating the relevant line integral, determine how much external work must be done to bring another point charge + q from infinity to the point r2= aŷ ? Consider the difference between external work and work done by the force. {2}
   
2. b) Both of these charges are held fixed and a third charge + q is now brought from infinity to position r= ax̂ + ay.̂ Repeat the same type of calculation as in a). What is the total stored potential energy of this three-charge system? {2}

Homework Equations

U=-W=∫ F.dr

The Attempt at a Solution

check the image[/B] for my full solution

My question is why do i need to include the sqrt(2) in the highlighted part, shouldn't it be sqrt(2)*a because that is the distance between those 2 points. Any explanation is much appreciated :)

 

[kuruman]

It is not clear to me what the force on the third charge is. Obviously it depends on the integration path that you choose. So what path did you choose? I would recommend coming in from infinity along the perpendicular bisector between the first two charges to exploit the symmetry.

 

[haruspex]

My question is why do i need to include the sqrt(2) in the highlighted part,

I am confused on a couple of fronts.
You describe this as your solution. If you don't know why you need a √2, why did you include it?
Secondly, it looks like these two terms are supposed to represent the force vector from the charge at the origin (1) on the third charge (3) when it is at (x,y). But I do not see factors converting the whole force, $\frac{kq^2}{x^2+y^2}$, into its x and y components.

A possible source of confusion is that the writing is too small to be clear in some places. Are those $\hat x, \hat y$ or $\vec x, \vec y$? And is that a $\hat y.dx$ where I would expect $\hat y.dy$?

 

[haruspex]

It is not clear to me what the force on the third charge is. Obviously it depends on the integration path that you choose. So what path did you choose? I would recommend coming in from infinity along the perpendicular bisector between the first two charges to exploit the symmetry.

The force depends on the location, not the path. The integral $\int \vec F.\vec{dr}$ can be performed without needing to specify a path.

 

[kuruman]

The force depends on the location, not the path. The integral $\int \vec F.\vec{dr}$ can be performed without needing to specify a path.

Perhaps I was not clear. The integrand is a typical element contributing to the sum of such elements and involves an expression for the field at a typical point along the path. The path is a collection of locations on which the field depends. Different path, different integrand. That's what I meant. Also, I don't see how the integral can be performed without needing to specify a path. The problem explicitly requires to calculate the "relevant line integral."

 

[haruspex]

Perhaps I was not clear. The integrand is a typical element contributing to the sum of such elements and involves an expression for the field at a typical point along the path. The path is a collection of locations on which the field depends. Different path, different integrand. That's what I meant. Also, I don't see how the integral can be performed without needing to specify a path. The problem explicitly requires to calculate the "relevant line integral."

Ah, yes - quite so. Apologies.

 

[Burhan Uddin]

I am confused on a couple of fronts.
You describe this as your solution. If you don't know why you need a √2, why did you include it?
Secondly, it looks like these two terms are supposed to represent the force vector from the charge at the origin (1) on the third charge (3) when it is at (x,y). But I do not see factors converting the whole force, $\frac{kq^2}{x^2+y^2}$, into its x and y components.

A possible source of confusion is that the writing is too small to be clear in some places. Are those $\hat x, \hat y$ or $\vec x, \vec y$? And is that a $\hat y.dx$ where I would expect $\hat y.dy$?

They are $\hat x, \hat y$ and its $\hat y.dx$ because dy=dx.
sqrt(2) was included because i knew the answer but wasn't sure on the solution.