Confused with this integral from Griffith's electrodynamics.

[jaded2112]

Homework Statement

The problem is to find the electric field due to a line charge of length 2L, above the mid point at a distance d.

Relevant Equations

This was pretty straightforward and I solved the resulting integral but came to a confusing part where the book plugged the limits from -L to L and came up with the first solution whereas i plugged it in from 0 to 2L and came up with a different solution. I was just wondering where i screwed up.

 

[mitochan]

Your solution shows electric field not at P but say at Q where QA=d, angle QAB is rectangle and, A and B are the ends of the line.

 

[jaded2112]

Your solution shows electric field not at P but say at Q where QA=d, angle QAB is rectangle and, A and B are the ends of the line.

Are you saying that evaluating it over 0 to 2L would not make it symmetric? If that is the case, then if i were to say, shift the line charge L units to the right and put the limits from 0 to 2L would it be correct? I'm abit confused as to why I need to evaluate the integral by specifying the position of dx with respect to the mid point of the line. Thanks.

 

[mitochan]

I'm abit confused as to why I need to evaluate the integral by specifying the position of dx with respect to the mid point of the line. Thanks.

You do not have to. And if your position is not above the mid point, as Q is not, you observe not only vertical but also horizontal component of E , coefficients of which are sin$\theta$ and do not vanish in integration.

 

[vela]

What's your reason for integrating from 0 to $2L$?

 

[jaded2112]

What's your reason for integrating from 0 to $2L$?

Well when i was working on my soln, i drew a figure where the length of the line charge extendend from 0 to 2L which is different from the book(Griffith drew it from - L to L as you can see from the figure in my post) . So i subdivided the line into two equal x's from the mid point and integrated it from there.

 

[jaded2112]

You do not have to. And if your position is not above the mid point, as Q is not, you observe not only vertical but also horizontal component of E , coefficients of which are sin$\theta$ and do not vanish in integration.

I understand it can be done in a way
which takes advantage of the symmetry of the problem, but i was wondering how to solve it by extending the line charge from 0 to 2L rather than - L to L. I apologize if this seems trivial.

 

[mitochan]

Say the line is -L to L, Px=0. The distance from the line element is
\sqrt{(x-0)^2+d^2}=\sqrt{x^2+d^2}
Integration is easy as you do.

Say the line is 0 to 2L, Px=L. The distance from the line element is
\sqrt{(x-L)^2+d^2}
It appears a litte bit more tedious to do integration but the same result, isn't it?

 

[jaded2112]

Say the line is -L to L, Px=0. The distance from the line element is
\sqrt{(x-0)^2+d^2}=\sqrt{x^2+d^2}
Integration is easy as you do.

Say the line is 0 to 2L, Px=L. The distance from the line element is
\sqrt{(x-L)^2+d^2}
It appears a litte bit more tedious to do integration but the same result, isn't it?

Yes, it does seem more tedious but I am still confused as to why we treat Px as the origin, i.e. why do we measure the distance from the length element on the x-axis (x-L) from Px as opposed to measuring it from (0,0), where we would

 

[jaded2112]

Yes, it does seem more tedious but I am still confused as to why we treat Px as the origin, i.e. why do we measure the distance from the length element on the x-axis (x-L) from Px as opposed to measuring it from (0,0), where we would

Never mind i think i got it now. Thanks for your time.