Electric Potential on the z-axis

[Rockstar47]

Homework Statement

Use direct integration over the charge on a wire segment to find the potential V(z) on the z-axis due to a finite segment of wire with linear charge density lambda extending along the x-axis from x = - a/2 to +a/2

Homework Equations

The electric potential dV at some point P due to the charge element dq is
dV = ke dq / r
where r is the distance from the charge element to point P.

Thus the total potential can be found by integrating:

V = ke integral of dq / r

The Attempt at a Solution

I'm sort of lost here, but I have a few thoughts. I'd appreciate some critique of my work here and hopefully some help...unless, of course, I'm right. Then letting me know that would surely help :D.

I assume the wire has length a (since it extends from - a/2 to +a/2). Using right-triangle geometry, I believe the distance from the x-axis to any element on the z-axis will be equal to the square root of a^2 + b^2 (where b is the distance on the z-axis). So, I'll integrate this over the limits x = o to x = a and I come up with V = ke lamdba integral of dx / square root of a^2 + b^2. Such an integral appears in the table in my book and so V becomes equal to
ke lamda ln(a + (square root of a^2 + b^2).

Am I close here...or way off? Thanks a bunch for any help!

 

[anathema]

Your approach is correct. The potential at point P due to the charge element dq is given by dV= ke dq/r, where r is the distance from the charge element to point P. In this case, the distance r is given by the Pythagorean theorem as you have correctly stated.

To find the total potential V, we need to integrate this expression over the entire length of the wire, from x=-a/2 to x=a/2. This gives us V=ke∫dq/r= ke∫lambda dx/√(a^2+b^2).

To evaluate this integral, we can use the substitution u=a^2+b^2, which gives us du/dx=2x. This means that dx=du/2x. Substituting this into our integral gives us V=ke∫lambda/2x du=ke/2∫lambda/u du.

Now, we can evaluate this integral as u goes from a^2 to a^2+b^2. This gives us V=ke/2 [lambda ln(a^2+b^2)-lambda ln(a^2)]. Simplifying this expression gives us the final result: V=ke lambda ln(a+√(a^2+b^2)).

So, your approach was correct and you were able to find the correct expression for the potential due to a finite segment of wire with linear charge density. Keep up the good work!

 

[m2003]

Your approach is on the right track. To find the potential at a point on the z-axis, we can use the equation V = ke integral of dq / r, where r is the distance from the charge element to the point on the z-axis.

Since we are only considering a finite segment of wire, we can break it up into small charge elements dq and integrate over the entire length of the wire. The distance r can be found using the Pythagorean theorem, as you mentioned.

Thus, the integral becomes V = ke integral of lambda dx / square root of (a^2 + b^2), where lambda is the linear charge density and dx is the small charge element.

Evaluating this integral from x = -a/2 to x = a/2, we get V = ke lambda ln(a + square root of (a^2 + b^2)) - ke lambda ln(a - square root of (a^2 + b^2)). Simplifying this expression, we get V = ke lambda ln((a + square root of (a^2 + b^2)) / (a - square root of (a^2 + b^2))).

This is the potential at a point on the z-axis due to a finite segment of wire with linear charge density lambda extending along the x-axis from x = -a/2 to x = a/2. I hope this helps clarify your approach.