Electric Field - Charged Rod Question

[Physicus2]

Homework Statement

A charged rod with uniform charge per length and total charge Q is placed along the z-axis with one end at the origin. The rod is located on the positive z-axis according the the diagram accompanying the problem (not shown). Find the electric field E(x,0,0) at any point along the x-axis.

Homework Equations

1) I have been given the integral x dx / (x^2 + a^2)^3/2.
2) I have been given a second integral dx / (x^2 + a^2)^3/2

The Attempt at a Solution

Since I know I need to use those two integrals, I solved them first:

1) = - 1 / square root of (x^2 + a^2)
2) = x / a * square root of (x^2 + a^2)

I realize that the total field will be the vector sum of all the segments of the rod (and so I imagine that even though I'm solving for only the x component that it has more than just that). I believe that I need to use the equation E = ke integral of dq / r^2 unit vector r. In this, r is the distance from the charge element to a point and the unit vector r is directed "from the element toward the point." I've played around with this a bit, but I'm at a loss. I realize that I will need to use the two integrals (otherwise they wouldn't have been given in the problem), but I'm not sure where exactly I need to utilize them. I'm not sure where to go with the equation for E that I expect to need, either. Any help would be very much appreciated.

 

[chaoseverlasting]

You take a small element dz at a distance z from the origin whose charge is $$dq=\lambda dz$$.
Then the electric field at a distance x on the x-axis is $$dE(x)=\frac{Kdq}{(x^2+z^2)}$$. Integrate this from 0 to $$\frac{Q}{\lambda}$$ and that I think is your answer. Here, $$\lambda$$ is charge per unit length.

 

[m2003]

Dear student,

Thank you for your question. It seems like you have made a good start in solving this problem. As you mentioned, the total electric field at a point due to a charged rod can be found by summing the contributions from all the segments of the rod. This can be done using the equation E = ke integral of dq / r^2 unit vector r, where r is the distance from the charge element to the point of interest and the unit vector r is directed from the charge element towards the point.

In this problem, we can consider each segment of the rod as a small charge element dq, with a charge of dq = λ dx, where λ is the linear charge density and dx is the length of the segment. Therefore, the total charge Q can be written as an integral of λ dx along the length of the rod.

Now, to find the electric field at a point (x,0,0) along the x-axis, we can consider the contribution from each segment of the rod, which can be written as E = ke λ dx / r^2 unit vector r. Here, r is the distance from the segment to the point (x,0,0). As you have already found the integrals for 1/r^2 and x/r^2, you can use them to find the electric field at a point on the x-axis by integrating over the length of the rod.

I hope this helps. Keep up the good work!