Total charge of semi-infinite rod?

[TwoTruths]

Homework Statement

A thin insulating rod of length L is placed such that its end point is distance R away from a semi-infinite insulating line. The line has linear charge density \lambda. The rod has charge Q. Both are distributed uniformly. What is the force exerted on the rod?

Define Q_o = charge of the insulating line

Homework Equations

Integrals
E=\frac{kq}{r^2}

The Attempt at a Solution

Putting aside my main question, I figured I would take the force exerted on an infinitesimally small segment on the rod at distance r. This gives:

\int \frac{k(Q/L)Q_o dr}{r^2} = k(Q/L)Q_o \int \frac{dr}{r^2}

My question is: what is the charge of a semi-infinite line with charge density \lambda? In other words, what is the length of something that is semi-infinite?

 

[kuruman]

The length of a line that is semi-infinite is, well, infinite. The charge on it is also infinite. That's not what should worry you because you are given that the semi-infinite line carries linear charge density λ.

What should worry you is the way you set up the integral. You cannot use Coulomb's Law that way. First you need to find the electric field produced by the semi-infinite line by doing an integral from zero to infinity - from one end of the rod to the other. Once you have the electric field, you need to put the rod in it and calculate the force on it. You need to do a second integral for that.

 

[RoyalCat]

This is a question we had in class a while ago. :)

It's easier to look at the force that the infinite wire exerts on the finite wire (Newton's third law: \vec F_{12}=-\vec F_{21})

There you only have to integrate over the finite length of the finite wire, which is simple since you can use Gauss' law to find the field of the infinite wire quite easily.

As for your second question, consider the following:

For a uniformly charged wire of length L and charge Q
\lambda\equiv \frac{Q}{L}

Since this charge is uniform, no matter what section we take, the ratio between the charge on it, and its length will remain constant at \lambda

\lambda=\frac{dQ}{dL}

 

[TwoTruths]

Thank you for the help. I figured out the answer after revisiting the problem a couple hours later. My mistake was taking the integral to find the charge of the semi-finite line before taking the integral for the electric field and thus the force. I simply needed to leave the integral as it was for the charge, after which I could easily put it into the electric field integral, then subsequently the force integral. I'll spare the details.

To RoyalCat, the rod and line are parallel, or on the same axis. That means there's no happy Gaussian surface in order to make Gauss's Law a viable "easy way out" solution, as I call it. There's no symmetry that can be found parallel for a line. I believe you are thinking of the symmetry you can find perpendicular to an infinite line.

 

[RoyalCat]

Thank you for the help. I figured out the answer after revisiting the problem a couple hours later. My mistake was taking the integral to find the charge of the semi-finite line before taking the integral for the electric field and thus the force. I simply needed to leave the integral as it was for the charge, after which I could easily put it into the electric field integral, then subsequently the force integral. I'll spare the details.

To RoyalCat, the rod and line are parallel, or on the same axis. That means there's no happy Gaussian surface in order to make Gauss's Law a viable "easy way out" solution, as I call it. There's no symmetry that can be found parallel for a line. I believe you are thinking of the symmetry you can find perpendicular to an infinite line.

Oh oh oh, I see now. My mistake. I thought they were parallel, and not on the same axis. I see now that Gauss' law is irrelevant here.

 

[TwoTruths]

Sorry for the vagueness of the question. The rod was both parallel and on the same axis.

 

[RoyalCat]

Sorry for the vagueness of the question. The rod was both parallel and on the same axis.

Gauss' law wouldn't have made things any more simple either, since we're dealing with an edge, so the field wouldn't be perpendicular to any easy to imagine surface, meaning an integral approach would still be necessary, and using regular old Coloumb's Law is much more simple here.