This question is an exact duplicate of:

How to find the net electric force exerted on rod 2 by rod 1, both being on the x-axis, both having the same length and constant linear charge density, being some distance apart?

More specifically, if rod 1 is on the x-axis from $-L$ to $L$ and rod 2 is on the x-axis from $L+x$ to $3L+x$, and both have the same lambda linear charge density, what is the net force of rod 1 on rod 2?

I understand that the electric field created by the first rod is not uniform over the space occupied by the second rod.

I know $F=qE$, where $F$ and $E$ are vectors, and the electric field for continuous charge distribution formula.

However, I don't know how to proceed with solving this problem using this information. In particular, there are two position vectors that vary (identify positions with different electric fields, not one point), instead of 1 as usual. Please help by identifying the strategy and starting point to solve this problem. I drew a diagram and wrote down the equations but further than this I've failed to go. Thank you for your help.


marked as duplicate by sammy gerbil, John Rennie homework-and-exercises Oct 2 '18 at 10:28

This question was marked as an exact duplicate of an existing question.


You will have to perform two integrals. One to get the field at an arbitrary point away from one rod, and then you have to integrate the field over the other rod to get the force.

To not get confused with the x-coordinte, let's say rod 2 has ends at $L+d$ and $3L+d$. Now let's consider the field produced by rod 1 at some arbitrary point $x$ to its right. All we need to do is integrate the contributions from all charges. In other words, using Coulomb's law, and using $x'$ to represent the coordinates of the charge distribution of rod 1, $$d\mathbf E=\frac{1}{4\pi\epsilon_0}\frac{dq}{(x-x')^2}\mathbf{\hat x}=\frac{1}{4\pi\epsilon_0}\frac{\lambda dx'}{(x-x')^2}\mathbf{\hat x}$$

$$\mathbf E(x)=\frac{1}{4\pi\epsilon_0}\int_{-L}^L \frac{\lambda dx'}{(x-x')^2}\mathbf{\hat x}$$

Once you have $\mathbf E(x)$, you can then use the relationship you mention $\mathbf F=q\mathbf E$ and integrate over the second rod. In other words $$d\mathbf F=dq\mathbf E(x)=\lambda \mathbf E(x) dx$$

$$\mathbf F=\int_{L+d}^{3L+d}\lambda \mathbf E(x) dx=\frac{\lambda^2 \mathbf{\hat x}}{4\pi\epsilon_0}\int_{L+d}^{3L+d}\int_{-L}^L \frac{1}{(x-x')^2}dx'\ dx$$

I will leave the integrals to you. This process is something you can always fall back on in principle: 1) determine the electric field in the region of space in question and 2) determine the force that each charge in question experiences based on that field.

  • 1
    $\begingroup$ Thank you very much have a great night! I understand everything now. $\endgroup$ – Arthur Alex Karapetov Oct 2 '18 at 2:35

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