# Electric field and electric scalar potential of two perpendicular wires

So, I've been trying to come up with an idea that seems plausible to me for the last half week or so, but I'm stuck and I can't seem to get anywhere with this problem.

The problem is as follows (probably crude translation of the task):

Consider two uniformly charged wires of a certain line charge density $\lambda$ that intersect at a 90 degree angle. Compute the electric field $\vec{E}(\vec{r})$ and the electric scalar potential $\phi(\vec{r})$.

Now, here's my progress (if you want to call it that) so far:

From Gauss' Law we know that the electric field of a single charged wire should be

$$\vec{E}_1(\vec{r})=\frac{Q}{2\pi\epsilon_0r_1l}\hat{r}_1$$

Using the line charge density $\lambda=\frac{Q}{l}$, I know that $Q=\lambda l$. Once I insert that into the equation for my electric field and cross out the $l$s I get

$$\vec{E}_1(\vec{r})=\frac{\lambda}{2\pi\epsilon_0r_1}\hat{r}_1$$

This should be my electric field for just one charged wire, right? Now if I add another wire that is exactly the same I could make use of the superposition principle and add them together. Then my total electric field should look something like this:

$$\vec{E}_{total}(\vec{r})=\frac{\lambda}{2\pi\epsilon_0}(\frac{1}{r_1}\hat{r_1}+\frac{1}{r_2}\hat{r_2})$$

With $\hat{r_1}$ and $\hat{r_2}$ being the unit vectors into the respective direction of a test charge and $r_1$ and $r_2$ being the distances from that test charge to my wires. Here's my first big problem: I do not know whether this is actually the correct answer to the problem and I lack the confidence to just assume that it is. I can't seem to find any means of checking my result either.

As the second part of the question relies heavily on the first any further results may already be wrong from the get-go. Here's my attempt at finding the electric scalar potential:

Instead of using the equation for $\vec{E}_{total}$ I tried to compute the two potentials individually and add them together afterwards relying on the superposition principle once more.

$$\phi_1(\vec{r})=-\int_\infty^r{\vec{E}_1(\vec{r})dr'}=-\int_\infty^r{\frac{\lambda}{2\pi\epsilon_0r'_1}dr'}=-\frac{\lambda}{2\pi\epsilon_0}\int_\infty^r{\frac{1}{r'_1}dr'}$$

So, here's my next big problem. If I compute this integral and form the anti-derivative of $\frac{1}{r'_1}dr'$ I end up with:

$$\phi_1(\vec{r})=-\frac{\lambda}{2\pi\epsilon_0}[\ln{r'_1}]_\infty^r=-\frac{\lambda}{2\pi\epsilon_0}(\ln{r}-\ln{\infty})=-\frac{\lambda}{2\pi\epsilon_0}\ln{\frac{r}{\infty}}$$

Now I have no idea where to go from there … if I continue and plug in the integration limits I'd end up with some infinite potential so I tried a different approach and used a different equation for the scalar potential:

$$\phi_1(\vec{r})=\frac{1}{2\pi\epsilon_0}\frac{Q}{r_1}=\frac{1}{2\pi\epsilon_0}\frac{\lambda l}{r_1}$$

In one way or another this does resemble the above result, which makes it seem pretty plausible to me, so I continued with this and after creating a "second" potential like this one I added them together relying on the superposition principle:

$$\phi_{total}(\vec{r})=\frac{\lambda l}{2\pi\epsilon_0}(\frac{1}{r_1}+\frac{1}{r_2})$$

Again, I stand (sit) here with little to no confidence in my result.

• Your third equations is correct in a sense ... but not very useful. The variables $r_1$ and $r_2$ are referred to different coordinate systems, as are the unit vectors involved. You need to use one system of coordinates for everything. Consider placing the wires on the $x$ and $y$ axes, and express the field in terms of Cartesian coordinates $x$, $y$, $z$, $\hat{x}$, $\hat{y}$ $\hat{z}$. – garyp Nov 10 '15 at 20:43
• As @garyp says, you must define a coordinate system. You must also choose an origin, or at least a location where the wires cross. Also, are these finite or infinitely long wires? It might be easier to find the function for the potential first, then take the gradient to find the E-field: $\vec{E}=-\nabla \phi$ – Bill N Nov 10 '15 at 23:05
• (1/2) O.K. So I tried to do what you two recommended, but I just can't see it. I put them in the same coordinate system, with each wire running along one of the axes and I picked my origin at the point where they cross. Now if I try to turn any of the above equations into that form I end up with something that looks really similar to what I already have and I wouldn't know where to go from there: $\vec{E}_x(\vec{r})=\frac{\lambda}{2\pi\epsilon_0\sqrt{y^2+z^2}}\hat{P}$. $\hat{P}$ is the unit vector of the position of my test charge. That should be the same for both wires, since they're in … – Sephi- Nov 11 '15 at 12:00
• (2/2) … the same coordinate system. $\sqrt{y^2+z^2}$ is just the above $r_1$, meaning the distance from my point charge to the wire. It's the same for the second E-field, just that $r_2$ is $\sqrt{x^2+z^2}$ in this case. So my total E-field looks like this: $\vec{E}_{total}(\vec{r})=\frac{\lambda}{2\pi\epsilon_0}\hat{P}(\frac{1}{\sqrt{y^2+z^2}}+\frac{1}{\sqrt{x^2+z^2}})$, yet I don't see how this helps me. :/ As for your question @BillN, the task itself says nothing about the length of the wires, so I have to assume they're infinitely long. – Sephi- Nov 11 '15 at 12:02

So, I think I got it thanks to the tips you two gave me. As you mentioned, my single electric fields weren't incorrect, but I have to put them into the same coordinate system. The electric field for the wire running along the x-axis should look something like this then: $$\vec{E}_x(\vec{r})=\frac{\lambda}{2\pi\epsilon_0\sqrt{y^2+z^2}}\frac{1}{\sqrt{y^2+z^2}}\begin{pmatrix} 0 \\ y \\ z\end{pmatrix}=\underline{\frac{\lambda}{2\pi\epsilon_0y^2+z^2}\begin{pmatrix} 0 \\ y \\ z\end{pmatrix}}.$$ Where $\sqrt{y^2+z^2}$ is the distance $r_x$ from the wire/the x-axis and $\frac{1}{\sqrt{y^2+z^2}}\begin{pmatrix} 0 \\ y \\ z\end{pmatrix}$ is the unit vector $\hat{r}_x$ in the direction of the wire's field. The field for the wire along the y-axis is similar, just that now I have to regard the x-value rather than y-value. So it should look like this: $$\vec{E}_y(\vec{r})=\underline{\frac{\lambda}{2\pi\epsilon_0x^2+z^2}\begin{pmatrix} x \\ 0 \\ z\end{pmatrix}}.$$ Putting them together while relying on the superposition principle, I get a total electric field of $$\vec{E}_{total}(\vec{r})=\underline{\underline{\frac{\lambda}{2\pi\epsilon_0}\left(\frac{1}{y^2+z^2}\begin{pmatrix} 0 \\ y \\ z\end{pmatrix}+\frac{1}{x^2+z^2}\begin{pmatrix} x \\ 0 \\ z\end{pmatrix}\right)}}.$$
Now, if I start with just one potential and pick a finite reference point, I end up with something fairly similar to my above ideas just without the infinite potential: $$\begin{eqnarray*}\phi_{x}(\vec{r})=-\int_a^{r_x}{\vec{E}_x(\vec{r})dr'} &=& -\int_a^{r_x}{\frac{\lambda}{2\pi\epsilon_0}\frac{1}{r}dr'} \\ &=&-\frac{\lambda}{2\pi\epsilon_0}\int_a^{r_x}{\frac{1}{r}dr'} \\ &=&-\frac{\lambda}{2\pi\epsilon_0}\left[\ln{r}\right]_a^{r_x} \\ &=&-\frac{\lambda}{2\pi\epsilon_0}\left(\ln{r_x}-\ln{a}\right) \\ &=&\underline{-\frac{\lambda}{2\pi\epsilon_0}\ln{\frac{r_x}{a}}} .\end{eqnarray*}$$ Now I can do the same thing for the "y-wire's" potential using the same finite reference point a to get $$\phi_{y}(\vec{r})=\underline{-\frac{\lambda}{2\pi\epsilon_0}\ln{\frac{r_y}{a}}}.$$ $r_x$ and $r_y$ I know from above, so if I add the two potentials relying on the superposition principle, my total potential for both wires should look like this: $$\begin{eqnarray*}\phi_{total}(\vec{r})=\phi_{x}(\vec{r})+\phi_{y}(\vec{r}) &=& -\frac{\lambda}{2\pi\epsilon_0}\ln{\frac{r_x}{a}}+\left(-\frac{\lambda}{2\pi\epsilon_0}\ln{\frac{r_y}{a}}\right) \\ &=& -\frac{\lambda}{2\pi\epsilon_0}\left(\ln{\frac{r_x}{a}}+\ln{\frac{r_y}{a}}\right) \\ &=& \underline{\underline{-\frac{\lambda}{2\pi\epsilon_0}\ln{\frac{r_x\cdot r_y}{a^2}}}}.\end{eqnarray*}$$ Now if I plug in $r_x$ and $r_y$ I get $$\phi_{total}(\vec{r})=-\frac{\lambda}{2\pi\epsilon_0}\ln{\frac{\sqrt{y^2+z^2}\cdot \sqrt{x^2+z^2}}{a^2}}=\underline{\underline{-\frac{\lambda}{2\pi\epsilon_0}\ln{\frac{\sqrt{(y^2+z^2)(x^2+z^2)}}{a^2}}}}.$$ Whether this is really correct, I'll have to wait until we check it in class next week, but it's the first time I actually feel confident in my results.