## Section 8.8 Potential due to an Infinite Line of Charge

In Section 8.7, we found the electrostatic potential due to a finite line of charge. The answer

was an unilluminating, complicated expression involving the logarithm of a fraction. Suppose, however, that the voltmeter probe were placed quite close to the charge. Then, to a fairly good approximation, the charge would look like an infinite line. Perhaps the expression for the electrostatic potential due to an infinite line is simpler and more meaningful. In principle, we should be able to get this expression by taking the limit of Equation (8.8.1) as \(L\) goes to infinity.

The denominator in this last expression goes to zero in the limit, which means that the potential goes to infinity. What has happened? Remember that we assumed that the ground probe was at infinity when we wrote our original integral expression for the potential, namely (4.1.1). Therefore, as we let the line charge become infinitely long, in the limit, it reaches the ground probe. So, of course, the potential difference between the ground probe and the active probe is infinite. One of the probes is touching the charge. This problem will occur whenever the (idealized) source extends all the way to infinity.

What is the resolution? We must move the ground probe somewhere else. Where else? Anywhere that's not touching the charge density. Let's choose to put the ground probe at

Now, we want to calculate the difference in potential between the active probe and the ground probe. Do we need to start all over again? No, we can use the expression for the potential due to a finite line, namely (8.8.1), if we are careful about the order in which we do various mathematical operations.

Notice that, even though we have written (8.8.1) as if it were the expression for \(V(s,0,0)\text{,}\) it is really the expression for the potential difference between the two probes, i.e. \(V(s,0,0)-V(\infty,0,0)\text{.}\) The potential difference that we want, i.e. \(V(s,0,0)-V(s_0,0,0)\) can be found by subtracting two expressions like (8.8.1), one evaluated at \(s\) and one evaluated at \(s_0\text{.}\)

Notice that each of the terms in the third line is separately infinite in the limit that \(L\rightarrow\infty\text{.}\) In effect, we are trying to subtract infinity from infinity and still get a sensible answer. We can do this by doing the subtraction *before* we take the limit, This process for trying to subtract infinity from infinity by first putting in a “cut-off,” in this case, the length of the source \(L\text{,}\) so that the subtraction makes sense and then taking a limit, is a process that is used often in advanced particle physics. In those cases, the process is called “renormalization.”

In the last line (8.8.8), we see that the troubling infinities have canceled. It is now safe to take the limit as \(L\rightarrow\infty\) to find the potential due to an infinite line.

### Question 8.8.1. Taking the limit.

In the second to the last line, we kept only the highest order term in each of the four Laurent series inside the logarithm. Why was it ok to do this?

All of the other terms in each Laurent series, including the terms that are not explicitly written, have factors of \(L\) in the denominator. Each of these terms goes to zero in the limit, so only the leading term in each Laurent series survives.

### Question 8.8.2. The z-dependence of \(V\).

What is the \(z\)-dependence of the potential?

The derivation in Section 8.7 for the potential due to a finite line of charge *assumed* that the point where the potential was evaluated was at \(z=0\text{.}\) So, technically we have only found the potential due to the infinite charge at \(z=0\text{.}\) However, once we take the limit that \(L\rightarrow\infty\text{,}\) we can no longer tell where the center of the line is. Therefore, the resulting potential in Equation (8.8.11) is valid for all \(z\text{.}\)

### Question 8.8.3. Sign of the potential.

Notice that if \(s>s_0\text{,}\) then the argument of the logarithm is less than one and the electrostatic potential is negative. If \(s\lt s_0\text{,}\) then the the electrostatic potential is positive. Why is this expected?

The potential is a continuous function which is infinity on the line of charge and decreases monotonically as you move away from the charge. Since we chose to put the zero of potential at \(s_0\text{,}\) the potential must change sign there.

### Question 8.8.4. Choosing other points for the zero of potential.

When we chose the potential at the point (8.8.2), we chose both \(\phi_0=0\) and \(z_0=0\text{.}\) What would have happened if we made different choices?

Notice that the formula for the potential due to a finite line of charge (8.8.1) does not depend on the angle \(\phi\text{.}\) This is expected because of the spherical symmetry of the problem. Therefore, the calculation would not change if we chose \(\phi_0\ne 0\text{.}\) However, the calculation in Section 8.7 for the potential due to a finite line of charge *assumed* that the point where the potential was evaluated was at \(z=0\text{.}\) We would have to redo the entire calculation from both that section and this one if we wanted to move \(z_0\) to a point other than zero. Terms involving \(z_0\) would appear in the calculation up until the time we take the limit that the length of the line \(L\) goes to infinity. In the limit, all of the terms involving \(z_0\) have to go to zero, because at that stage, the problem gains a translational symmetry along the \(z\)-axis. We leave this latter calculation as a not very illuminating exercise for the energetic reader.