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2.3: Computing Potential Fields for Known Charge Distributions

  • Page ID
    17356
  • Electrostatic Potential Field of Straight Line Segment of Uniform Density

    We return to the problem we first solved way back in Section 1.3. Actually, we are going to take it a step further than we did before, which will show the power of approaching the problem of computing fields with potentials. In our direct calculation of the electric field, we punted the computation of the \(z\)-component due to the lack of symmetry. It's not that it was impossible to do, but the vector element of the calculation was daunting, and in any case we now have a different approach that is easier to use. So the physical set-up is the same as before – a uniform linear segment of charge located along the \(z\)-axis from \(z=-a\) to \(z=+a\). But now we seek the potential field at a position a distance \(r\) from the \(z\)-axis, and a distance \(z\) from the \(\left(x,y\right)\) plane. We still have cylindrical symmetry, which means we don't have to specify the polar angle of the position of the point in space.

    Figure 2.3.1a – Electrostatic Potential Field of a Uniform Line Segment

    line_of_charge_potential.png

    We start the same way as we did for the electric field – identifying an element of charge, setting up a coordinate system, and determining the distance from the point charge to the point in space:

    Figure 2.3.1b – Electrostatic Potential Field of a Uniform Line Segment

    line_of_charge_potential2.png

    To start the math, we make several notes: First, unlike the electric field, we are not dealing with vectors, so we don't have to track components (yay!). Second, we need to reference the zero point of potential, and we will do so with the usual \(V\left(\infty\right)=0\). Finally, it should be noted that it is a bit awkward to integrate over the \(z\) variable, while our final answer is a function of the \(z\) component of the position in space, so we have named our integration variable \(z'\).

    \[\left. \begin{array}{l} dq=\lambda \;dz' \\dV=\dfrac{dq}{4\pi \epsilon_o R} \end{array} \right\} \;\;\; \Rightarrow \;\;\; V\left(r,z\right) = \int\limits_{-a}^{+a} \dfrac{\lambda\;dz'}{4\pi \epsilon_o \sqrt{r^2+\left(z'-z\right)^2}} \]

    With a substitution of \(u\equiv z'-z\), the integral becomes easy to look up in an integral table:

    \[V\left(r,z\right) = \dfrac{\lambda}{4\pi \epsilon_o} \int\limits_{-z-a}^{-z+a} \dfrac{du}{\sqrt{r^2+u^2}} = \dfrac{\lambda}{4\pi \epsilon_o}\left[\ln\left(\sqrt{r^2+u^2}+u\right)\right]_{-z-a}^{-z+a} = \dfrac{\lambda}{4\pi \epsilon_o} \ln\left(\dfrac{\sqrt{r^2+\left(z-a\right)^2}+a-z}{\sqrt{r^2+\left(z+a\right)^2}-a-z}\right)\]

    Okay, so that's quite a mess, but consider this: The set-up and the math to get here was not all that tough, and we have so much more information here. This is the potential field throughout all of space. When we calculated the electric field for this charge distribution, we were somewhat daunted by the lack of symmetry that occurs off the \(x\)-axis, making dealing with the vector components off the \(x\)-axis problematic (not impossible, but certainly not much fun). And now if we want the electric field at any point in space (the \(\widehat z\) component as well as the \(\widehat r\) component!), we only need to take derivatives, since the potential field is related to the electric field through the gradient.

    Electrostatic Potential Field of Long Straight Line of Uniform Density

    As with the case of the electric field, the potential for a long ("infinite") line of charge is of importance to us. So the natural step to take is to simply use the result above, and take the limit as \(a\rightarrow \infty\). But when we try to do this, we find that the limit diverges. This is puzzling, as the electric field comes out right from the gradient, and it doesn't diverge in the limit. The answer has to do with our choice of where the potential is chosen to be zero, and we can fix the problem by simply choosing another position for the zero potential. Let's choose the positionfor zero potential to be: \(r=r_o,\;z=0\). If we do this, we get a revised version of our potential function for the line segment:

    \[V\left(r,z\right) = \dfrac{\lambda}{4\pi \epsilon_o} \left[\ln\left(\dfrac{\sqrt{r^2+\left(z-a\right)^2}+a-z}{\sqrt{r^2+\left(z+a\right)^2}-a-z}\right)-\ln\left(\dfrac{\sqrt{r_o^2+a^2}+a}{\sqrt{r_o^2+a^2}-a}\right)\right]\]

    The reader can easily confirm that in fact \(V\left(r_o,0\right)=0\). It's not immediately clear how this rescues our limit of \(a\rightarrow\infty\). To see how this happens requires a bit of math, but it is useful to see, so here goes... First, it's clear that when the line becomes infinitely-long, the value of \(z\) is irrelevant, and might as well be set to zero. Plugging this in, and dividing the numerator and denominator of both natural log arguments by \(a\) simplifies our potential to:

    \[V\left(r\right) = \dfrac{\lambda}{4\pi \epsilon_o} \left[\ln\left(\dfrac{\sqrt{1+\frac{r^2}{a^2}}+1}{\sqrt{1+\frac{r^2}{a^2}}-1}\right)-\ln\left(\dfrac{\sqrt{1+\frac{r_o^2}{a^2}}+1}{\sqrt{1+\frac{r_o^2}{a^2}}-1}\right)\right]\]

    Now we make use of the following expansion to first order for very small \(\epsilon\):

    \[\sqrt{1+\epsilon}\approx 1+\frac{1}{2}\epsilon \]

    In our case, of course the ratios with \(a^2\) in the denominator are very small, so we get:

    \[V\left(r\right) = \dfrac{\lambda}{4\pi \epsilon_o} \left[\ln\left(\dfrac{1+\frac{r^2}{2a^2}+1}{1+\frac{r^2}{2a^2}-1}\right)-\ln\left(\dfrac{1+\frac{r_o^2}{2a^2}+1}{1+\frac{r_o^2}{2a^2}-1}\right)\right] = \dfrac{\lambda}{4\pi \epsilon_o} \left[\ln\left(4\dfrac{a^2}{r^2}+1\right)-\ln\left(4\dfrac{a^2}{r_o^2}+1\right)\right] = \dfrac{\lambda}{4\pi \epsilon_o}\ln\left(\dfrac{\frac{1}{r^2}+\frac{1}{4a^2}}{\frac{1}{r_o^2}+\frac{1}{4a^2}}\right)\]

    Now take our limit of \(a\rightarrow\infty\) at last, to obtain:

    \[V\left(r\right) = \dfrac{\lambda}{4\pi \epsilon_o} \ln\left(\dfrac{r_o^2}{r^2}\right) = \dfrac{\lambda}{2\pi \epsilon_o} \ln\left(\dfrac{r_o}{r}\right)\]

    We will find this quite useful for cylindrical symmetries. It clearly still vanishes at \(r=r_o\) and its negative gradient gives the correct field for the long line of uniform charge (Equation 1.3.23).

    An Identity from Vector Calculus

    We derived the negative-gradient relationship between the potential and the electric field from the same relation between a conservative force and its associated potential energy (Equation 2.2.7). For a force to have an associated potential energy, it is necessary that it be conservative. We have been assuming all along that the electric force is conservative. As we will see later, this is actually not always the case. It turns out that the electromagnetic field is conservative, but it is possible for the magnetic field to transfer energy to/from the electric field, making the electric field by itself not conservative. For this to be the case, the source charges need to be moving, and since we are still discussing only electrostatics, we can safely continue to use the electrostatic potential and the negative gradient relation.

    The existence of a potential energy function is sufficient to prove that a force is conservative, though proving this can be troublesome, without the tools provided by vector calculus. The simple test for whether a force is conservative is if its curl vanishes:

    \[\overrightarrow \nabla \times \overrightarrow F = 0 \;\;\; \leftrightarrow \;\;\; F\;is\;conservative\;\;\; \leftrightarrow \;\;\; \overrightarrow F=-\overrightarrow \nabla U\]

    The reason this works as a test is that the geometry of the curl and gradient are such that the curl of a vector field that comes from a gradient of a scalar field is always identically zero:

    \[\overrightarrow \nabla \times \left[\overrightarrow \nabla \left(anything\right)\right] \equiv 0\]

    So if the force can be written as the negative gradient of a potential energy function, then its curl must vanish, and this corresponds to a conservative force. Extending this to electrostatics, we see that if the electric field can be expressed as the negative gradient of a potential, then its curl vanishes. So we have:

    \[\overrightarrow \nabla \times \overrightarrow E = 0 \;\;\; \leftrightarrow\;\;\; electrostatics \;\;\; \leftrightarrow \;\;\; \overrightarrow E=-\overrightarrow \nabla V\]