10.3: Fields by Superposition
10.3.1 Electric field of a continuous charge distribution
Charge really comes in discrete chunks, but often it is mathematically convenient to treat a set of charges as if they were like a continuous fluid spread throughout a region of space. For example, a charged metal ball will have charge spread nearly uniformly all over its surface, and for most purposes it will make sense to ignore the fact that this uniformity is broken at the atomic level. The electric field made by such a continuous charge distribution is the sum of the fields created by every part of it. If we let the “parts” become infinitesimally small, we have a sum of an infinitely many infinitesimal numbers: an integral. If it was a discrete sum, as in example 3 on page 564, we would have a total electric field in the \(x\) direction that was the sum of all the \(x\) components of the individual fields, and similarly we'd have sums for the \(y\) and \(z\) components. In the continuous case, we have three integrals. Let's keep it simple by starting with a onedimensional example.
\(\triangleright\) What is the voltage associated with a point charge?
\(\triangleright\) As derived previously in selfcheck A on page 563, the field is
The difference in voltage between two points on the same radius line is
In the general discussion above, \(x\) was just a generic name for distance traveled along the line from one point to the other, so in this case \(x\) really means \(r\).
The standard convention is to use \(r_1=\infty\) as a reference point, so that the voltage at any distance \(r\) from the charge is
The interpretation is that if you bring a positive test charge closer to a positive charge, its electrical energy is increased; if it was released, it would spring away, releasing this as kinetic energy.
Example 9: Field of a uniformly charged rod 

\(\triangleright\) A rod of length \(L\) has charge \(Q\) spread uniformly along it. Find the electric field at a point a distance \(d\) from the center of the rod, along the rod's axis. \(\triangleright\) This is a onedimensional situation, so we really only need to do a single integral representing the total field along the axis. We imagine breaking the rod down into short pieces of length \(d z\), each with charge \(d q\). Since charge is uniformly spread along the rod, we have \(d q=\lambda d z\), where \(\lambda= Q/ L\) (Greek lambda) is the charge per unit length, in units of coulombs per meter. Since the pieces are infinitesimally short, we can treat them as point charges and use the expression \(kd q/ r^2\) for their contributions to the field, where \(r= d z\) is the distance from the charge at \(z\) to the point in which we are interested. \[\begin{align*} E_{z} &= \int \frac{ kd q }{ r^2} \\ &= \int_{ L/2}^{+ L/2} \frac{ k\lambda d z }{ r^2} \\ &= k \lambda \int_{ L/2}^{+ L/2} \frac{d z}{( d z)^2} \end{align*}\] The integral can be looked up in a table, or reduced to an elementary form by substituting a new variable for \(d z\). The result is \[\begin{align*} E_{z} &= k\lambda\left(\frac{1}{ d z}\right)_{ L/2}^{+ L/2} \\ &= \frac{ kQ}{ L} \left(\frac{1}{ d L/2}\frac{1}{ d+ L/2}\right) . \end{align*}\] For large values of \(d\), this expression gets smaller for two reasons: (1) the denominators of the fractions become large, and (2) the two fractions become nearly the same, and tend to cancel out. This makes sense, since the field should get weaker as we get farther away from the charge. In fact, the field at large distances must approach \( kQ/ d^2\) (homework problem 2). It's also interesting to note that the field becomes infinite at the ends of the rod, but is not infinite on the interior of the rod. Can you explain physically why this happens? 
Example 9 was onedimensional. In the general threedimensional case, we might have to integrate all three components of the field. However, there is a trick that lets us avoid this much complication. The voltage is a scalar, so we can find the voltage by doing just a single integral, then use the voltage to find the field.
Example 10: Voltage, then field 

\(\triangleright\) A rod of length \(L\) is uniformly charged with charge \(Q\). Find the field at a point lying in the midplane of the rod at a distance \(R\). \(\triangleright\) By symmetry, the field has only a radial component, \(E_R\), pointing directly away from the rod (or toward it for \(Q\lt0\)). The bruteforce approach, then, would be to evaluate the integral \(E=\int d\mathbf{E}\text{cos}\ \theta\), where \(d\mathbf{E}\) is the contribution to the field from a charge \(dq\) at some point along the rod, and \(\theta\) is the angle \(d\mathbf{E}\) makes with the radial line. It's easier, however, to find the voltage first, and then find the field from the voltage. Since the voltage is a scalar, we simply integrate the contribution \(dV\) from each charge \(dq\), without even worrying about angles and directions. Let \(z\) be the coordinate that measures distance up and down along the rod, with \(z=0\) at the center of the rod. Then the distance between a point \(z\) on the rod and the point of interest is \(r=\sqrt{ z^2+ R^2}\), and we have \[\begin{align*} V &= \int \frac{ kdq}{ r} \\ &= k\lambda \int_{ L/2}^{+ L/2}\frac{dz}{ r} \\ &= k\lambda \int_{ L/2}^{+ L/2}\frac{dz}{\sqrt{ z^2+ R^2}} \\ \end{align*}\] The integral can be looked up in a table, or evaluated using computer software: \[\begin{align*} V &= \left. k\lambda\: \text{ln}\left( z+\sqrt{ z^2+ R^2}\right)\right_{ L/2}^{+ L/2} \\ &= k\lambda\: \text{ln}\left(\frac{ L/2+\sqrt{ L^2/4+ R^2}}{ L/2+\sqrt{ L^2/4+ R^2}}\right) \\ \end{align*}\] The expression inside the parentheses can be simplified a little. Leaving out some tedious algebra, the result is \[\begin{equation*} V = 2 k\lambda\: \text{ln}\left(\frac{ L}{2 R}+\sqrt{1+\frac{ L^2}{4 R^2}}\right) \end{equation*}\]\ This can readily be differentiated to find the field: \[\begin{align*} E_{R} &= \frac{dV}{dR} \\ &= (2 k\lambda)\frac{ L/2 R^2 +(1/2)(1+ L^2/4 R^2)^{1/2}( L^2/2 R^3) }{L/2 R+(1+ L^2/4 R^2)^{1/2} } , \\ \text{or, after some simplification,} E_{R} &= \frac{ k\lambda L}{ R^2\sqrt{1+ L^2/4 R^2}} \end{align*}\] For large values of \(R\), the square root approaches one, and we have simply \(E_{R}\approx k\lambda L/ R^2= k Q/ R^2\). In other words, the field very far away is the same regardless of whether the charge is a point charge or some other shape like a rod. This is intuitively appealing, and doing this kind of check also helps to reassure one that the final result is correct. 
The preceding example, although it involved some messy algebra, required only straightforward calculus, and no vector operations at all, because we only had to integrate a scalar function to find the voltage. The next example is one in which we can integrate either the field or the voltage without too much complication.
Example 11: Onaxis field of a ring of charge 

\(\triangleright\) Find the voltage and field along the axis of a uniformly charged ring. \(\triangleright\) Integrating the voltage is straightforward. \[\begin{align*} V &= \int \frac{ kdq}{ r} \\ &= k \int \frac{dq}{\sqrt{ b^2+ z^2}} \\ &= \frac{ k}{\sqrt{ b^2+ z^2}} \int dq \\ &= \frac{ kQ}{\sqrt{ b^2+ z^2}} , \end{align*}\] where \(Q\) is the total charge of the ring. This result could have been derived without calculus, since the distance \(r\) is the same for every point around the ring, i.e., the integrand is a constant. It would also be straightforward to find the field by differentiating this expression with respect to \(z\) (homework problem 10). Instead, let's see how to find the field by direct integration. By symmetry, the field at the point of interest can have only a component along the axis of symmetry, the \(z\) axis: \[\begin{align*} E_{x} &= 0 \\ E_y &= 0 \end{align*}\] To find the field in the \(z\) direction, we integrate the \(z\) components contributed to the field by each infinitesimal part of the ring. \[\begin{align*} E_{z} &= \int dE_z \\ &= \int d\mathbf{E}\:\text{cos}\:\theta , \end{align*}\] where \(\theta\) is the angle shown in the figure. \[\begin{align*} E_{z} &= \int \frac{ kdq}{ r^2}\:\text{cos}\:\theta \\ &= k \int \frac{dq}{ b^2+ z^2}\:\text{cos}\:\theta \end{align*}\] Everything inside the integral is a constant, so we have \[\begin{align*} E_{z} &= \frac{ k}{ b^2+ z^2}\:\text{cos}\:\theta \int dq \\ &= \frac{ kQ}{ b^2+ z^2}\:\text{cos}\:\theta \\ &= \frac{ kQ}{ b^2+ z^2}\:\frac{ z}{ r} \\ &= \frac{ kQz}{\left( b^2+ z^2\right)^\text{3/2}} \end{align*}\]

In all the examples presented so far, the charge has been confined to a onedimensional line or curve. Although it is possible, for example, to put charge on a piece of wire, it is more common to encounter practical devices in which the charge is distributed over a twodimensional surface, as in the flat metal plates used in Thomson's experiments. Mathematically, we can approach this type of calculation with the divideandconquer technique: slice the surface into lines or curves whose fields we know how to calculate, and then add up the contributions to the field from all these slices. In the limit where the slices are imagined to be infinitesimally thin, we have an integral.
Example 12: Field of a uniformly charged disk 

\(\triangleright\) A circular disk is uniformly charged. (The disk must be an insulator; if it was a conductor, then the repulsion of all the charge would cause it to collect more densely near the edge.) Find the field at a point on the axis, at a distance \(z\) from the plane of the disk. \(\triangleright\) We're given that every part of the disk has the same charge per unit area, so rather than working with \(Q\), the total charge, it will be easier to use the charge per unit area, conventionally notated \(\sigma\) (Greek sigma), \(\sigma= Q/\pi b^2\). Since we already know the field due to a ring of charge, we can solve the problem by slicing the disk into rings, with each ring extending from \(r\) to \(r+dr\). The area of such a ring equals its circumference multiplied by its width, i.e., \(2\pi rdr\), so its charge is \(dq=2\pi\sigma rdr\), and from the result of example 11, its contribution to the field is \[\begin{align*} dE_{z} &= \frac{ kzdq}{\left( r^2+ z^2\right)^\text{3/2}} \\ &= \frac{2\pi\sigma kzrdr}{\left( r^2+ z^2\right)^\text{3/2}} \\ \end{align*}\] The total field is \[\begin{align*} E_{z} &= \int dE_{z} \\ &= 2\pi\sigma kz \int_0^{b} \frac{ rdr}{\left( r^2+ z^2\right)^\text{3/2}} \\ &= 2\pi\sigma kz \left. \frac{1}{\sqrt{ r^2+ z^2}} \right_{ r=0}^{ r=\text{b}} \\ &= 2\pi\sigma k\left(1\frac{ z}{\sqrt{ b^2+ z^2}}\right) \end{align*}\] 
The result of example 12 has some interesting properties. First, we note that it was derived on the unspoken assumption of \(z>0\). By symmetry, the field on the other side of the disk must be equally strong, but in the opposite direction, as shown in figures e and g. Thus there is a discontinuity in the field at \(z=0\). In reality, the disk will have some finite thickness, and the switching over of the field will be rapid, but not discontinuous.
At large values of \(z\), i.e., \(z\gg b\), the field rapidly approaches the \(1/r^2\) variation that we expect when we are so far from the disk that the disk's size and shape cannot matter (homework problem 2).
A practical application is the case of a capacitor, f, having two parallel circular plates very close together. In normal operation, the charges on the plates are opposite, so one plate has fields pointing into it and the other one has fields pointing out. In a real capacitor, the plates are a metal conductor, not an insulator, so the charge will tend to arrange itself more densely near the edges, rather than spreading itself uniformly on each plate. Furthermore, we have only calculated the onaxis field in example 12; in the offaxis region, each disk's contribution to the field will be weaker, and it will also point away from the axis a little. But if we are willing to ignore these complications for the sake of a rough analysis, then the fields superimpose as shown in figure f: the fields cancel the outside of the capacitor, but between the plates its value is double that contributed by a single plate. This cancellation on the outside is a very useful property for a practical capacitor. For instance, if you look at the printed circuit board in a typical piece of consumer electronics, there are many capacitors, often placed fairly close together. If their exterior fields didn't cancel out nicely, then each capacitor would interact with its neighbors in a complicated way, and the behavior of the circuit would depend on the exact physical layout, since the interaction would be stronger or weaker depending on distance. In reality, a capacitor does create weak external electric fields, but their effects are often negligible, and we can then use the lumpedcircuit approximation, which states that each component's behavior depends only on the currents that flow in and out of it, not on the interaction of its fields with the other components.
10.3.2 The field near a charged surface
From a theoretical point of view, there is something even more intriguing about example 12: the magnitude of the field for small values of \(z\) (\(z\ll b\)) is \(E=2\pi k\sigma\), which doesn't depend on \(b\) at all for a fixed value of \(\sigma\). If we made a disk with twice the radius, and covered it with the same number of coulombs per square meter (resulting in a total charge four times as great), the field close to the disk would be unchanged! That is, a flea living near the center of the disk, h, would have no way of determining the size of her flat “planet” by measuring the local field and charge density. (Only by leaping off the surface into outer space would she be able to measure fields that were dependent on \(b\). If she traveled very far, to \(z\gg b\), she would be in the region where the field is well approximated by \(\mathbf{E}\approx kQ/z^2=k\pi b^2\sigma/z^2\), which she could solve for \(b\).)
What is the reason for this surprisingly simple behavior of the field? Is it a piece of mathematical trivia, true only in this particular case? What if the shape was a square rather than a circle? In other words, the flea gets no information about the size of the disk from measuring \(E\), since \(E=2\pi k\sigma\), independent of \(b\), but what if she didn't know the shape, either? If the result for a square had some other geometrical factor in front instead of \(2\pi\), then she could tell which shape it was by measuring \(E\). The surprising mathematical fact, however, is that the result for a square, indeed for any shape whatsoever, is \(E=2\pi\sigma k\). It doesn't even matter whether the surface is flat or warped, or whether the density of charge is different at parts of the surface which are far away compared to the flea's distance above the surface.
This universal \(E_\perp=2\pi k\sigma\) field perpendicular to a charged surface can be proved mathematically based on Gauss's law^{1} (section 10.6), but we can understand what's happening on qualitative grounds. Suppose on night, while the flea is asleep, someone adds more surface area, also positively charged, around the outside edge of her diskshaped world, doubling its radius. The added charge, however, has very little effect on the field in her environment, as long as she stays at low altitudes above the surface.
As shown in figure i, the new charge to her west contributes a field, T, that is almost purely “horizontal” (i.e., parallel to the surface) and to the east. It has a negligible upward component, since the angle is so shallow. This new eastward contribution to the field is exactly canceled out by the westward field, S, created by the new charge to her east. There is likewise almost perfect cancellation between any other pair of opposite compass directions.
A similar argument can be made as to the shapeindependence of the result, as long as the shape is symmetric. For example, suppose that the next night, the tricky real estate developers decide to add corners to the disk and transform it into a square. Each corner's contribution to the field measured at the center is canceled by the field due to the corner diagonally across from it.
What if the flea goes on a trip away from the center of the disk? The perfect cancellation of the “horizontal” fields contributed by distant charges will no longer occur, but the “vertical” field (i.e., the field perpendicular to the surface) will still be \(E_\perp=2\pi k\sigma\), where \(\sigma\) is the local charge density, since the distant charges can't contribute to the vertical field. The same result applies if the shape of the surface is asymmetric, and doesn't even have any welldefined geometric center: the component perpendicular to the surface is \(E_\perp=2\pi k\sigma\), but we may have \(E_\parallel\neq0\). All of the above arguments can be made more rigorous by discussing mathematical limits rather than using words like “very small.” There is not much point in giving a rigorous proof here, however, since we will be able to demonstrate this fact as a corollary of Gauss' Law in section 10.6. The result is as follows:
At a point lying a distance \(z\) from a charged surface, the component of the electric field perpendicular to the surface obeys
where \(\sigma\) is the charge per unit area. This is true regardless of the shape or size of the surface.
Example 13: The field near a point, line, or surface charge 

\(\triangleright\) Compare the variation of the electric field with distance, \(d\), for small values of \(d\) in the case of a point charge, an infinite line of charge, and an infinite charged surface. \(\triangleright\) For a point charge, we have already found \(E\propto d^{2}\) for the magnitude of the field, where we are now using \(d\) for the quantity we would ordinarily notate as \(r\). This is true for all values of \(d\), not just for small \(d\)  it has to be that way, because the point charge has no size, so if \(E\) behaved differently for small and large \(d\), there would be no way to decide what \(d\) had to be small or large relative to. For a line of charge, the result of example 10 is \[\begin{equation*} E = \frac{ k\lambda L}{ d^2\sqrt{1+ L^2/4 d^2}} . \end{equation*}\] In the limit of \(d\ll L\), the quantity inside the square root is dominated by the second term, and we have \(E\propto d^{1}\). Finally, in the case of a charged surface, the result is simply \(E=2\pi\sigma k\), or \(E\propto d^{0}\). Notice the lovely simplicity of the pattern, as shown in figure j. A point is zerodimensional: it has no length, width, or breadth. A line is onedimensional, and a surface is twodimensional. As the dimensionality of the charged object changes from 0 to 1, and then to 2, the exponent in the nearfield expression goes from 2 to 1 to 0. 
Contributors
Benjamin Crowell (Fullerton College). Conceptual Physics is copyrighted with a CCBYSA license.