2.2  Applications of Gauss' Law

2.2.1  Electric Field Due to a Line Charge - Cylindrical Symmetry

Realistically, all line charges are finite, but, first, we have done the finite length problem explicitly, and second, a line of charge which is quite long compared to the distance from it at which we would like to know the E field is not an uncommon problem. We can deal with the approximate answer as easily as with the true solution and compare the differences. Physicists will often engage in thoughts of this kind in theoretical research as it is important to know where our ideas ``break down''. Those ideas which can hold up under the most extreme extrapolations without delivering clearly nonsensical results indicate something deep about our understanding.

The above being said, we still need to do the calculation. Consider the figure below which shows a view of the line charge and a point P a distance h away from it. We wish to find the electric field at point P. To set up the integral, we do, as before, the trick of taking infinitesimally small line segments of charge in pairs so that their horizontal components cancel and the vertical (i.e. radial) components add.

Figure 2.4: Calculation of the electric field at the midpoint of a line charge of length l.

Hence we only need to change the definition of dq, the charge on an infinitesimal segment with length dx, otherwise our approach for the finite line charge is unchanged for the infinite length case. Now, dq = ldx, so, as with the finite length charge, we use the angle, q with respect to the vertical to identify the radial component, r, as the distance from the infinitesimal charge to point P, i.e.

r =   ______ Öx2 + h2

(2.2.1.7)

and thus

cosq = h r = h   ______ Öx2 + h2 .

(2.2.1.8)

We calculate the magnitude of E as follows.

Enet = ó õ 2 dE cosq = ó õ ¥ 0  é ê ë 2 dq 4pe0r2 cosq ù ú û = 1 2pe0 ó õ ¥ 0  l h dx (x2 + h2)3/2 = lh 2pe0 é ê ë x h2   ______ Öx2 + h2   ù ú û ¥ 0  Enet = l 2pe0h (2.2.1.9)

To see a 3-D picture of what the electric field from an infinite line charge looks like, consider this digital animation.

We can also approach this problem using Gauss' Law.

Figure 2.5: Gaussian surface with cylindrical symmetry surrounding a line charge.

The Gaussian surface should be a closed, hollow can. The sides of the can are perpendicular to the E field (since we just determined previously that the E field is radial only, i.e. there are no components perpendicular to the radius). Hence E·dA is just E dA since the electric field also only depends on h as we just showed with the explicit integration done earlier. If we make the Gaussian can of finite length (we should since we will need to find the amount of enclosed charge), say, L as in figure 1.16, then the charge enclosed is l L. The calculation of the E field proceeds in this way:

ó (ç) õ E·dA = qenc e0 E ó (ç) õ dA = lL e0 E(2ph L) = lL e0 Þ E = l 2pe0 h (2.2.1.10)

We get the same result, but with somewhat less work. Gauss' Law is useful for calculation of the electric field for situations which have a high degree of symmetry. In fact, for freshman physics, we can generally only apply it in cases in which one of the following symmetries are exhibited:

• Spherical symmetry (point charges, spherical charge distributions)
• Cylindrical symmetry (infinitely long lines of charge, infinitely long cylindrical charge distributions)
• Planar symmetry (infinite sheets or blocks of charge)

We have discussed examples of the first two of these. We can proceed now with the latter.

2.2.2  Planar Symmetry

In the case of charge distributions which exhibit planar symmetry, i.e. that have a charge distribution which extends in two dimensions, we wish to find the E field due to a large, assumed infinite, sheet of charge. The infinite extent applies to only two of a possible three spatial dimensions. We talk of the sheet being infinitely wide and infinitely long, but it may be of finite or infinite thickness. The character of the electric field depends not only on the arrangement of the charges on the sheet but on whether the sheet is composed of conducting or non-conducting material as we will see.

Before beginning this examination of planar symmetry though, it's important to understand some implications of Gauss' Law (or more importantly of the symmetries implicit in the use of Gauss' Law) for charges on conductors. First, note that conducting material has the property of allowing charges (electrons are almost always the charges we are talking about for real materials, but we must keep in mind that real materials are not perfect conductors like those we will be discussing) to move freely, i.e. without resistance to their motion, through them. The mobility of the charges means that they respond to any electric field in their vicinity by moving according to F = qE. If we think about the consequences of this, you'll quickly come to the conclusion that the electric field inside the bulk of any conducting material must quickly go to and stay at zero. The reasoning is simple: if charges move in response to electric fields, then the charges will only stop moving once the electric field goes to zero. However, since the negative and positive charges move in opposite directions they will themselves create an electric field. A simple picture suffices to show that the electric field created always opposes the external electric field creating the charge separation.

Figure 2.6: Charges separating within a perfectly conducting material.

The charges keep separating until they reach the surface of the material. More and more charges accumulate at the surface and create a net electric field that exactly cancels that of the external field. We assume that there are enough charges in the material so that any electric field can be neutralized. Thus, the initial electric field in the material is zero since the positive and negative charges are "combined" so as to leave no net charge to create an electric field. Once an external field is turned on, the net electric field inside the bulk material is zero since the charges separate to produce an electric field that opposes that of the external field. Later, we will see how to modify the above picture for "perfect" conductors so that we can determine what happens for real conductors.

What we have shown above works fine for conductors that have no net charge. We can also apply similar reasoning to the case of a perfect conductor with a non-zero net charge. In the absence of any external electric field, the net charge on the conductor winds up on its outside surface. This maximizes the separation between the charges and minimizes the forces acting on them.

Figure 2.7: A perfect conductor with a non-zero electric charge.

Note that the charges inside the conductor don't separate, but instead orient themselves so as to still generate a net field that opposes that due to the external charges, thereby neutralizing them so that net electric field inside the bulk material remains zero. Again, it is assumed that there are enough charges in the conducting bulk to neutralize the field due to any amount of external charge placed on the conducting material. Again, we will talk more of this later, but for real materials we know that the number of internal charges is not infinite. So, the electric field can penetrate a certain very small distance into the conducting material. We refer to this as the skin depth of the material. We can assume the skin depth to be of zero width. This is an acceptable approximation for many "good" conductors.

To return to our original discussion of planar symmetry, we go first to the picture of what happens to the net field produced by charges confined to a sheet of infinitesimal thickness. The figure below shows the result of adding up the E fields from charges at various points on the sheet. As in the spherical and cylindrical cases, we see a symmetry that allows us to cancel components of the net E field that are parallel to the sheet and only keep those that are perpendicular to the sheet.

Figure 2.8: The electric field near an infinite sheet of charge.

Before going through this problem with Gauss' Law, we can take the ''old'' approach of Coulomb's Law. At least we can now make use of a previous result to reduce our workload. Remember that we have already calculated the magnitude and direction of the E field due to a wire. We now make use of this result. Consider the sheet as being composed of an infinite number of infinitesimally thin wires, each with width dx as shown in the figure below. We wish to find the electric field at a point P located a distance b from the sheet.

Figure 2.9: Calculating the E field near an infinite sheet of charge.

As usual, we note that the symmetry, planar symmetry in this case, allows us to ignore components of the net E field as these cancel. The components perpendicular to the sheet add, so, since the sheet is infinite in extent width and length, the E field is perpendicular to the sheet no matter what point P we pick. Let's choose, as a reference point, the spot on the charged sheet which is closest to the point P where we want to know the field magnitude (marked as point O in figure 2.9. This point lies on a line through P which is perpendicular to the sheet. If we choose "wires" which are equidistant from point P, then the contributions to the field at point P from components which are parallel to the charged sheet will exactly cancel. The perpendicular components are each dE cosq where q is the angle between a line from the wire to point P and the line from point O to point P. If we state the wires as being a distance r from point P, then we have

r =   ______ Öb2 + x2         and      cosq = b   ______ Öb2 + x2

(2.2.2.11)

According to the result derived previously, the electric field at point P due to a single wire is

E = s dx 2pe0b

(2.2.2.12)

Note that we have replaced the linear charge density of the previous lecture with s dx. The distance from the wire to point P, h, has been replaced by the distance r. Now the integration over all the wires can occur.

Enet = ó õ 2 dE cosq = ó õ ¥ 0  é ê ë 2s dx 2pe0r cosq ù ú û = s pe0 ó õ ¥ 0  b r2 dx = sb pe0 ó õ ¥ 0  dx x2 + b2 = sb pe0 1 b é ê ë tan-1 æ ç è x b ö ÷ ø ù ú û ¥ 0  = s pe0 é ê ë p 2 - 0 ù ú û Enet = s 2e0 (2.2.2.13)

where you should note that the integral can be evaluated using Maple or by consulting a table of integrals. The above result works for an infinitesimally thin sheet of conducting material or for any finite thickness sheet of non-conducting material since the E field configuration will be the same.

If we apply Gauss' Law to the same case, we can arrive at the solution considerably faster. Imagine a Gaussian surface in the form of a block with a rectangular cross-section.

Figure 2.10: Gaussian pillbox for a finite thickness planar charge distribution.

The electric field through the Gaussian surface is parallel to the sides and perpendicular to both the endcaps. The result is that the flux through the sides is zero while the flux through the endcaps is

EA + EA

(2.2.2.14)

with A being the cross-sectional area and where we take advantage of our previous calculation with Coulomb's Law for the infinitesimally thick conducting sheet (or finite thickness non-conducting sheet) to state that the electric field should be constant in direction and magnitude on either side of the sheet. Hence, Gauss' Law states that

flux = F = 2EA = qenc e0

(2.2.2.15)

Since the charge enclosed by the Gaussian surface is

qenc = s·AÞ E = s·A 2Ae0 = s 2e0 .

(2.2.2.16)

This is the same result as was derived using Coulomb's Law as we expect. The difference in effort to get this result is impressive, although we did depend on having seen the symmetry that allowed us to relate the flux to the electric field in this case.

The real payoff of using Gauss' Law comes when we apply the same thought process to the case for a conducting sheet of semi-infinite thickness as shown in figure 2.11.

Figure 2.11: Gauss' Law applied to a slab of semi-infinite thickness.

Using Gauss' Law with the same arrangement as before, we note that now only one endcap receives a net flux since the endcap inside the conducting material sits in a zero field region. Hence the magnitude of the E field in this case is

flux = F = EA = qenc e0 = s e0

(2.2.2.17)

so the field is twice the magnitude of the non-conducting sheet of finite thickness or the conducting sheet of infinitesimal thickness.

2.2.3  The Finite Thickness Infinite Sheet

Even though we've used planar symmetry to derive results for a couple of different examples of infinite sheets of charge, we should consider one more example, not for derivation, but strictly for application of the ideas learned so far. Consider the following problem from the 1991 midterm I.

An infinite slab of insulating material has a uniform volume charge density of 2.0×10^-6 C/m³. The surfaces of the slab are located at x = +1 cm and x = -1 cm. (See figure 2.12.)

Figure 2.12: An infinite slab of insulating material with a uniform charge density throughout.

Find the electric field at:

1. x = 2.0 cm
2. x = 0
3. x = 0.7 cm

Solution:

1. We need to draw a Gaussian surface which reflects the symmetry. From what we just covered in the last lecture, we know that the electric field should be constant in magnitude and direction outside the nonconducting material. For positive charge, the electric field is directly perpendicularly away from the surface of the material. We can use a Gaussian surface of any shape we desire, but let's choose one of rectangular cross-section as we have thus far in our derivations. That means we have a situation that looks as follows:
   
   Figure 2.13: A rectangular Gaussian surface in the vicinity of a charged slab of infinite extent.

   Note that the Gaussian surface on the +x side extends only to the 2.0 cm mark for which we desire the field magnitude (direction is already known to be along +x, i.e. perpendicular to the surface at +1.0 cm). On the -x side we need only note that the Gaussian surface must extend <i>outside</i> the non-conductor to preserve the symmetry of the electric field. The reason is that we know the field has constant magnitude and direction outside the nonconducting material, hence it does not matter where we place the endcap of our Gaussian surface as long as it is at x less than -1 (i.e. x = -3 cm or x = -10 cm work equally well). Since the field has the same magnitude on either side of the nonconducting material and we have drawn the surface so the sides of the Gaussian "box" are parallel to the direction of the field for either +x or -x, the net flux through the Gaussian surface is 2E·A where A is the cross-sectional area of the Gaussian box.

   To get the charge enclosed by the box, we use the volume charge density, which we call rho, and the volume contained by the part of the Gaussian box which is inside the nonconducting material. This total charge is
   

   qenclosed = r·A·t

   (2.2.3.18)

   where t is the thickness of the non-conducting material. From the information given, t is obviously 2 cm. Hence, it is quick to apply Gauss' Law:
   

   2EA = qenclosed e0 2EA = rA t e0 Þ E = rt 2e0 = (2.0×10-6 C/m3)(0.02 m) 2(8.85×10-12 C2/(N·m2)) E = 2260 N/C (2.2.3.19)

2. By symmetry, at the center of the non-conducting material must have zero net electric field. Fields from each half of the slab would cancel.
3. Now we have a choice of how to draw the Gaussian surface. We can choose a symmetric arrangement in which the two endcaps lie at -0.7 cm and +0.7 cm or one in which the two endcaps lies at 0 and +0.7 cm as shown in the figure below.
   
   Figure 2.14: A Gaussian surface partially embedded in a charged slab of infinite extent.

   The reason either of these works is because we can evaluate the flux and separate out the contribution of E to the flux in either case. For the first case, the net flux through the surface is just 2EA since, by symmetry, the E field will be the same magnitude at the same distance from the symmetry point of x = 0. In the second case, the flux is just through the right endcap since the magnitude of E at x = 0 (location of the left endcap in this case) is zero. Hence the flux would be EA. Now note that we get the same result for either case. For example, in the first case
   

   2EA = qenclosed e0 = r·(volume enclosed) e0 = r·A·t0.7 e0 E = r·t0.7 2e0 = (2.0×10-6 C/m3)(2*0.007 m) 2(8.85×10-12 C2/(N·m2)) E = 1582 N/C (2.2.3.20)

   You can (and should) easily verify that the second case gives exactly the same result. Remember that the flux is one-half of what it is in the case just shown, but that the volume enclosed and therefore charge enclosed in the second case is also half of what we get for the first case.

Send comments to larryg@upenn5.hep.upenn.edu.
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