HW 02 - The Law of Gauss

1 1 1 Equation Chapter 1 Section 1 HW 02 - the law of Gauss : chapter 23: 2, 11, 14, 26, 35, 39, 41, 52

•• chapter 23, problem 2: An electric field given by where ,

, and pierces a Gaussian1 cube of edge length and positioned as shown in the Figure. What is the electric flux through the (a) top face, (b) bottom face, (c) left face, and (d) back

face? (e) What is the net electric flux through the cube?

(a) Looking at the figure, we see that the x-bounds of the cube are . We have to carefully think about what the label “top” means:

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We combine the statements in 12, and we get,

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(b) We “rinse-and-repeat” the set of steps in 12 and 13.



515\* MERGEFORMAT (.)1 A “Gaussian” surface is a closed surface in three-dimensional space. Hence, bubbles, hollow-cylinders, hollow-cubes are Gaussian surfaces, while cups, jars, toilet-paper-rolls, etc. are non-Gaussian surfaces. The notion of a surface being Gaussian or not is critical to the notion of flux of a vector field (e.g., gravitational, magnetic, electric, etc.) through the surface.

(c) the left-face has,

616\* MERGEFORMAT (.)We combine the statements in 16, and we get,


(d) the back-face has,




(e) the front-face has,




In retrospect, it’s easy to see that , because both point in either of the

directions, while the electric field it is dotted with is contained in the xy-plane.

The right face has,




Hence,


Going further: We note that we have not used Gauss’s Law; we have just calculated “electric flux”, a thing that appears in Gauss’s Law. Gauss’s Law, formally, states that the total electric flux2 through a Gaussian surface is

proportional to the charge it encloses, where the factor of proportionality3 is , where ,


Combining 114 and 115, we see that we can calculate the total charge enclosed by4 the Gaussian cube,


Chapter 23, problem 11: The Figure shows a Gaussian (closed) surface in the shape of a cube of edge length

, with one corner at , as indicated. The cube lies in a

2 Looking at 115, we see that the electric flux is, in fact, a placeholder variable that disappears on combining the statements

, , and .

3 The quantity is called the “permittivity” or “dielectric constant” of free space, and is a more fundamental

quantity than Coulomb’s constant ( ). As indicated, its units are , but they are equivalently “capacitance per meter” (helpful in other (later) contexts). Some of the useful ways of writing its units are:

WARNING: In PHYS 2049 units (which you should always retain and manage!) will be more difficult to keep track of.4 Nobody said the cube didn’t have any charge enclosed in it! Indeed, it is required to, otherwise Gauss’s Law would be violated.

region where the electric field vector is given by with ,

, and . What is the net charge contained by the cube?

The net charge in the cube is given as , where has six contributions (one from each face of the cube, as before). Setting up the flux-integrals,


Carrying out the integrals in 117, and using the dot-product operation to simplify things,




Hence, the total charge in the cube is easily calculated as,


Afterword: Note that the enclosed charge is the total value of the charge! We have no idea how the charge is

distributed! That is, if was the charge density inside of the cube, 121 would be the value of the integral5,


Triple integrals can appear formidable, but they are really just necessitated by the 3 dimensions of a cube. Often, spheres rather than cubes are considered in electromagnetism problems, because electromagnetic properties in vacuum and homogeneous materials on the meso-scale or larger are spherically-isotropic, and calculating integrals over spheres then requires a single integral6.

•• chapter 23, problem 26: The Figure shows a narrow charged solid cylinder that is coaxial with a larger charged cylindrical shell. Both are non-conducting and thin and have uniform surface charge densities on their outer surfaces. The Figure also gives the radial component of the electric field versus radial distance r from the common axis,

and .What is the shell’s linear

charge density?

Because the shell (bearing charge ) and inner-cylinder (bearing charge , and hereafter referred to as a “wire”) are said to be “thin”, the respective shell-thickness and wire-radius are each negligible. The shell has a

radial coordinate of , as shown by the plot. Let be the radial coordinate at which the electric field

5 We notice that has units of length-cubed (volume), and has units of Coulombs per unit volume, so the units in 122 work out.6 Example: suppose a density of charge is smeared throughout a sphere of radius (note the use of spherical

coordinates rather than ). Furthermore, suppose the charge density was isotropic; this means that only (i.e.,

the density is independent of . Then, the total-charge-triple-integral quickly simplifies to a single integral, as,

A mass-analogue: the density of planet-Earth doesn’t depend on latitude or longitude, only your (subterranean) -coordinate).

in one radial dimension is measured. Gauss’s Law7 can be used for and , and on a length8 of the cylinder,


Carrying out an integral over9 ,


The shell’s linear charge density is , and the wire’s linear charge density is ; dividing the 2nd expression of 124 through by , we have,


This 125 is the simplest exact form of the answer in “givens” (one reads off from the plot of vs.

shown above, with the constraints and , as mentioned before). Let’s pick ,

which allows us to read off and , yielding10,


• chapter 23, problem 35: The Figure shows three plastic sheets that are large, parallel, and uniformly charged. The second Figure gives the component of the net electric field along an x-axis through the sheets. The scale of the vertical axis is set by

. What is the ratio of the charge density on sheet 3 to that on sheet 2?

7 Although we refer to Gauss’s Law here, we do not calculate , but rather go straight to . That is, is just a placeholder variable.8 Strictly speaking, the cylinder and the shell are infinite in length, but to use in the would mean that infinite charge is enclosed, and Gauss’s Law would collapse to the tautology , in which there are no useful results.9 That is, the integral is over the perimeter of the Gaussian-cylinder-surface of radius- .

10 Actually, maybe only after plugging in , , and do we have SEF truly in terms of “givens”.

Gauss’s Law applied to a pillbox of area enclosing charge and of sheets 2 and 3 respectively is as

follows, where the left and right faces make disparate contributions with normal-surface-vectors,


In 127, the electric fields and the vector-signed area-elements are


Combining the statements in 127 and 128 so as to calculate , we have,


•• chapter 23, problem 39: In the Figure, a small,

non-conducting ball of mass and

charge (distributed uniformly through its volume) hangs from an insulating thread that makes an angle with a vertical, uniformly charged non-conducting sheet (shown in cross section). Considering the gravitational force on the ball and assuming the sheet extends far vertically and into and out of the page, calculate the surface charge density of the sheet.

The electric field felt by the massive, charged sphere is given by Gauss’s Law as,

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Newton’s laws in the x and y directions are,


Calculating the ratio from 131 and solving for ,

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Plugging in numbers to 132,


•• chapter 23, problem 41: An electron is shot directly toward the center of a large metal plate that has surface

charge density . If the initial kinetic energy of the electron is and if the electron is to stop (due to electrostatic repulsion from the plate) just as it reaches the plate, how far from the plate must the launch point be?

Recalling the kinematical relation between initial and final velocities due to acceleration acting over displacement ,11 we see that the problem falls apart once we know the acceleration , which is

constant since the electric field due to a very large plate is uniform; indeed, Gauss’s Law for a metal12 plate is,


Combining the statements 134 in the expression for given by the abovementioned kinematical statement, and

noting that the definitions and naturally occur,

35135\* MERGEFORMAT (.)Chapter 23, problem 52: The Figure shows a spherical shell with uniform volume charge density11 Remember what I said: to understand an equation, translate it into an English sentence! This is a prime example taken from material which should be very familiar to you (kinematics).12 For a metal plate, charge density lies at the surface (even so the metal sheet is a sheet, it has some thickness), and so the Gaussian pillbox is “halfway” through the thickness. However, there is zero field inside the conductor; hence, the term in 134! In contrast, for a non-conducting plate,

, inner radius , and outer radius . What is the magnitude of the electric field at radial distances (a) ; (b)

, (c) , (d) , (e) , and (f) ?

There are three regions where we may be interested in the electric field : within the center-sphere , within the shell , and outside of the shell . The electric field is purely-radial, and a

function of . In all three regions, ,


In 136, we have,


Combining 136 and 137, we immediately have,


Solving the expressions in 138 for and recalling that , we finally have,


Hence, the electric field at (a) ; (b) , (c) are each zero. Notice, furthermore, that the 1st and 2nd expressions in 139 each predict the same electric field (zero) when is plugged in. For (d) and (e) , we use the 2nd expression in 139,



Finally, the electric field at is given by the 3rd expression of 139,


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HW 02 - The Law of Gauss