Applications of Gauss ’ s Law

4.3 Gauss ’ s Law

4.3.1 Applications of Gauss ’ s Law

There are two basic forms in which Gauss’s law may be used:

(1) Calculation of the electric field intensity or electric flux density from known charge configurations.

(2) Calculation of the equivalent charge in a volume, provided the electric field is known everywhere in space, but, in particular, at the location Gauss’s law is applied.

The evaluation of electric field intensities and equivalent charge densities is discussed next and in examples.

4.3.1.1 Calculation of the Electric Field Intensity

To use Gauss’s law for the calculation of the electric field intensityE, there is a slight difficulty: the unknown value in Eq. (4.13)is inside the integral sign and, in addition, the scalar product between the electric field intensityEand the vector dSmust be evaluated. This certainly cannot be done in general configurations. However, under two special conditions, the evaluation of the electric field intensity is very simple. These are as follows:

(1) The angle between the electric field and the surfacesis constant anywhere on the surface. This allows the evaluation of the scalar product and, therefore, of the integral. In particular, if the electric field intensity is in the direction ofds (perpendicular to the surface), the scalar product is equal to the product of the magnitudes ofEandds. IfEis in the direction oppositeds, the scalar product is the negative of the product of the magnitudes ofEandds. If the electric field intensity is parallel to the surface, the scalar productE·dsis zero (seeFigures4.3aand4.3b).

(2) The electric field intensity is constant in magnitude over the surfaces. This means that the electric field intensity can be taken out of the integral after evaluating the scalar product inEq. (4.13).

A surface that satisfies these conditions is called a Gaussian surface. With the above conditions, we can write on the Gaussian surface

þ

s

E

^ds¼Eþ

s

ds¼Q ε0

or E¼ Q

ε0

þ

s

ds N

C ð4:14Þ

Similarly, for the electric flux densityD, we can write D¼ Q

þ

s

ds

C m²

ð4:15Þ Q

ds ds1

ds2

ds3

ds4

E0

E0

E0

E0

E0

E0

E0

s₂ s₁

a b

Figure 4.3 Conditions necessary for the application of Gauss’s law:

(a)E₀·ds¼E0ds.

(b)E₀·ds1¼E0ds1, E₀·ds2¼–E0ds2, E0·ds3¼E0·ds4¼0

144 4 Gauss’s Law and the Electric Potential

Note that the vector notation has been lost since the scalar product was evaluated beforehand. However, from knowledge of location and signs of charges, we can easily restore the direction of the fields, as will be shown shortly.

Before continuing, we must note that although the simplifications afforded by the assumptions used here restrict its application to a small set of simple configurations, the idea of Gauss’s law is completely general. Also, the charge in Eqs. (4.14)and(4.15)may be replaced by a line, surface, or volume charge distribution by using one of the relations

Q¼ ð

l⁰

ρldl⁰ Q¼ ð

s⁰

ρsds⁰ Q¼ ð

v⁰

ρvdv⁰ ½ C ð4:16Þ

as long as the basic assumptions of a constant, perpendicular (or parallel) field are preserved. Also, note that the surface in Eq. (4.16)is not necessarily the same as inEq. (4.14)or(4.15)although it can be. The surfaces⁰, contourl⁰, and volumev⁰ indicate the domains in which charge densities exist. The integrals inEq. (4.16) are over the sources enclosed by the Gaussian surface, whereas the surface integrals inEqs. (4.14)and(4.15)are over the Gaussian surface itself.

For the conditions of a Gaussian surface to be satisfied, the charges or charge distributions must be highly symmetric, as we shall see in examples. In fact, there are only four general classes of problems that can be solved under these conditions:

(1) Point charges.

(2) Symmetrical spherical charge distributions and charged spherical layers.

(3) Infinite lines of charges or symmetrically charged cylindrical objects.

(4) Infinite charged surfaces or infinite charged layers.

These classes of problems (see examples that follow) are rather limited, but their usefulness can be extended through the use of superposition of solutions and by using Gauss’s law on geometries, which are approximately the same as one of the classes above. For example, a square cross-sectional object may be approximated as an assembly of thinner, round objects, whereas a finite length line may be solved as if it were infinite in extent, provided the line is“long.”

Before demonstrating the use of Gauss’s law, we wish to outline the method in terms of the basic steps required for solution:

(1) Inspect the charges or charge distributions. Careful study of symmetries and charge distributions can provide hints on how to approach the problem or, in fact, if the problem is solvable using Gauss’s law.

(2) Sketch the electric field (direction and magnitude) to see if a surface can be identified on which the electric field intensity is both constant in magnitude and directed perpendicular to the surface, or a surface to which the electric field intensity is parallel. In some cases, sections of a surface may be found on which the electric field intensity is perpendicular and constant, whereas on other sections, the electric field intensity is parallel to the surface. For example, for a point charge, the surface will be spherical, whereas for a straight line, a cylindrical surface will be required. The Gaussian surface must pass through the location at which the field is evaluated and must be a closed surface.

(3) If a Gaussian surface cannot be found, try to separate the charges into two or more charge configurations (without modifying the problem) for each of which a Gaussian surface can be obtained. The electric field intensity of the configuration will then be the superposition of the individual fields.

Calculate the electric fields on the Gaussian surface or surfaces.

Example 4.2 Electric Field of Spherical Charge Distributions Two very thin, spherical, conducting shells are charged as inFigure4.4a. The inner shell has a total charge –Q, whereas the outer shell has a total charge +Q. Charges are distributed uniformly on the surfaces of the corresponding shells:

(a) Calculate the electric field intensity everywhere in space.

(b) A point charge of magnitude +Qis now inserted at the center of the two shells as shown inFigure4.4b. Calculate the electric field intensity everywhere in space.

(c) Plot the two solutions.

Solution: Because the electric field intensity for this symmetric charge distribution is always radial and constant at constant distances from the center, the Gaussian surfaces required are spheres of various radii. Here, we calculate the field in three

areas: inside the inner shell, between the two shells, and outside the outer shell. The three surfaces are shown as dotted lines inFigures4.4aand4.4b.

(a) Draw a Gaussian surface inside the inner shell inFigure 4.4a. Since this shell encloses no charge, the electric field intensity is zero:

E¼0, 0R<a

The field between the two shells is calculated ons2. BecauseEis perpendicular to this surface and the Gaussian surface encloses a total charge –Q, we can write

Es₂¼E4πR²¼Q ε0

! E¼ Q

4πε0R² or E¼ R^ Q 4πε0R²

N

C , aRb

where the fact that the field is known to be radial was used to write the vector form of the field. Note also that this solution is the same as if the charge were located at the center of the sphere as a point charge.

Outside the outer shell, the Gaussian surface iss3. This encloses both charged shells symmetrically. The total charge enclosed bys3is zero. Thus, the electric field intensity outside the outer shell is zero:

E¼0, b<R<1

(b) The solution in part(a)can be used here as well, but the charges enclosed by the various Gaussian surfaces are different.

Inside surfaces1, there is a point charge +Q. Gauss’s law gives Es₁¼E4πR²¼Q

ε0

! E¼ Q

4πε0R² or E¼R^ Q 4πε0R²

N

C , 0<Ra

−Q

+Q a

b R

R

R

−Q

+Q

a b R

R

R s₁ s₂

s₃ s3

s₂ s₁ Q

a b

E

a b

c d e

R

−Q/4πε0b²

−Q/4πε0a²

a b

E

R Q/4πε⁰b² Q/4πε⁰a²

a b

E

R

−ρ^sa²/ε⁰b²

−ρ^s/ε⁰

ρ^s(b²−a²)/ε⁰b²

Figure 4.4 (a) Two uniformly charged spherical shells. (b) A point charge surrounded by two uniformly charged spherical shells. (c) Plot of the electric field intensity in (a). (d) Plot of the electric field intensity in (b). (e) Plot of the electric field intensity inExercise 4.2

146 4 Gauss’s Law and the Electric Potential

Between the two shells, surface s2 now encloses zero charge density (a point charge +Q and a surface charge totaling –Q). Thus, the electric field intensity between the two shells is zero:

E¼0, a<R<b

Surfaces3includes a net positive charge +Q. Thus, the electric field intensity outside the outer shell is E¼R^ Q

4πε0R² N

C , bR<1

(c) The two solutions are drawn in Figures 4.4c and 4.4d. Note the jump in the electric field intensity at the shells themselves. This is due to the charge located at this point. Also, in(b), the electric field intensity atR¼0 tends to +1, because of the existence of a point charge at this location.

Exercise 4.2 Consider the geometry inFigure4.4a. Assume a uniform surface charge density –ρs[C/m²] is placed on the inner shell and a uniform surface charge density +ρs[C/m²] is placed on the outer shell. Calculate the electric field intensity everywhere in space. Plot the solution.

Answer

E¼0, 0R<a, E¼ R^ a²ρs

ε0R²

N

C , aR<b,

E¼R^ b²a² ρs

ε0R²

N

C , bR<1 SeeFigure4.4efor a plot of the solution.

Example 4.3Application: Electric Field of Surface Layers The electric field of flat layers is very important, especially in design of capacitors and capacitor-like devices. Here, we show how Gauss’s law can be used to calculate the electric field intensity of simple charged layers.

A charge densityρs[C/m²] is uniformly distributed over a very large surface (such as a large, thin aluminum foil), as inFigure4.5. Calculate the electric field intensity everywhere in space.

ρ_s

x y z

−z ds2

ds1

ds3

E1

E2

s

s₁ s₃ s₂ h

Figure 4.5 A large (infinite) charged layer and the Gaussian surface required for calculation of the electric field intensity

Solution: From symmetry considerations, the electric field must be perpendicular to the layer and constant in magnitude on any surface parallel to the layer. The electric field points in the negativezdirection below the layer and in the positivez direction above the layer. The appropriate Gaussian surface is shown inFigure4.5. Note thatdsis positive in the direction out of the volume. Assumings1¼s2¼s, the Gaussian surface cuts a section of the charged layer equal tos. The charge enclosed by the surface is, therefore,ρss.

The scalar productE ·dsis nonzero ons1ands2but is zero on the lateral surface (E ·ds3¼0). Therefore, þ

s

E

^ds¼ð

s₁

E1

^ds1þð

s₂

E2

^{ds2¼E1s1þE2s2¼2Es}

whereE1¼E2¼Efrom symmetry considerations. Thus, 2Es¼ρss

ε0

! E¼ ρss 2sε0

¼ ρs

2ε0

N C

The electric field intensity is in the positivezdirection above the layer and in the negativezdirection below the layer (recall that the electric field intensity always points away from positive charges). The solution is

E¼z^ ρs

2ε0

N

C ðz>0Þ and E¼ ^z ρs

2ε0

N

C ðz<0Þ

Example 4.4Application: Electric Field of a DC Overhead Transmission Line—Gauss’s Law

and Superposition Overhead transmission lines generate an electric field everywhere in space. The electric field intensity must be kept low enough for safety reasons. The following example gives an idea of how this electric field behaves and introduces the idea of superposition of solutions using Gauss’s law.

Consider a very long (infinite) line, located at a distanced ¼10 m above ground and charged with a uniform, line charge densityρl¼10^–7C/m as shown inFigure4.6a. Neglect the ground’s influence:

(a) Calculate the electric field intensity everywhere in space.

(b) What is the magnitude of the electric field intensity at ground level, directly below the line?

(c) A second, identical line is now placed at a distancea¼2 m below the first line as shown inFigure4.7aand is charged with a line charge densityρl[C/m]. Calculate the electric field intensity at a general point in space as well as at ground level directly below the lines. Compare with(a)and(b).

Solution: To solve the problem, we need to identify a Gaussian surface that encloses the line of charge symmetrically.

Because the line is infinite, we choose a Gaussian surface in the shape of a concentric cylinder of arbitrary lengthLand radiusrand calculate the electric field intensity at the surface of the cylinder. In(c), we repeat the solution for the second line and superimpose the two solutions.

ρl

ρl

L E

a b ^E

E

r d=10 mm

ground

dsa

dsb

dsc

. Figure 4.6 (a) A single line

above ground. (b) The Gaussian surface for a long line of charge. The surface is a cylinder of radiusr, arbitrary lengthL, and coaxial with the line

148 4 Gauss’s Law and the Electric Potential

(a) From Gauss’s law applied to the line inFigure4.6a, þ

s

E

^ds¼ρ_ε^lL

0

whereLis an arbitrary length of the line of charge. The Gaussian surface is shown inFigure4.6b. The electric field intensity at a distancerfrom the line is therefore

þ

s

E

^ds¼ð

s_a

Edsa¼2πrLE¼ρlL ε0

! E¼ ρl

2πrε0

N C In vector notation (the electric field is in the positiverdirection),

E¼r^ ρl

2πrε0

N C

Note that in this case, dsa¼r^dsa and E¼r^E. Also, the scalar products E·dsband E ·dsc are zero since E is perpendicular to dsb and dsc. Therefore, the only nonzero contribution to the scalar product is E·dsa and this equals E ·dsa. BecauseEis constant in magnitude on the surface, it is taken outside the integral.

(b) At ground level (r¼10 m, directly below the line), the field points downward and its magnitude is E¼ ρl

2πrε0

¼ ρl

2πdε0

¼ 10⁷

2π108:85410¹²¼179:75 N C

(c) Now, we have two lines, each at a different location in space. Therefore, we will use a Cartesian coordinate system and place the negatively charged line at (x2¼0, y2¼d – a¼8 m) and the positively charged line at (x1¼0, y1¼d¼10 m). Ground level is aty¼0. We use the result in(a)and write(seeFigure4.7b)

r1¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x²þðyy1Þ² q

, r2¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x²þðyy2Þ² q

The electric field intensity atP(x, y) due to each line is E^þ¼r^1

ρl 2πε0

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x²þðyy1Þ²

q , E¼r^2

ρl 2πε0

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x²þðyy2Þ²

q N

C

The total electric field intensity atP(x, y) is the superposition of the two fields. To calculate the vector sum of the two electric field intensities atP(x, y), we separate the two fields into their components. FromFigure4.7b, the components of the two electric fields are

E^þ¼x^E^þcosθ1þy^E^þsinθ1, E¼ ^xEcosθ2y^Esinθ2

N C

ground d=10m

θ1

θ2

ground

P(x,y)

a b

d ρl

ρ^l

−ρ^l

−ρ^l

(x₁,y₁)

(x2,y2) r₁ r₂

E⁺

E⁻ Ex−

Ey

−

Ex+

Ey +

a=2m a

x y

Figure 4.7 (a) A transmission line. (b) The electric field intensities produced by the two lines atP(x,y)

Also fromFigure4.7b,

cosθ1¼ x r1

¼ x

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x²þðyy₁Þ²

q , sinθ1¼yy1

r1

¼ yy1

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x²þðyy₁Þ² q

cosθ2¼ x r2

¼ x

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x²þðyy₂Þ²

q , sinθ2¼yy² r2

¼ yy₂ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi x²þðyy₂Þ² q

Substituting these intoE⁺andE^–and summing the components gives the electric field intensity atP(x, y) E¼x^ ρlx

2πε0

1

x²þðyy₁Þ² 1 x²þðyy₂Þ²

" #

þy^ ρl 2πε0

yy₁

x²þðyy₁Þ² yy₂ x²þðyy₂Þ²

" #

N C

At ground level, immediately below the two lines (x¼0,y¼0) and writingy1¼dandy2¼d–a, the magnitude of the electric field intensity is

Eð0;0Þ ¼ ρl

2πε0

1 y₂ 1

y₁ 2 4

3 5¼ ρl

2πε0

a

d dð aÞ¼ 10⁷

2π8:85410¹² 2

10ð102Þ¼44:94 N C

This electric field intensity is much lower than that of a single line. It can be reduced further by decreasing the distance between the two lines.

4.3.1.2 Calculation of Equivalent Charges

The second application of Gauss’s law is for the calculation of charge distributions from known electric field intensities.

This relies on Gauss’s law in the following form:

Q¼ε0

þ

s

E

^ds¼þ

s

D

^ds _m^C2

ð4:17Þ

Thus, as long as the electric field intensity is known, the equivalent charge can be evaluated. We note, however, that only the equivalent charge is evaluated, not the exact distribution. To see that this is the case, consider againFigure4.1. The point chargeQinFigure4.1aand the volume charge densityρvinFigure4.1bgenerate an identical electric field intensityEat a distanceRif the total charge inFigure4.1bis the same. Thus, even though the two charge distributions are quite different, the equivalent charges calculated from Gauss’s law are the same and equal toQ. This fact is not very surprising because of the integral relation used, but in practical terms, it also means that, in general, it is not possible to distinguish between the fields of point and distributed charges from knowledge of the electric field intensity alone. Sometimes, there may be additional information on the charge which allows us to distinguish between point charges and other distributions and their corresponding electric field intensities. The following two examples show applications of Gauss’s law to calculation of equivalent charges usingEq. (4.17).

Example 4.5 Application: Electric Field Intensity and Charge Density In and Around a Thundercloud A spherical cloud of charged particles of radius R0¼1 m produces a known electric field intensity inside the cloud (R R0), given asE¼R^R² [N/C]. Calculate:

(a) The total charge in the cloud.

(b) The charge density in the cloud.

150 4 Gauss’s Law and the Electric Potential

Solution: The total charge in the cloud is found from Gauss’s law from the fact that outside the charge distribution, the electric field intensity only depends on the total charge enclosed by the Gaussian surface, not on its distribution. From this, both the charge distribution and the electric field intensity may be found:

(a) Enclosing the sphere by a Gaussian surface of radiusR0, we write from Gauss’s law, Q¼4πR²₀ε0E¼4πR²0ε0R²₀¼4πε0 ½ C whereE¼R₀²¼1 was used as given.

(b) One method of calculating the charge density is to assume a radius-dependent charge density of general form, integrate over the volume of the sphere, and equate the result to that found in(a). A simpler method is to calculate the divergence of the electric field in the charge distribution. We show both.

Method A Assume a charge distribution of the formρv(R) [C/m³] and calculate the total charge in the sphere:

ð

v

ρvð ÞdvR ¼ ðR0

R¼0

ð_π

θ¼0

ð2π

ϕ¼0ρvð ÞRR ²sinθdRdθdϕ ½ C Although the charge distribution is not known, we can still integrate overθandϕ. This gives

ð

v

p_vð ÞdvR ¼4πðR₀

R¼0p_vð ÞRR ²dR¼4πR⁴0ε0 ½ C where the right-hand side is the result in(a). Clearly, for this to be satisfied, we must have

ρvð Þ ¼R 4Rε0

C m³

Method B The divergence of the electric field intensity [Eq. (4.6)] gives

∇

^{E¼ 1}

R²

∂ R²ER

∂R ¼ 1 R²

∂ R⁴

∂R ¼4R¼ρv

ε0

! ρv¼4Rε0

C m³

which, of course is the same as in(a).

Example 4.6Application: Generation of a Uniform Electric Field Intensity It is necessary to generate a uniform electric field intensity of magnitude E0¼10⁵ N/C between two very large parallel plates, separated a distance d ¼10 mm as shown inFigure4.8a. The material between the plates is free space and the electric field intensity outside the plates is zero. Calculate the required charge densities on the surface of each plate.

.

ε⁰

very large

E

a b

ds2

ds1

ε⁰

s s1

s2

d Figure 4.8 (a) Two very

large charged plates. (b) The Gaussian surface necessary to calculate the electric charge density on the upper plate

Solution: A Gaussian surface is drawn in the form of an arbitrarily shaped volume symmetric about one of the plates, as shown inFigure4.8b. This volume has surfaces either parallel to the plate or perpendicular to it to facilitate calculation of the scalar product in Gauss’s law. Also, we assume the electric field points from the upper to the lower plate. From Gauss’s law, we have

þ

s

E

^ds¼Eð

s2

ds2¼Es2¼Q ε0

The only charge is on the surfacesin the form of a uniform charge densityρs[C/m²]. The total charge on surfacesis ρss[C]. Sinces¼s1¼s2, we get

Es2¼ρss2

ε0

! ρs¼ε0E¼8:85410¹²10⁵¼8:85410⁷ C m²

The charge density on the lower plate must be identical and negative (sinceEanddsare in opposite directions andE points from upper to lower plate). Thus, the solution is

ρs¼8:85410⁷ ½C=m²on the upper plate and

ρs¼ 8:85410⁷ ½C=m²on the lower plate