Every problem in the chapter beside the book’s answer, a chapter at a time as you scroll.

Chapter 6: Gauss's Law

54 problems · 21 figures

Check Your Understanding 6.1 §6.1 source ↗

What angle should there be between the electric field and the surface shown in Figure 6.11 in the previous example so that no electric flux passes through the surface?
Figure 6.11 in the book ↗
The electric field produces a net electric flux through the surface S.

Answer

Place it so that its unit normal is perpendicular to E→.

Check Your Understanding 6.2 §6.1 source ↗

If the electric field in Example 6.4 is E→=mxk^, what is the flux through the rectangular area?

Answer

m a b 2 / 2

Check Your Understanding 6.3 §6.2 source ↗

Calculate the electric flux through the closed cubical surface for each charge distribution shown in Figure 6.20.
Figure 6.20 in the book ↗
A cubical Gaussian surface with various charge distributions.

Answer

a. 3.4×105N·m2/C; b. −3.4×105N·m2/C; c. 3.4×105N·m2/C; d. 0

Check Your Understanding 6.4 §6.3 source ↗

Check that the electric fields for the sphere reduce to the correct values for a point charge.

Answer

In this case, there is only E→out.So, yes.

Check Your Understanding 6.5 §6.3 source ↗

A thin straight wire has a uniform linear charge density λ0. Find the electric field at a distance d from the wire, where d is much less than the length of the wire.

Answer

E→= λ02πε01dr^; This agrees with the calculation of Example 5.5 where we found the electric field by integrating over the charged wire. Notice how much simpler the calculation of this electric field is with Gauss’s law.
Example 5.5 Electric Field of a Line Segment in the book ↗

Check Your Understanding 6.6 §6.4 source ↗

How will the system above change if there are charged objects external to the sphere?

Answer

If there are other charged objects around, then the charges on the surface of the sphere will not necessarily be spherically symmetrical; there will be more in certain direction than in other directions.

6-1 §6.1 source ↗

Discuss how to orient a planar surface of area A in a uniform electric field of magnitude E0 to obtain (a) the maximum flux and (b) the minimum flux through the area.

Answer

a. If the planar surface is perpendicular to the electric field vector, the maximum flux would be obtained. b. If the planar surface were parallel to the electric field vector, the minimum flux would be obtained.

6-3 §6.1 source ↗

The net electric flux crossing a closed surface is always zero. True or false?

Answer

False. The net electric flux crossing a closed surface is always zero if and only if the net charge enclosed is zero.

6-5 §6.2 source ↗

Two concentric spherical surfaces enclose a point charge q. The radius of the outer sphere is twice that of the inner one. Compare the electric fluxes crossing the two surfaces.

Answer

Since the electric field vector has a 1r2 dependence, the fluxes are the same since A=4πr2.

6-7 §6.2 source ↗

(a) If the electric flux through a closed surface is zero, is the electric field necessarily zero at all points on the surface? (b) What is the net charge inside the surface?

Answer

a. no; b. zero

6-9 §6.2 source ↗

Discuss the similarities and differences between the gravitational field of a point mass m and the electric field of a point charge q.

Answer

Both fields vary as 1r2. Because the gravitational constant is so much smaller than 14πε0, the gravitational field is orders of magnitude weaker than the electric field. Also, the gravitational flux through a closed surface is zero or positive; however, the electric flux is positive, negative, or zero, depending on the definition of flux for the given situation.

6-11 §6.2 source ↗

Is the term E→ in Gauss’s law the electric field produced by just the charge inside the Gaussian surface?

Answer

No, it is produced by all charges both inside and outside the Gaussian surface.

6-13 §6.3 source ↗

Would Gauss’s law be helpful for determining the electric field of two equal but opposite charges a fixed distance apart?

Answer

No, since the situation does not have symmetry, making Gauss’s law challenging to simplify.

6-15 §6.3 source ↗

Discuss the restrictions on the Gaussian surface used to discuss planar symmetry. For example, is its length important? Does the cross-section have to be square? Must the end faces be on opposite sides of the sheet?

Answer

Any shape of the Gaussian surface can be used. The only restriction is that the Gaussian integral must be calculable; therefore, a box or a cylinder are the most convenient geometrical shapes for the Gaussian surface.

6-17 §6.4 source ↗

Under electrostatic conditions, the excess charge on a conductor resides on its surface. Does this mean that all the conduction electrons in a conductor are on the surface?

Answer

No. If a metal was in a region of zero electric field, all the conduction electrons would be distributed uniformly throughout the metal.

6-19 §6.4 source ↗

The conductor in the preceding figure has an excess charge of –5.0µC. If a 2.0-µC point charge is placed in the cavity, what is the net charge on the surface of the cavity and on the outer surface of the conductor?

Answer

Since the electric field is zero inside a conductor, a charge of −2.0μC is induced on the inside surface of the cavity. This will put a charge of +2.0μC on the outside surface leaving a net charge of −3.0μC on the surface.

6-21 §6.1 source ↗

Calculate the flux through the sheet of the previous problem if the plane of the sheet is at an angle of 60° to the field. Find the flux for both directions of the unit normal to the sheet.

Answer

Φ= E→·A→→ EAcosθ= 3.8×104N·m2/C electric field in direction of unit normal; Φ= E→·A→→ EAcosθ= −3.8×104N·m2/C electric field opposite to unit normal

6-23 §6.1 source ↗

The electric flux through a square-shaped area of side 5 cm near a large charged sheet is found to be 3×10−5N·m2/C when the area is parallel to the plate. Find the charge density on the sheet.

Answer

3 × 10 −5 N · m 2 / C ( 0.05 m ) 2 = E⇒σ= 2.12× 10 −13 C/ m 2

6-25 §6.1 source ↗

A square surface of area 2cm2 is in a space of uniform electric field of magnitude 103N/C. The amount of flux through it depends on how the square is oriented relative to the direction of the electric field. Find the electric flux through the square, when the normal to it makes the following angles with electric field: (a) 30°, (b) 90°, and (c) 0°. Note that these angles can also be given as 180°+θ.

Answer

a. Φ= 0.17N·m2/C;
b. Φ=0; c. Φ= EAcos0°= 1.0×103N/C(2.0×10−4m)2cos0°= 0.20N·m2/C

6-27 §6.1 source ↗

Consider the uniform electric field E→= (4.0j^+3.0k^)×103N/C. What is its electric flux through a circular area of radius 2.0 m that lies in the xy-plane?

Answer

Φ= 3.8× 10 4 N· m 2 /C

6-29 §6.1 source ↗

An infinite charged wire with charge per unit length λ lies along the central axis of a cylindrical surface of radius r and length l. What is the flux through the surface due to the electric field of the charged wire?

Answer

E → (z)= 1 4 π ε 0 2 λ z k ^ , ∫ E → · n ^ dA= λ ε 0 l

6-31 §6.2 source ↗

Find the electric flux through the closed surface whose cross-sections are shown below.

Figure a shows an irregular shape with a positive charge inside it labeled 3 into 10 to the power minus 8 C. There is a negative charge outside it, labeled minus 2 into 10 to the power 8 C. Figure b shows an irregular shape with three charges outside it. These are plus 4 into 10 to the power minus 6 C, plus 5 into 10 to the power minus 6 C and minus three into 10 to the power minus 6 C. Figure c shows a square with the length of each side equal to a. There is a charge minus 2 into 10 to the power minus 6 C within it. Figure d shows a shaded strip with plus signs near the inside edges. It is labeled conductor. An arrow points outward from either end of the strip. These arrows are labeled infinity. A small rectangle is attached to one side of the strip, covering one plus sign. It is labeled end cap of area, 4 into 10 to the power minus 4 m squared. The strip is labeled sigma equal to 2 into 10 to the power minus 6 C by m squared.Figure a shows an irregular shape with a positive charge inside it labeled 3 into 10 to the power minus 8 C. There is a negative charge outside it, labeled minus 2 into 10 to the power 8 C. Figure b shows an irregular shape with three charges outside it. These are plus 4 into 10 to the power minus 6 C, plus 5 into 10 to the power minus 6 C and minus three into 10 to the power minus 6 C. Figure c shows a square with the length of each side equal to a. There is a charge minus 2 into 10 to the power minus 6 C within it. Figure d shows a shaded strip with plus signs near the inside edges. It is labeled conductor. An arrow points outward from either end of the strip. These arrows are labeled infinity. A small rectangle is attached to one side of the strip, covering one plus sign. It is labeled end cap of area, 4 into 10 to the power minus 4 m squared. The strip is labeled sigma equal to 2 into 10 to the power minus 6 C by m squared.

Answer

a. Φ= 3.39×103N·m2/C; b. Φ=0;
c. Φ= −2.25×105N·m2/C;
d. Φ= 90.4N·m2/C

6-33 §6.2 source ↗

A point charge of 10μC is at an unspecified location inside a cube of side 2 cm. Find the net electric flux though the surfaces of the cube.

Answer

Φ= 1.13× 10 6 N· m 2 /C

6-35 §6.2 source ↗

A charge q is placed at one of the corners of a cube of side a, as shown below. Find the magnitude of the electric flux through the shaded face due to q. Assume q>0.

Figure shows a cube with length of each side equal to a. The back surface of it is shaded. One front corner has a small circle on it labeled q.

Answer

Make a cube with q at the center, using the cube of side a. This would take four cubes of side a to make one side of the large cube. The shaded side of the small cube would be 1/24th of the total area of the large cube; therefore, the flux through the shaded area would be
Φ=124qε0.

6-37 §6.2 source ↗

The electric flux through a spherical surface is 4.0×104N·m2/C. What is the net charge enclosed by the surface?

Answer

q= 3.54× 10 −7 C

6-39 §6.2 source ↗

Repeat the previous problem, assuming that the electric field is directed along a body diagonal of the cube.

Answer

zero, also because flux in equals flux out

6-41 §6.3 source ↗

Recall that in the example of a uniform charged sphere, ρ0= Q/(43πR3). Rewrite the answers in terms of the total charge Q on the sphere.

Answer

r> R, E= Q 4 π ε 0 r 2 ; r< R, E= q r 4 π ε 0 R 3

6-43 §6.3 source ↗

A very long, thin wire has a uniform linear charge density of 50μC/m. What is the electric field at a distance 2.0 cm from the wire?

Answer

EA= λ l ε 0 ⇒E= 4.50× 10 7 N/C

6-45 §6.3 source ↗

Repeat your calculations for the preceding problem, given that the charge is distributed uniformly over the surface of a spherical conductor of radius 10.0 cm.

Answer

a. 0; b. 0; c. E→= 6.74×106N/C(−r^)

6-47 §6.3 source ↗

When a charge is placed on a metal sphere, it ends up in equilibrium at the outer surface. Use this information to determine the electric field of +3.0μC charge put on a 5.0-cm aluminum spherical ball at the following two points in space: (a) a point 1.0 cm from the center of the ball (an inside point) and (b) a point 10 cm from the center of the ball (an outside point).

Answer

a. 0; b. E= 2.70×106N/C

6-49 §6.3 source ↗

Determine if approximate cylindrical symmetry holds for the following situations. State why or why not. (a) A 300-cm long copper rod of radius 1 cm is charged with +500 nC of charge and we seek electric field at a point 5 cm from the center of the rod. (b) A 10-cm long copper rod of radius 1 cm is charged with +500 nC of charge and we seek electric field at a point 5 cm from the center of the rod. (c) A 150-cm wooden rod is glued to a 150-cm plastic rod to make a 300-cm long rod, which is then painted with a charged paint so that one obtains a uniform charge density. The radius of each rod is 1 cm, and we seek an electric field at a point that is 4 cm from the center of the rod. (d) Same rod as (c), but we seek electric field at a point that is 500 cm from the center of the rod.

Answer

a. Yes, the length of the rod is much greater than the distance to the point in question. b. No, The length of the rod is of the same order of magnitude as the distance to the point in question. c. Yes, the length of the rod is much greater than the distance to the point in question. d. No. The length of the rod is of the same order of magnitude as the distance to the point in question.

6-51 §6.3 source ↗

The electric field at 2 cm from the center of long copper rod of radius 1 cm has a magnitude 3 N/C and directed outward from the axis of the rod. (a) How much charge per unit length exists on the copper rod? (b) What would be the electric flux through a cube of side 5 cm situated such that the rod passes through opposite sides of the cube perpendicularly?

Answer

a. E→= Rσ0ε01rr^⇒σ0= 5.31×10−11C/m2,
λ= 3.33×10−12C/m;
b. Φ= qencε0= 3.33×10−12C/m(0.05m)ε0= 0.019N·m2/C

6-53 §6.3 source ↗

Charge is distributed uniformly with a density ρ throughout an infinitely long cylindrical volume of radius R. Show that the field of this charge distribution is directed radially with respect to the cylinder and that

E = ρ r 2 ε 0 ( r ≤ R ) ; E = ρ R 2 2 ε 0 r ( r ≥ R ) .

Answer

E2πrl= ρπr2lε0⇒E= ρr2ε0(r≤R);
E2πrl= ρπR2lε0⇒E= ρR22ε0r(r≥R)

6-55 §6.3 source ↗

The electric field 10.0 cm from the surface of a copper ball of radius 5.0 cm is directed toward the ball’s center and has magnitude 4.0×102N/C. How much charge is on the surface of the ball?

Answer

Φ= q enc ε 0 ⇒ q enc = −1.0× 10 −9 C

6-57 §6.3 source ↗

Charge is distributed throughout a spherical volume of radius R with a density ρ=αr2, where α is a constant. Determine the electric field due to the charge at points both inside and outside the sphere.

Answer

qenc= 45παr5,
E4πr2= 4παr55ε0⇒E= αr35ε0(r≤R),
qenc= 45παR5, E4πr2= 4παR55ε0⇒E= αR55ε0r2(r≥R)

6-59 §6.3 source ↗

The volume charge density of a spherical charge distribution is given by ρ(r)= ρ0e−αr, where ρ0 and α are constants. What is the electric field produced by this charge distribution?

Answer

integrate by parts: qenc= 4πρ0[−e−αr((r)2α+2rα2+2α3)+2α3]⇒E= ρ0r2ε0[−e−αr((r)2α+2rα2+2α3)+2α3]

6-61 §6.4 source ↗

An uncharged spherical conductor S of radius R has two spherical cavities A and B of radii a and b, respectively as shown below. Two point charges +qa and +qb are placed at the center of the two cavities by using non-conducting supports. In addition, a point charge +q0 is placed outside at a distance r from the center of the sphere. (a) Draw approximate charge distributions in the metal although metal sphere has no net charge. (b) Draw electric field lines. Draw enough lines to represent all distinctly different places.

Figure shows a sphere with two cavities. A positive charge qa is in one cavity and a positive charge qb is in the other cavity. A positive charge q0 is outside the sphere at a distance r from its center.

Answer


Figure shows a sphere with two cavities. A positive charge qa is in one cavity and a positive charge qb is in the other cavity. A positive charge q0 is outside the sphere at a distance r from its center.

6-63 §6.4 source ↗

A long cylinder of copper of radius 3 cm is charged so that it has a uniform charge per unit length on its surface of 3 C/m. (a) Find the electric field inside and outside the cylinder. (b) Draw electric field lines in a plane perpendicular to the rod.

Answer

a. Outside: E2πrl= λlε0⇒E= 3.0C/m2πε0r; Inside Ein=0; b.

A shaded circle is shown with plus signs around its edge. Arrows from the circle radiate outwards.

6-65 §6.4 source ↗

A long cylinder of aluminum of radius R meters is charged so that it has a uniform charge per unit length on its surface of λ. (a) Find the electric field inside and outside the cylinder. (b) Plot electric field as a function of distance from the center of the rod.

Answer

a. E2πrl= λlε0⇒E= λ2πε0rr≥R E inside equals 0; b.

A graph of E versus r is shown. The curve rises up in a vertical line from a point R on the x axis. It then drops gradually and evens out just above the x axis.

6-67 §6.4 source ↗

Two parallel plates 10 cm on a side are given equal and opposite charges of magnitude 5.0×10−9C. The plates are 1.5 mm apart. What is the electric field at the center of the region between the plates?

Answer

E= 5.65× 10 4 N/C

6-69 §6.4 source ↗

The surface charge density on a long straight metallic pipe is σ. What is the electric field outside and inside the pipe? Assume the pipe has a diameter of 2a.

Figure shows a pipe, with a cylindrical section highlighted. An arrow pointing up and one pointing down along the pipe from the cylinder are labeled infinity. There are plus signs inside the walls of the cylinder.

Answer

λ= λlε0⇒E= aσε0r for r≥a, E=0 inside since qenclosed=0

6-71 §6.4 source ↗

A solid cylindrical conductor of radius a is surrounded by a concentric cylindrical shell of inner radius b. The solid cylinder and the shell carry charges +Q and –Q, respectively. Assuming that the length L of both conductors is much greater than a or b, determine the electric field as a function of r, the distance from the common central axis of the cylinders, for (a) r<a; (b) a< r< b; and (c)r>b.

Answer

a. E=0; b. E2πrL= Qε0⇒E= Q2πε0rL; c. E=0 since r would be either inside the second shell or if outside then q enclosed equals 0.

6-73 §6.AP source ↗

Repeat the preceding problem, with E→= 2xi^+ 3x2k^.

Answer

∫ E → · n ^ d A = a 4

6-75 §6.AP source ↗

(a) Calculate the electric flux through the open hemispherical surface due to the electric field E→=E0k^ (see below). (b) If the hemisphere is rotated by 90° around the x-axis, what is the flux through it?

A hemisphere with radius R is shown with its base in the xy plane and center of base at the origin. An arrow is shown beside it, labeled vector E equal to E0 k hat.

Answer

a. ∫E→·n^dA=E0r2π; b. zero, since the flux through the upper half cancels the flux through the lower half of the sphere

6-77 §6.AP source ↗

The electric field in a region is given by E→= a/(b+cx)i^, where a=200N·m/C,b=2.0m, and c=2.0. What is the net charge enclosed by the shaded volume shown below?

Figure shows a cuboid with one corner on the origin of the coordinate axes. Its length along the x axis is 2 m, along y axis is 1.5 m and along z axis is 1 m. An arrow outside the cuboid points along the x axis. It is labeled vector E.

Answer

Φ=qencε0; There are two contributions to the surface integral: one at the side of the rectangle at x=0 and the other at the side at x=2.0m;
− E(0)[1.5m2]+ E(2.0m)[1.5m2]= qencε0= −100Nm2/C
where the minus sign indicates that at x=0, the electric field is along positive x and the unit normal is along negative x. At x=2, the unit normal and the electric field vector are in the same direction: qenc= ε0Φ= −8.85×10−10C.

6-79 §6.AP source ↗

A fellow student calculated the flux through the square for the system in the preceding problem and got 0. What went wrong?

Answer

didn’t keep consistent directions for the area vectors, or the electric fields

6-81 §6.AP source ↗

Two 10cm×10cm pieces of aluminum foil of thickness 0.1 mm face each other with a separation of 5 mm. One of the foils has a charge of +30μC and the other has −30μC. (a) Find the charge density at all surfaces, i.e., on those facing each other and those facing away. (b) Find the electric field between the plates near the center assuming planar symmetry.

Answer

a. σ= 3.0×10−3C/m2, +3×10−3C/m2 on one and −3×10−3C/m2 on the other; b. E= 3.39×108N/C

6-83 §6.AP source ↗

The infinite slab between the planes defined by z=−a/2 and z=a/2 contains a uniform volume charge density ρ (see below). What is the electric field produced by this charge distribution, both inside and outside the distribution?

Figure shows a cuboid with its center at the origin of the coordinate axes. Arrows perpendicular to the surfaces of the cuboid point outward. The arrows along positive x and y axes are labeled infinity and the arrows along the negative x and y axes are labeled minus infinity. The cuboid is labeled rho. Its top surface is labeled z equal to plus a by 2 and its bottom surface is labeled z equal to minus a by 2.Figure shows a cuboid with its center at the origin of the coordinate axes. Arrows perpendicular to the surfaces of the cuboid point outward. The arrows along positive x and y axes are labeled infinity and the arrows along the negative x and y axes are labeled minus infinity. The cuboid is labeled rho. Its top surface is labeled z equal to plus a by 2 and its bottom surface is labeled z equal to minus a by 2.

Answer

Construct a Gaussian cylinder along the z-axis with cross-sectional area A.
|z|≥a2qenc= ρAa, Φ= ρAaε0⇒E= ρa2ε0,
|z|≤a2qenc= ρA2z, E(2A)= ρA2zε0⇒E= ρzε0

6-85 §6.AP source ↗

A non-conducting spherical shell of inner radius a1 and outer radius b1 is uniformly charged with charged density ρ1 inside another non-conducting spherical shell of inner radius a2 and outer radius b2 that is also uniformly charged with charge density ρ2. See below. Find the electric field at space point P at a distance r from the common center such that (a) r>b2, (b) a2< r< b2, (c) b1< r< a2, (d) a1< r< b1, and (e) r<a1.

Figure shows two concentric circular shells. The inner and outer radii of the inner shell are a1 and a2 respectively. The inner and outer radii of the outer shell are a2 and b2 respectively. The distance from the center to a point P between the two shells is labeled r.

Answer

a. r> b2E4πr2= 43π[ρ1(b13−a13)+ρ2(b23−a23)ε0⇒E=ρ1(b13−a13)+ρ2(b23−a23)3ε0r2;
b. a2< r< b2  E4πr2= 43π[ρ1(b13−a13)+ρ2(r3−a23)]ε0⇒E= ρ1(b13−a13)+ρ2(r3−a23)3ε0r2;
c. b1< r< a2  E4πr2= 43πρ1(b13−a13)ε0⇒E= ρ1(b13−a13)3ε0r2;
d. a1< r< b1  E4πr2= 43πρ1(r3−a13)ε0⇒E= ρ1(r3−a13)3ε0r2; e. 0

6-87 §6.AP source ↗

A disk of radius R is cut in a non-conducting large plate that is uniformly charged with charge density σ (coulomb per square meter). See below. Find the electric field at a height h above the center of the disk. (h>>R,h<<lorw). (Hint: Fill the hole with ±σ.)

A plate with length l and width w has a hole in the center. A point P above the plate is at a distance h from its center.A plate with length l and width w has a hole in the center. A point P above the plate is at a distance h from its center.

Answer

Electric field due to plate without hole: E=σ2ε0.
Electric field of just hole filled with −σE= −σ2ε0(1−zR2+z2).
Thus, Enet= σ2ε0hR2+h2.

6-89 §6.AP source ↗

Shown below are two concentric conducting spherical shells of radii R1 and R2, each of finite thickness much less than either radius. The inner and outer shell carry net charges q1 and q2, respectively, where both q1 and q2 are positive. What is the electric field for (a) r<R1; (b) R1< r< R2; and (c) r>R2? (d) What is the net charge on the inner surface of the inner shell, the outer surface of the inner shell, the inner surface of the outer shell, and the outer surface of the outer shell?

Figure shows section of two concentric spherical shells. The inner one has radius R1 and the outer one has radius R2.

Answer

a. E=0; b. E=q14πε0r2; c. E= q1+q24πε0r2; d. 0, q1, − q1, q1+ q2

6-91 §6.CP source ↗

The Hubble Space Telescope can measure the energy flux from distant objects such as supernovae and stars. Scientists then use this data to calculate the energy emitted by that object. Choose an interstellar object which scientists have observed the flux at the Hubble with (for example, Vega3), find the distance to that object and the size of Hubble’s primary mirror, and calculate the total energy flux. (Hint: The Hubble intercepts only a small part of the total flux.)

Answer

Given the referenced link, using a distance to Vega of 237×1015 m1 and a diameter of 2.4 m for the primary mirror,2 we find that at a wavelength of 555.6 nm, Vega is emitting 2.44×1024J/s at that wavelength. Note that the flux through the mirror is essentially constant.

6-93 §6.CP source ↗

An infinite plate sheet of charge of surface charge density σ is shown below. What is the electric field at a distance x from the sheet? Compare the result of this calculation with that of worked out in the text.

Figure shows a plane with plus signs on it. It goes through a cylinder S perpendicularly at the center, dividing the cylinder into two halves of length x each. The top surface of the cylinder is labeled A. Arrows labeled E emerge from both ends of the cylinder, along its axis.Figure shows a plane with plus signs on it. It goes through a cylinder S perpendicularly at the center, dividing the cylinder into two halves of length x each. The top surface of the cylinder is labeled A. Arrows labeled E emerge from both ends of the cylinder, along its axis.

Answer

The symmetry of the system forces E→ to be perpendicular to the sheet and constant over any plane parallel to the sheet. To calculate the electric field, we choose the cylindrical Gaussian surface shown. The cross-section area and the height of the cylinder are A and 2x, respectively, and the cylinder is positioned so that it is bisected by the plane sheet. Since E is perpendicular to each end and parallel to the side of the cylinder, we have EA as the flux through each end and there is no flux through the side. The charge enclosed by the cylinder is σA, so from Gauss’s law, 2EA=σAε0, and the electric field of an infinite sheet of charge is
E=σ2ε0, in agreement with the calculation of in the text.

6-95 §6.CP source ↗

Find the electric field of a large conducting plate containing a net charge q. Let A be area of one side of the plate and h the thickness of the plate (see below). The charge on the metal plate will distribute mostly on the two planar sides and very little on the edges if the plate is thin.

A cylinder goes through a hole at the center of a plate of thickness h. The plate has plus signs on its edge. The cylinder is labeled Gaussian pill box. The portion of the cylinder on the right side of the plate is of length z. Two arrows from its flat surface point outward, perpendicular to the surface. These are labeled vector E subscript P and delta vector A. The flat surface of the cylinder to the left of the plate has two arrows perpendicular to it, pointing outward. These are labeled mod vector E equal to mod vector E subscript P and delta vector A.

Answer

There is Q/2 on each side of the plate since the net charge is Q: σ=Q2A,
∮SE→·n^dA= 2σΔAε0⇒EP= σε0= Qε02A

Loading chapter 7…

Problems and answers are adapted from University Physics Volume 2 by Samuel J. Ling, Jeff Sanny and William Moebs, OpenStax, and are used under the CC BY-NC-SA 4.0 licence. Access for free at https://openstax.org/books/university-physics-volume-2/pages/1-introduction. Illustrations: Copyright Rice University, OpenStax, under CC BY-NC-SA 4.0 license. This adaptation is released under the same licence, is non-commercial, and is neither affiliated with nor endorsed by OpenStax or Rice University.