Chapter 23

What is the x coordinate for the patch (and any other patch on the left face)?

x = 1.0 m

What is the x coordinate for the patch (and any other patch on the right face)?

x = 3.0 The points on the right face differ in their y and z coordinates but they are all at coordinate x=3.0 m.

What is the flux psi total in dot product form?

∫ (3.0xi + 4.0j) dot (dAj)

We enclose the particle in a Gaussian sphere (or G-sphere) that is centered on the particle. The remaining integral is just an instruction to sum the areas of all the patches on our G-sphere with radius r. What is the sum of those areas?

∫dA = 4πr² This is the surface area of the sphere with radius r.

A spherical metal shell contains a central particle with charge +Q. The shell has a net charge of +3Q. (d) What should be substituted for the symbol q if the point is between the shell's inner surface and outer surface?

0 By Gauss' law, an isolated metal object cannot have an internal electric field.

How much charge is enclosed by S1?

0 S1 only encloses the empty interior of the shell.

On either end cap, what is the angle θ between the field vector E→ and the area vector dA→?

0 degrees

Gauss' law contains the dot product E→⋅dA→, which can be expanded as E⁢ dA⁢ cos⁢θ, where θ is the angle between the two vectors. What is the angle in this case?

0 degrees since the angle vector is radially outward.

For any patch on the surface of the G-sphere, what is the angle θ between field vector E→ piercing the patch and the patch's area vector dA→?

0 degrees. Both vectors are radially outward.

What is the flux Φt through the top face?

16 N * m^2/C

In terms of the cylinder's cross-sectional area A, what is the total area through which there is flux?

2A. Recall that only the end caps are pierced.

In this next figure, the same G-cylinder extends through the same metal plate with the same surface charge density. In terms of the cylinder's cross-sectional area A, what is the area through which there is flux?

2A. Both the left end cap and the right end cap are pierced by the field.

What is the volume Vball of the ball (radius R) with charge q?

4/3 pi R^3

A spherical metal shell contains a central particle with charge +Q. The shell has a net charge of +3Q. (e) What should be substituted for the symbol q if the point is outside the shell's outer surface?

4Q Imagine a concentric Gaussian sphere enclosing the shell and thus also the particle. The net charge enclosed by the G-sphere is +4Q.

In terms of e, what is the charge qenc enclosed by the Gaussian sphere?

5e. The Gaussian sphere does not enclose the charge on the plastic shell.

In this second case, in terms of the cylinder's cross-sectional area A, what is the area through which there is flux?

Area pierced by E field = A Recall that we cannot have an electric field inside an isolated piece of metal. So, there is no field piercing through the end cap that is buried inside the metal.

What now is the angle θ between field vector E→ and the patch area vector dA→?

Because the net charge is negative, the electric field vectors on the sphere's surface pierce inward, the angle between E and dA is 180.

When we give the shell that net positive charge, what happens to the amount of negative charge on the outer wall?

Decreases With the same amount of positive charge on the inner wall but now with a need of a positive net charge on the shell, there must be less negative charge on the outer wall. The outer charge is now −2.0 μC, so that the net charge on the shell is −2.0 μC + 5.0 μC = +3.0 μC.

What is the field magnitude at the center of the nucleus?

E = 0

What is the electric field magnitude E inside the metal and thus on the internal cap?

E = 0 N/C An isolated conductor cannot have an internal electric field.

Which statement is true about the magnitude E of the electric field at points on the Gaussian sphere?

E has the same value. Because the Gaussian sphere is centered on the particle, all points on its surface are at the same distance r1 from the particle and thus have the same magnitude of electric field.

For points between the center and the surface, which describes how the field magnitude depends on radial distance r

E is a linear function of r. Inside the nucleus, the field magnitude E increases linearly as r increases.

If we extend the G-cylinder through the metal plate instead of burying one end cap inside the metal, our results for the _____ change but not the _______.

Enclosed charge and the flux through the cylinder Result for the field magnitude

Which component pierces the patch?

Ex=3.0⁢ x The y component only skims the patch (and all the other patches on the left face) and thus does not contribute any flux.

Which component of the field vector E→=3.0xi^+4.0j^ pierces the top face?

Ey = 4.0

The figure here shows a Gaussian cube of face area A immersed in a uniform electric field that has the positive direction of the z axis. In terms of E and A, what is the flux through (a) the front face (which is in the xy plane), (b) the rear face, (c) the top face, and (d) the whole cube?

Front. +EA. The field pierces outward. Rear. -EA. The field pierces inward. Top. 0. The field skims. Same with the faces at the bottom, left side, and right side. Whole. 0. Net flux is zero.

The figure shows three situations in which a Gaussian cube sits in an electric field. The arrows and the values indicate the directions of the field lines and the magnitudes (in N·m2/C) of the flux through the six sides of each cube. (The lighter arrows are for the hidden faces.) In which situation does the cube enclose (a) a positive net charge, (b) a negative net charge, and (c) zero net charge?

Ingoing arrows indicate negative flux; outgoing arrows indicate positive flux. For each cube, find the net flux. By Gauss' law, the sign of the net flux Φ is the sign of the net enclosed charge qenc. Cube 1: 14 ingoing, 14 outgoing, zero Φ , zero qenc Cube 2: 13 ingoing, 18 outgoing, positive Φ, positive qenc Cube 3: 18 ingoing, 15 outgoing, negative Φ , negative qenc

Is the sign of the flux through the right face consistent with the piercing component of the electric field or do we need to insert a minus sign?

It is consistent. The flux has an implicit plus sign, which is consistent with the x component of the field piercing outward from the interior of the cube.

Which is true about the field magnitude E at various points on the surface of the G-sphere?

Magnitude E is the same over the surface.

What are the sign and direction of the electric flux on the surface of the Gaussian sphere?

Negative and radially inward Because the net charge enclosed by the new Gaussian sphere is negative, the electric flux through the surface must be negative and thus inward.

Does our result for field magnitude E just outside a conductor's surface contain any dependence on the distance from the conductor?

No

When the Gaussian surface is centered at x=R/2, y=0 is the magnitude of the electric field |E→| the same (uniform) over the surface of the sphere?

No

Does our result for field magnitude E contain any dependence on the distance from the sheet?

No There is no dependence on distance as long as the point of field measurement is near a large nonconducting sheet with uniform charge density.

When the Gaussian surface is centered at the origin, is the magnitude of the electric field |E→| the same (uniform) over the surface of the sphere?

No.

When you calculate the electric flux through a Gaussian surface, of what are you determining the flow through the surface?

Not charge, electric current, electric energy, electric field. None of the above answers are correct.

Which gives the SI unit of flux?

N·m2/C Think in terms of flux being EA and thus the unit is (N/C)(m2).

Where does the electric field pierce the cylinder?

Only on the end caps The field vectors skim the curved side but pierce directly through the end caps.

Where does the electric field pierce the cylinder?

Only on the external end cap The field vectors skim the curved side but pierce directly through the end cap that is outside the conductor.

What is the magnitude of the electric field at point S, which lies on the x axis at x=0.100 m?

Point S has coordinate x=0.100 m. This means the point is inside the shell. Remember the shell's radius is R=0.250 m, and the center of the shell lies at the origin. The shell theorem tells you that the electric field everywhere inside a uniformly-charged shell is equal to zero. So the electric field magnitude at S is zero (ANSWER).

Which expression gives the area vector for a patch on the bottom face?

Points down

What is the flux Φbot through the bottom face?

Points down, opposite of top

The next figure shows, in cross section, a plastic, spherical shell with uniform charge Q = −16e and radius R = 10 cm. A particle with charge q = +5e is at the center. We want the electric field (magnitude and direction) at (a) point P1 at radial distance r1 = 6.00 cm and (b) point P2 at radial distance r2 = 12.0 cm. What are the sign and direction of the electric flux on the surface of the Gaussian sphere?

Positive and radially outward. Because the charge enclosed by the Gaussian sphere is positive, the electric flux through the surface must be positive and outward.

A spherical metal shell contains a central particle with charge +Q. The shell has a net charge of +3Q. (c) We want to find the electric field magnitude at a point that is a radial distance r from the central particle (charge +Q) by using the generic equation E = k (q/r^2). What should be substituted for the symbol q if the point is between the particle and the shell's inner surface?

Q

What is the direction of the area vector dA→ for a patch element on the Gaussian sphere at P1?

Radially outward.

In the example, the metal shell is electrically neutral. If, instead, we give it a net positive charge of +3.0 μC, what happens to the amount of positive charge on the inner wall?

Remains the same The magnitude of the positive charge on the inner wall must still match the magnitude of the negative charge of the enclosed particle. By Gauss' law, that is the only way that the electric field inside the metal can be zero.

What is the flux Φb through the back face?

Same reasoning as before

Is flux a vector quantity or a scalar quantity?

Scalar. The expression Φv=|v→|cosθA involves the scalar product of two vectors, so the result must be a scalar.

The simulation shows that ΦE=1.13×105 Nm2/C, just like for a spherical Gaussian surface centered at the origin. Which of the following is the best explanation for this observation?

The cubical surface encloses the same amount of charge as the spherical surface.

Which is true?

The field magnitude is uniform across the external end cap. The cap is tiny; there is nothing that would make the field magnitude greater at, say, the top of the cap than at the bottom.

The figure shows a cross section of a spherical metal shell o dinner radius R. A particle with a charge of -5.0 uC is located at a distance R/2 from the center of the shell. If the shell is electrically neutral, what are the induced charges on its inner and outer surfaces? Are those charges uniformly distributed? What is the field patter inside and outside the shell?

The figure shows a cross section of a spherical Gaussian surface within the metal, just outside the inner wall of the shell. The electric field must be zero inside the metal (and thus on the Gaussian surface inside the metal). This means that the electric flux through the Gaussian surface must also be zero. Gauss' law then tells us that the net charge enclosed by the Gaussian surface must be zero.

What is the charge qenc that is enclosed by the cylinder (cross-sectional area A=2.00×10-2 m2) if the charge is only on one surface?

This is the same result as previously, because the surface charge density and enclosed area are unchanged.

In this second case, what is the total flux through the surface of the G-cylinder?

This too is the same result, because the amount of charge enclosed by the cylinder is unchanged.

If the charge on the conductor is negative, in which direction is the electric field?

Toward the conductor

Here is almost the same question. Which describes the field outside the sheet?

Uniform The field has the same magnitude anywhere to the right or left of the sheet.

Two concentric spherical shells are positioned with their common center at the origin of an xyz coordinate system. One shell has radius R=0.825 m and uniform surface charge density −σ where σ=0.375 μC/m2. The second shell has radius 2R and area charge density 2σ. (See Fig. 23.9.2 and the +z axis directed radially away from the common center.) Is the net electric field at point T equal to zero? Point T is located at z=R/2.

Yes

Let's bring an enormous charge Q up close to surface S4. (a) Will it change the pattern of the field lines? (b) Will it change the net flux through the four Gaussian surfaces?

Yes No. The value of Q would not enter Gauss' law in any way, because Q lies outside all four of the Gaussian surfaces.

On the external end cap, what is the angle θ between the field vector E→ and an area vector dA→?

Zero.

A spherical metal shell contains a central particle with charge +Q. The shell has a net charge of +3Q. (a) What is the total charge qint on the shell's interior surface? (b) What is the total charge qext on the shell's external surface?

a) -Q b) 4Q

The figure here shows a Gaussian cube of face area A immersed in a uniform electric field E that has the positive direction of the z axis. In terms of E and A, what is the flux through a) the front face (which is in the xy plane), b) the rear face, c) the top face, and d) the whole cube?

a) E * A= Ek * Ak = EA b) E*A= Ek*A(-k) = -EA c) Top face (parallel to xz plane) E*A = Ek*Aj = 0 (k*j = cos90 = 0) d) ∅cube = 0

Which expression gives the area vector for a patch on the left face?

dA-> = -dAi hat The area vector must be perpendicular to the patch and pointing away from the cube's interior.

A nonuniform electric field given by E→=3.0xi^+4.0j^ pierces the Gaussian cube shown in this next figure. We will first consider the right face, which consists of tiny patches, each with area dA. Which expression gives the area vector for a patch on the right face?

dA-> = dAi hat

Which expression gives the area vector for a patch on the top face?

dA-> = dAj hat Since dA now points in the positive direction of the y axis.

In terms of that last answer and dA, what is the flux dΦ through the patch?

dΦ=-3.0x⁢ dA This is the same result as when we used a dot product in unit-vector notation.

What is the dot product?

dθ = 3.0x dA This is the same result as when we used a dot product in unit-vector notation.

The next figure shows, in cross section, a plastic, spherical shell with uniform charge Q = −16e and radius R = 10 cm. In terms of e, what is the charge qenc enclosed by the new Gaussian sphere?

qenc = q + Q = 5e + (-16e) = -11e

How much charge is enclosed by Gaussian sphere S2?

qenq = q