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Week 3 - Gauss's Law

Due: 10:00pm on Monday, October 15, 2012

Note: You will receive no credit for late submissions. To learn more, read your instructor's Grading Policy

Gauss's Law

Learning Goal:

To understand the meaning of the variables in Gauss's law, and the conditions under which the law is applicable. Gauss's law is usually written

where is the permittivity of vacuum.

Part A

How should the integral in Gauss's law be evaluated? ANSWER:

In the integral for Gauss's law, the vector represents an infinitesimal surface element. The magnitude of is the area of the surface element. The direction of is normal to the surface element, pointing out of the enclosed volume.

Part B

In Gauss's law, to what does refer? ANSWER:

Physics 210 Q1 2012

( PHYSICS210BRIDGE )

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Week 3 - Gauss's Law

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around the perimeter of a closed loop over the surface bounded by a closed loop over a closed surface

the net charge inside the closed surface

the charge residing on insulators inside the closed surface all the charge in the physical system

any charge inside the closed surface that is arranged symmetrically

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The major use of Gauss's law is to determine an electric field when the charge distribution, both inside and outside the Gaussian surface, is symmetric. Of course, the electric field can always be found by adding up (or integrating) the contributions of all the charge in the problem. In highly symmetric situations, however, Gauss's law is much simpler computationally than dealing with all such contributions, and it provides better physical insight, too.

A Conducting Shell around a Conducting Rod

An infinitely long conducting cylindrical rod with a positive charge per unit length is surrounded by a conducting cylindrical shell (which is also infinitely long) with a charge per unit length of and radius , as shown in the figure.

Part A

What is , the radial component of the electric field between the rod and cylindrical shell as a function of the distance from the axis of the cylindrical rod?

Express your answer in terms of , , and , the permittivity of free space.

Hint 1. The implications of symmetry

Because the cylinder and rod are cylindrically symmetric, the magnitude of the electric field cannot vary as a function of angle around the rod, nor as a function of longitudinal position along the rod (typically represented by the spatial variables and ). By symmetry, the magnitude of the electric field can only depend on the distance from the axis of the rod (the spatial variable ).

Hint 2. Apply Gauss' law

Gauss's law states that , where is the electric flux through a Gaussian surface, and is the total charge enclosed by the surface. Construct a cylindrical Gaussian surface with radius and length coaxial with the rod, with .

Hint 3. Find the charge inside the Gaussian surface What is the total charge enclosed by the surface?

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ANSWER:

Hint 4. Find the flux

What is , the electric flux through the Gaussian surface?

Express your answer in terms of the magnitude of the electric field and given variables.

ANSWER:

Part B

What is , the surface charge density (charge per unit area) on the inner surface of the conducting shell?

Hint 1. Apply Gauss's law

The magnitude of the net force on charges within a conductor is always zero. This implies that the magnitude of the electric field within the conductor is zero. Think about a cylindrical Gaussian surface of length whose radius lies at the middle of the outer cylindrical shell. Since the electric field inside a conductor is zero and the Gaussian surface lies within the conductor, the electric flux across the Gaussian surface must be zero. What, then, must , the total charge inside this Gaussian surface, be?

ANSWER:

Hint 2. Find the charge contribution from the surface

What is , the total charge on the inner surface of the cylindrical shell that is contained within the Gaussian surface?

Express your answer in terms of and .

ANSWER:

To obtain the charge density per unit area, divide by the area of the inner surface of the conducting shell that is contained within the Gaussian surface.

ANSWER:

Part C

What is , the surface charge density on the outside of the conducting shell? (Recall from the problem statement that the conducting shell = = = = 0 = =

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has a total charge per unit length given by .)

Hint 1. What is the charge on the cylindrical shell?

What is , the total surface charge (the sum of charges on the inner and outer surfaces) of a portion of the shell of length ? ANSWER:

Since the charge on the inner surface of the cylinder is and the total charge on the cylinder is , it is now easy to obtain the charge on the outer surface of the cylinder. Then divide this result by the surface area of the portion of the cylinder that you took to obtain your result.

ANSWER:

Part D

What is the radial component of the electric field, , outside the shell?

Hint 1. How to approach the problem

Apply Gauss's law as you did to find the field between the rod and the shell. Again, choose the Gaussian surface to be a cylinder, with length and radius , coaxial with the rod. This time, you need to take .

Hint 2. Find the charge within the Gaussian surface

What is , the total charge contained within the Gaussian surface? ANSWER:

Now apply Gauss' law, , using for the enclosed charge.

Hint 3. Find the flux in terms of the electric field What is , the electric flux through the Gaussian surface?

Express your answer in terms of the magnitude of the electric field and given variables.

ANSWER: ANSWER: = = = =

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Charged Insulating Spheres

Two small insulating spheres with radius 5.50×10−2 are separated by a large center-to-center distance of 0.480 . One sphere is negatively charged, with net charge -2.50 , and the other sphere is positively charged, with net charge 3.95 . The charge is uniformly distributed within the volume of each sphere.

Part A

What is the magnitude of the electric field midway between the spheres?

Take the permittivity of free space to be = 8.85×10−12 .

Hint 1. How to approach the problem

Draw a diagram of the system to keep track of the directions of the fields. Calculate the electric field at the point midway between the charged spheres separately for each sphere, using Gauss's law, and use vector addition to determine the net electric field.

Hint 2. Using Gauss's law

You need to construct two separate Gaussian surfaces, each of which will surround one of the charged spheres. These surfaces also need to contain the point midway between the spheres so that the electric field at that point can be found. Therefore, for convenience, you should use a spherical Gaussian surface with one of the charged spheres in the center, since that way the electric field will be uniform and perpendicular to the surface at all points, so that the flux will be given by . Also, be careful about the radii of the spherical surfaces.

Hint 3. Calculate the field due to the negatively charged sphere

Calculate , the magnitude of the electric field at the midway point due to the sphere with charge -2.50 only.

Hint 1. Using the flux to calculate the field

Since the flux through the surface will be (due to Gauss's law), you can solve for the electric field using the equation for the surface area of the (Gaussian) sphere, , and the value of the enclosed charge.

ANSWER:

Hint 4. Determine the direction of the electric field from the first sphere

What will be the direction of the electric field due to the negatively charged sphere only? ANSWER:

=

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Note that the electric field will point toward a negatively charged sphere, just as it would point toward a negative point charge.

Hint 5. Calculate the field due to the positively charged sphere

Calculate , the magnitude of the electric field at the midway point due to the sphere of charge 3.95 only.

Hint 1. Using the flux to calculate the field

Since the flux through the surface will be (due to Gauss's law), you can solve for the electric field using the equation for the surface area of the (Gaussian) sphere, , and the value of the enclosed charge.

ANSWER:

Hint 6. Determine the direction of the electric field from the positively charged sphere What will be the direction of the electric field due to the positively charged sphere only? ANSWER:

Note that the electric field will point away from a positively charged sphere, just as it would point away from a positive point charge.

Hint 7. Vector addition

Keep in mind that you need to use vector addition in adding the electric fields from the two spheres. Also, keep in mind that the two fields point along the same line.

ANSWER:

Part B

toward the center of the negatively charged sphere away from the center of the negatively charged sphere

upward perpendicular to the line connecting the centers of the spheres downward perpendicular to the line connecting the centers of the spheres

= = 6.17×105

toward the positively charged sphere away from the positively charged sphere

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What is the direction of the electric field midway between the spheres? ANSWER:

Since the electric field will point toward a negative charge and away from a positive charge, the electric field from each sphere separately will point toward the negatively charged sphere, and so the total field will also point in that direction.

Exercise 22.21

A hollow, conducting sphere with an outer radius of 0.246 and an inner radius of 0.207 has a uniform surface charge density of +6.24×10−6 . A charge of -0.830 is now introduced into the cavity inside the sphere.

Part A

What is the new charge density on the outside of the sphere? ANSWER:

Part B

Calculate the strength of the electric field just outside the sphere. ANSWER:

Part C

What is the electric flux through a spherical surface just inside the inner surface of the sphere? ANSWER:

Exercise 22.22

A point charge of -2.00 is located in the center of a spherical cavity of radius 6.54 inside an insulating spherical charged solid. The charge density in the solid is 7.35×10−4 .

Part A

Calculate the magnitude of the electric field inside the solid at a distance of 9.49 from the center of the cavity.

Express your answer using two significant figures.

ANSWER:

toward the positively charged sphere toward the negatively charged sphere

= = 5.15×10−6

= = 5.82×105

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Part B

Find the direction of the electric field inside the solid at a distance of 9.49 from the center of the cavity. ANSWER:

Problem 22.34

A cube has sides of length = 0.300 . It is placed with one corner at the origin as shown in the figure . The electric field is not uniform but is

given by -4.93 2.49 .

Part A

Find the electric flux through each of the six cube faces , and .

Enter your answers in ascending order separated by commas.

ANSWER:

Part B

Find the total electric charge inside the cube. ANSWER:

Problem 22.46

A conducting spherical shell with inner radius and outer radius has a positive point charge located at its center. The total charge on the shell is , and it is insulated from its surroundings .

= = 2.3×105

The electric field points radially outward. The electric field points radially inward. The electric field is zero.

= , , , , , = 0, 6.72×10−2, 0, 0, -0.133, 0

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Part A

Derive the expression for the electric field magnitude in terms of the distance from the center for the region .

Express your answer in terms of some or all of the variables , , , and appropriate constants.

ANSWER:

Part B

ANSWER:

Part C

ANSWER:

Part D

What is the surface charge density on the inner surface of the conducting shell?

ANSWER: Part E = Also accepted: _{= 0} = Also accepted: =

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What is the surface charge density on the outer surface of the conducting shell?

ANSWER:

Problem 22.58

A nonuniform, but spherically symmetric, distribution of charge has a charge density given as follows: for for where is a positive constant.

Part A

Find the total charge contained in the charge distribution.

Express your answer in terms of the variables , , , and appropriate constants.

ANSWER:

Part B

Obtain an expression for the electric field in the region .

ANSWER:

Part C

Obtain an expression for the electric field in the region .

ANSWER:

Part D

Find the value of at which the electric field is maximum.

ANSWER: = = 0 _{= 0} = Also accepted: =

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Find the value of that maximum field.

ANSWER: =