PHYS 2212 Module 2 Self Assessment Practice Problems

Module 2 Self Assessment Practice Problems

2.1
To solve this problem, first derive an expression for the electric field  as a function of radial distance r using Gauss’s Law: .
(a) How many excess electrons must be distributed uniformly within the volume of an isolated plastic sphere 25.0 cm in diameter to produce an electric field of 1150 N/C just outside the surface of the sphere?
(b) What is the electric field at a point 12.5 cm outside the surface of the sphere?
Answer: (a) 1.25 x 1010 electrons (b) 288 N/C  (Click and drag over the invisible text to highlight and view the answers) 
2.2
A very long, solid cylinder with radius R has positive charge uniformly distributed throughout it, with charge per unit volume .
(a) Starting from Gauss’s Law: , derive the expression for the electric field inside the volume at a distance r from the axis of the cylinder in terms of the charge density .
(b) Starting from Gauss’s Law: , derive the expression for the electric field at a point outside the volume in terms of the charge per unit length λ in the cylinder.
Answer: (a) (b)
2.3
The electric field at one face of a parallelepiped is uniform over the entire face and is directed out of the face. At the opposite face, the electric field is also uniform over the entire face and is directed into that face. The two faces in question are inclined at 30.0° from the horizontal, while and are both horizontal; has a magnitude of 2.90 x 104 N/C , and has a magnitude of 8.60 x 104 N/C.
(a) Assuming that no other electric field lines cross the surfaces of the parallelepiped, determine the net charge contained within.
(b) Is the electric field produced only by the charges within the parallelepiped, or is the field also due to charges outside the parallelepiped? How can you tell?
Answer:  -7.57 x 10-10 C
2.4
To solve this problem, first derive an expression for the electric field  as a function of radial distance r using Gauss’s Law: .

The electric field 0.335 m from a very long uniform line of charge is 840 N/C. How much charge is contained in a section of the line of length 1.80 cm?
Answer:  2.82 x 10-10 C
2.5
To solve this problem, first derive an expression for the electric field  as a function of radial distance r using Gauss’s Law: .

The nuclei of large atoms, such as uranium, with 92 protons, can be modeled as spherically symmetric spheres of charge. The radius of the uranium nucleus is approximately 7.4 x 10-15 m.
(a) What is the electric field this nucleus produces just outside its surface?
(b) What magnitude of electric field does it produce at the distance of the electrons, which is about 1.1 x 10-10 m?
(c) The electrons can be modeled as forming a uniform shell of negative charge. What net electric field do they produce at the location of the nucleus?
Answer:   (a) 2.4 x 1021 N/C  (b) 1.1 x 1013 N/C  (c) 0 N/C
2.6
Human nerve cells have a net negative charge and the material in the interior of the cell is a good conductor.
(a) If a cell has a net charge of -8.65 pC, what is the magnitude of the net flux through the cell boundary?
(b) What is the sign of the net flux through the cell boundary?
Answer:   (a) 0.977 N•m2/C  (b) negative
2.7
An infinitely long cylindrical conductor has radius R and uniform surface charge density σ.
(a) In terms of σ and R, what is the charge per unit length λ for the cylinder?
(b) To solve this problem, first derive an expression for the electric field  as a function of radial distance r using Gauss’s Law: .

In terms of σ, what is the magnitude of the electric field produced by the charged cylinder at a distance r > R from its axis?
(c) Express the result of part b in terms of λ.
Answer: (a) (b) (c)
2.8
To solve this problem, first derive an expression for the electric field  as a function of radial distance r using Gauss’s Law: .

A very long uniform line of charge has charge per unit length λ1 = 4.74 µC/m and lies along the x-axis. A second long uniform line of charge has charge per unit length λ2 = -2.20 µC/m and is parallel to the x-axis at y1 = 0.400 m.
(a) What is the magnitude of the net electric field at point y2 = 0.200 m on the y-axis?
(b) What is the direction of the net electric field at point y2 = 0.200 m on the y-axis?
(c) What is the magnitude of the net electric field at point y3 = 0.600 m on the y-axis?
(d) What is the direction of the net electric field at point y3 = 0.600 m on the y-axis?
Answer:  (a) 6.24 x 105 N/C  (b) upward  (c) 5.57 x 104 N/C  (d) downward
2.9
To solve this problem, first derive an expression for the electric field  as a function of radial distance r using Gauss’s Law: .

Some planetary scientists have suggested that the planet Mars has an electric field somewhat similar to that of Earth, producing a net electric flux of -3.69 x 1016 N•m2/C at the planet’s surface.
(a) Calculate the total electric charge on the planet.
(b) Calculate the magnitude of the electric field at the planet’s surface. (RMars =3.39 x 106 m)
(c) Find the direction of the electric field at the planet’s surface.
(d) Calculate the charge density σ on Mars, assuming all the charge is uniformly distributed over the planet’s surface.
Answer:   (a) -3.27 x 105 C  (b) 256 N/C  (c) downward, toward the planet  (d) -2.26 x 10-9 C/m2
2.10
A negative charge –Q is placed inside the cavity of a hollow metal solid. The outside of the solid is grounded by connecting a conducting wire between it and the earth.
(a) Is there any excess charge induced on the inner surface of the piece of metal? If so, what is it?
(b)  Is there any excess charge on the outside of the piece of metal? Why or why not?
(c) Is there an electric field in the cavity? Explain.
(d) Is there an electric field within the metal? Why or why not?
(e) Is there an electric field outside the piece of metal? Why or why not?
Answer:  (a) Yes  (b) No  (c) Yes  (d) No  (e) No