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 plastic sphere is isolated and has a diameter of 25.0 cm. Excess electrons are distributed uniformly throughout its volume.
(a) How many excess electrons are needed to produce an electric field of 1150 N/C just outside the sphere’s surface?
(b) What is the electric field at a point 12.5 cm beyond the sphere’s surface (measured from the surface, not the center)?
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 of radius R carries a uniform positive charge density (charge per unit volume) throughout its volume.
(a) Using Gauss’s Law: , derive an expression for the electric field at a point inside the cylinder, a distance r from its central axis (r < R). Express your answer in terms of .
(b) Using Gauss’s Law: , derive an expression for the electric field at a point outside the cylinder, a distance r from its central axis (r > R). Express your answer in terms of the charge per unit length, λ.
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: .

A very long, uniform line of charge creates an electric field in the surrounding space. At a perpendicular distance of 0.335 m from the line, this field has a magnitude of 840 N/C. Use this information to determine how much charge is on a 1.80-cm length of the line.
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 nucleus of a large atom, such as uranium, can be modeled as a spherically symmetric sphere of charge. Uranium’s nucleus contains 92 protons and has a radius of approximately 7.4 × 10⁻¹⁵ m.
(a) What is the electric field this nucleus produces just outside its surface?
(b) What magnitude of electric field does the nucleus produce at the distance of the surrounding electrons, about 1.1 × 10⁻¹⁰ m away?
(c) The electrons can be modeled as forming a uniform spherical shell of negative charge, centered on the nucleus. What net electric field does this shell produce at its own center?
Answer:   (a) 2.4 x 1021 N/C  (b) 1.1 x 1013 N/C  (c) 0 N/C
2.6
Human nerve cells carry a net negative charge, and the material inside each cell is a good conductor.
(a) For a nerve cell with a net charge of −8.65 pC, what is the magnitude of the net flux through its 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 of radius R has a uniform surface charge density σ. Imagine this conductor as a charged wire: viewed up close, it clearly has a cylindrical shape with a definite radius, but viewed from far away, it appears simply as a line of charge with no width at all.
(a) In terms of σ and R, what is the charge per unit length λ for the cylinder?
(b) Up close, where the cylinder’s radius R is apparent, what is the magnitude of the electric field at a distance r > R from the axis, in terms of σ?
(c) From far away, where the cylinder appears only as a line of charge, express this same electric field in terms of λ instead.
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 lies along the x-axis and has a charge per unit length λ₁ = 4.74 µC/m. A second, very long, uniform line of charge is parallel to the x-axis at y₁ = 0.400 m and has a charge per unit length λ₂ = −2.20 µC/m.
(a) What is the magnitude of the net electric field at the point y₂ = 0.200 m on the y-axis?
(b) What is the direction of the net electric field at that same point?
(c) What is the magnitude of the net electric field at the point y₃ = 0.600 m on the y-axis?
(d) What is the direction of the net electric field at that same point?
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 Mars has an electric field somewhat similar to Earth’s. The net electric flux through the planet’s surface is −3.69 × 10¹⁶ N·m²/C, and the radius of Mars is RMars = 3.39 × 10⁶ m.
(a) Calculate the total electric charge on the planet.
(b) Calculate the magnitude of the electric field at the planet’s surface.
(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 a cavity within a solid piece of metal. The outside of the metal is grounded by a conducting wire connected to the earth. For each part below, explain your reasoning, using Gauss’s law to support your answer where it applies.
(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 piece of 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