<< Chapter < Page Chapter >> Page >

A positive point charge is placed at the angle bisector of two uncharged plane conductors that make an angle of 45 ° . See below. Draw the electric field lines.

An acute angle is shown. Its bisector is a dotted line. A positive charge q is shown on the dotted line.
Got questions? Get instant answers now!

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.

a. Outside: E 2 π r l = λ l ε 0 E = 3.0 C / m 2 π ε 0 r ; Inside E in = 0 ; b.
A shaded circle is shown with plus signs around its edge. Arrows from the circle radiate outwards.

Got questions? Get instant answers now!

An aluminum spherical ball of radius 4 cm is charged with 5 μ C of charge. A copper spherical shell of inner radius 6 cm and outer radius 8 cm surrounds it. A total charge of −8 μ C is put on the copper shell. (a) Find the electric field at all points in space, including points inside the aluminum and copper shell when copper shell and aluminum sphere are concentric. (b) Find the electric field at all points in space, including points inside the aluminum and copper shell when the centers of copper shell and aluminum sphere are 1 cm apart.

Got questions? Get instant answers now!

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.

a. E 2 π r l = λ l ε 0 E = λ 2 π ε 0 r r 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.

Got questions? Get instant answers now!

At the surface of any conductor in electrostatic equilibrium, E = σ / ε 0 . Show that this equation is consistent with the fact that E = k q / r 2 at the surface of a spherical conductor.

Got questions? Get instant answers now!

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

E = 5.65 × 10 4 N / C

Got questions? Get instant answers now!

Two parallel conducting plates, each of cross-sectional area 400 cm 2 , are 2.0 cm apart and uncharged. If 1.0 × 10 12 electrons are transferred from one plate to the other, what are (a) the charge density on each plate? (b) The electric field between the plates?

Got questions? Get instant answers now!

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 2 a .

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.

λ = λ l ε 0 E = a σ ε 0 r r a , E = 0 inside since q enclosed = 0

Got questions? Get instant answers now!

A point charge q = −5.0 × 10 −12 C is placed at the center of a spherical conducting shell of inner radius 3.5 cm and outer radius 4.0 cm. The electric field just above the surface of the conductor is directed radially outward and has magnitude 8.0 N/C. (a) What is the charge density on the inner surface of the shell? (b) What is the charge density on the outer surface of the shell? (c) What is the net charge on the conductor?

Got questions? Get instant answers now!

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 .

a. E = 0 ; b. E 2 π r L = Q ε 0 E = Q 2 π ε 0 r L ; c. E = 0 since r would be either inside the second shell or if outside then q enclosed equals 0.

Got questions? Get instant answers now!
Practice Key Terms 1

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'University physics volume 2' conversation and receive update notifications?

Ask