PHYSICS II -DISCUSSION SECTION

I. Coulomb’s Law

An electrically charged object will experience a force due to a second charged object. If the objects are small, the force between them may be calculated from Coulomb’s Law: F = k q₁q₂/r².

Where k = coulomb’s constant = 9 x 10⁹ n-m²/C², q1 and q2 are the charges, and r is the distance between the centers of the two charges.

1. In the diagram shown below, determine:

a) The resultant electrostatic force (magnitude and direction) on charge q1 due to q2.

b) The resultant electrostatic force (magnitude and direction) on charge q1 due to q3.

c) Determine the resultant electrostatic force (magnitude and direction) on charge q1 due to charges q2 and q3. (Remember forces are vectors and must be added by vector addition methods.)

2. In the diagram shown below, determine:

a) The resultant electrostatic force (magnitude and direction) on charge q1 due to q2.

b) The resultant electrostatic force (magnitude and direction) on charge q1 due to q3.

c) The resultant electrostatic force (magnitude and direction) on charge q1 due to q4.

c) Determine the resultant electrostatic force (magnitude and direction) on charge q1 due to charges q2, q3, q4

Proton charge = 1.6 x 10^-19 C. Electron charge = -1.6 x 10^-19 C. Mass_p =1.67 x 10^-27 kg. Mass_e = 9.11 x 10^-31 kg

II. Electric Fields:

An electrically charged object will experience a force acting on it if it is in an electric field. (Just as an object with mass will experience a force acting on it if it is in a gravitational field: F_g = mg.) The electric field is designated by the capital letter E. It is measured in Newton/Coulomb. The size of the electrical charge is measured in Coulombs and can be either positive or negative and is designated by the letter q. The force acting on a charge in an electric field is the product of the electric field and the charge: F_E = qE.

1. An electron is in an electric field of 45,000 Newton/Coulomb directed in the positive x-direction. If the electron starts from rest and the field is constant, determine: a) the force on the electron. b) the acceleration of the electron.

2. Another common charged particle is a proton (the nucleus of a hydrogen atom). If a proton is in an electric field of the same strength as in problem 1, determine: a) the force on it. b) the acceleration it experiences.

3. A point charge generates its own electric field given by E = kQ/r², where k is coulomb's constant, Q is the charge in coulombs, and r is the radial distance from the charge to the point is space where we wish to calculate the field. In the diagram below:

a) Determine the Electric Field (magnitude and direction) at point A due to charge q1.

b) Determine the Electric Field (magnitude and direction) at point A due to charge q2.

c) Determine the resultant Electric Field (magnitude and direction) at point A due to charges q1 and q2.

Proton charge = 1.6 x 10^-19 C. Electron charge = -1.6 x 10^-19 C. Mass_p =1.67 x 10^-27 kg. Mass_e = 9.11 x 10^-31 kg