Dr. Jie Zou PHY 1361 1

Chapter 23

Electric fields (cont.)

Dr. Jie Zou PHY 1361 2

Outline Electric force

Coulomb’s law: the magnitude of the electric force

Direction of the electric force Examples!

Electric field Definition Electric field of a point charge Electric field of a group of charges Examples!

Dr. Jie Zou PHY 1361 3

Coulomb’s law Coulomb’s law: an equation giving the

magnitude of the electric force between two point charges.

An inverse-square law. SI unit for charge: Coulomb (C). ke= Coulomb constant = 8.99×109 N·m2/C2; or

ke= 1/40, where 0= the permittivity of free space = 8.85 × 10-12 C2/N·m2.

Elementary charge e = 1.60 × 10-19 C. Question: 1 C of charge is approximately equal

to the charge of how many electrons or protons?

221

r

qqkF ee

Dr. Jie Zou PHY 1361 4

Example 23.1 The hydrogen atom

The electron and proton of a hydrogen atom are separated (on average) by a distance of approximately 5.3 × 10-11 m. Find the magnitude of the electric force and the gravitational force between the two particles.

Fe = ke|q1||q2|/r2; Fg = Gm1m2/r2. ke= 8.99×109 N·m2/C2; G = 6.67×10-11 N·m2/kg2. Answer: Fe = 8.2×10-8 N; Fg = 3.6×10-47 N.

Dr. Jie Zou PHY 1361 5

Direction of the electric force

Electric force is a vector quantity. The electric force is directed along

the line joining the two point charges. The electric force is attractive if the

charges are of opposite sign and repulsive if the charges have the same sign.

Vector form of the coulomb’s law:

: a unit vector directed from q1 to q2. F12: the electric force exerted by a

charge q1 on a second charge q2. Newton’s 3rd law: F21 = - F12.

rF ˆ221

12 r

qqke

r

Dr. Jie Zou PHY 1361 6

The resultant force

Find the resultant force: when more than two charges are present, the resultant force on any one of them equals the vector sum of the forces exerted by the various individual charges. For example: if four charges are

present, F1 = F21 + F31 + F41.

Dr. Jie Zou PHY 1361 7

Example 23.2 Find the resultant force

Consider three point charges located at the corners of a right triangle , where q1 = q3 = 5.0 C, q2 = -2.0 C, and a = 0.10 m. Find the resultant force

exerted on q3 in unit-vector form.

Magnitude = ? Direction ? Answer: F3 = (-1.1 + 7.9 ) N.

i j

Dr. Jie Zou PHY 1361 8

Example 23.3 Where is the resultant force zero?

Three point charges lie along the x axis. The positive charge q1 = 15.0 C is at x = 2.00 m, the positive charge q2 = 6.00 C is at the origin, and the resultant force acting on q3 is zero. What is the x coordinate

of q3?

Dr. Jie Zou PHY 1361 9

The electric field Definition of electric field: the electric

field vector E at a point in space is defined as the electric force Fe acting on a positive test charge q0 placed at that point divided by the test charge, E Fe/q0

The electric field does not depend on the existence of the test charge-it is established solely by the source charge.

The test charge serves as a detector of the electric field.

Fe = qE: the force on a charged particle placed in an electric field.

If q is “+”, Fe is in the same direction as the field E.

If q is “-”, Fe is in the opposite direction to the field E.

Dr. Jie Zou PHY 1361 10

Electric field of a point charge

Electric field of a point charge q at point P:

r: distance of point P from the source charge q.

: unit vector directed from q toward point P.

rE ˆ2r

qke

r

Dr. Jie Zou PHY 1361 11

Electric field at a point P due to a group of point charges

Superposition principle: A any point P, the total electric field due to a group of source charges equals the vector sum of the electric fields of all the charges.

Example 23.5 Electric field due to two charges: A charge q1 = 7.0 C is located at the origin, and a second charge q2 = -5.0 C is located on the x axis, 0.30 m from the origin.

Find the electric field at the point P, which has coordinates (0, 0.40) m. Express in the unit-vector form.

Magnitude = ? and Direction of the field?

Dr. Jie Zou PHY 1361 12

Homework

Ch. 23, P. 734, Problems: #7, 8, 15, 16.