The electric field

0qy

x

1q 2q

3q 4q

5q

000

limq

FE

q

E

F

has the same direction as

C

N

Coulomb

NewtonsE

Michael Faraday

1791-1867

“The best experimentalist in the history of science”

Electric field lines

These are fictitious lines we sketch which point in the direction of the electric field.

1) The direction of at any point is tangent to the line of force at that point.

2) The density of lines of force in any region is proportional to the magnitude of in that regionE

E

Lines never cross.

How to calculate ?E

0q1) Put a “test charge” at some point and do not allow it to move any other charges

2) Calculate the electric force on and divide by to obtain

0q 0qE

The force that N charges exert on a test charge : 0q

.ˆ

4

1

12

0

00

N

i i

iiq r

rqqF

We also calculated the force that a blob of charged material with charge density exerts on a test charge:

V r

rdVqF

20

0

ˆ

4

1

We wrote the similar formulas if the charge is smeared out over a surface with surface density , or over a line with line density .

0q

FE

In all of these cases, the force ends up proportional to the test charge . We might factor it out. This is the electric field!

0q

N

i i

ii

r

rq

12

0

ˆ

4

1

V r

rdV2

0

ˆ

4

1

0q

FE

(N point charges)

(Charge continuum)

Given an electric field, we can calculate force exerted on some point charge :q

EqF

Example 1: Electric field of a point charge is directly radially away from or toward the charge.

rr

qE ˆ

4

12

0

Example 2: Electric field of a dipole

Example 3: Electric field at the center of a charged ring

Think first!(before you start to do calculations)

0E

Example 4: Find the electric field at the center of a semi-circle of radius R, if a charge Q is uniformly spread over the semi-circle.

OxiR

QE

2

022

Have a great day!

Hw: All Chapter 2 problems and exercisesReading: Chapter 2