A charged particle with positive charge q1 is fixed at the point x=a, y=b. What are the x and y components of the force on a particle with positive charge q2 which is fixed at the point x=c, y=d?

Quiz

y

x

q1

q2

The negative charge of electron has exactly the same magnitude as the positive charge of the proton.

Neutral atom

Positive ionNegative ion

Charging of neutral objects1) By contact:

2) By induction (we’ll talk about it a little bit later)

a)

0;0 21 qq q q

b)

Qq 1 02 q Qqq 21 2Q

2Q

How you can make a balloon stick to the wall?

Principle of Superposition(revisited)

The presence of other charges does not change the force exerted by point charges. One can obtain the total force by adding or superimposing the forces exerted by each particle separately.

Suppose we have a number N of charges scattered in some region. We want to calculate the force that all of these charges exert on some test charge .0q

1q2q

3q

4q

5q

6q

7q

8q

0q

.ˆ

4

1...

ˆˆ

4

1

12

0

02

2

2202

1

110

00

N

i i

iiq r

rqq

r

rqq

r

rqqF

.ˆ

4

1...

ˆˆ

4

1

12

0

02

2

2202

1

110

00

N

i i

iiq r

rqq

r

rqq

r

rqqF

NWe introduce the charge density or charge per unit volume

q

How do we calculate the total force acting on the test charge ? q

We chop the blob up into little chunks of volume ; each chunk contains charge . Suppose there are N chunks, and we label each of them with some index .

VVq

i

Let be the unit vector pointing from th chunk to the test charge; let be the distance between chunk and test charge. The total force acting on the test charge is

ir̂ iir

N

i i

ii

r

rVqF

12

0

ˆ)(

4

1

This is approximation!

q

ir̂

The approximation becomes exact if we let the number of chunks go to infinity and the volume of each chunk go to zero – the sum then becomes an integral:

V r

rdVqF

20

ˆ

4

1

If the charge is smeared over a surface, then we integrate a surface charge density over the area of the surface A:

A r

rdAqF

20

ˆ

4

1

If the charge is smeared over a line, then we integrate a line charge density over the area of the length:

L r

rdlQF

20

ˆ

4

1

Problem 6 page 10

Suppose a charge were fixed at the origin and an amount of charge Q were uniformly distributed along the x-axis from x=a to x=a+L. What would be the force on the charge at the origin?

q

Another example on force due to a uniform line charge

A rod of length L has a total charge Q smeared uniformly over it. A test charge q is a distance a away from the rod’s midpoint. What is the force that the rod exerts on the test charge?

2

122

32 )()( cxc

x

cx

dx

2

122

32 )(

1

)( cxcx

xdx

The electric field

0qy

x

1q 2q

3q4q

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.

Have a great day!

Please don’t forget your pictures

Hw: All Chapter 1 problems and exercisesReading: Chapter 2