Field Definition And Coulomb’s Law

Coulomb’s Law

• Gives us the rule for dealing with two point charges.

(in practice for two charges whose separation is much greater than the radius of the charges.)

r

Charge Q1

Charge Q2

221

r

QQF

The field strength of an electric field is defined by

Q

FE

That is the force exerted by the field on unit charge

(ie a charge of 1 Coulomb)

placed at that point.

+1C

F

(Unit NC-1)

With a Radial Field

204

1

r

QE

Q+

Test Charge

The two variable quantities are the charge Q and the distance r

The electric field strength take sthe form of Coulombs law. Why?

The electric field strength formula is just coulombs law

applied to a test charge of 1C !!

Uniform electric fields

d

VE

In a uniform field the field strength at any point is given by

Remember this is just the force on a unit charge in the field

+ + +

_ _ _

Test charge

Remember this only applies where the field lines are parallel

Together with these relationships:

Q

WV

d

VE

204

1

r

QE

The definition of the volt

The electric field strength due to a point charge

The electric field strength in a uniform field

So the unit of E is Vm-1 as well as NC-1

Form the basis of any solution to the electric fields questions you will be asked

Electric Potential• The electrical potential of any point in the field is the

work done to bring a (+) charge of 1 coulomb from infinity (i.e. beyond the influence of the field) to that point in the field.

Q

So the electric potential at point P

P 1 coulomb positive charge

Implications:

1.The electrical potential of any point beyond the field is zero

2. The electric potential is

the potential energy change for 1C of charge

r

QV

04

1

The electric potentila is given by:

Calculations

A

AA r

QV

04

1

100mm

4.0μC -6.0μC

Two charges with the values shown are placed along are separated by a distance of 100mm. At what distance from A along the line AB does the electric potential reach 0V?

A B

When the potential along AB reaches zero

VA=VB

i.e. VA +VB = 0

B

BB r

QV

04

1

Now:

B

B

A

A

r

Q

r

Q

00 4

1

4

10

B

B

A

A

r

Q

r

Q

00 4

1

4

1

B

B

A

A

r

Q

r

Q

A

B

A

B

Q

Q

r

r

4

6

100.4

100.66

6

A

B

r

r

4.0μC -6.0μC

A B

100mm

60mm40mm

V=0

This ratio tells us that V=0 40mm from A

Where the numbers are not as straightforward you can continue as follows:

mmr

r

r

rr

rr

rr

rr

A

B

B

BB

BA

BA

BA

4060100

605

300

1003

5

1003

23

26

4

100

As the total distance between the charges is 100mm1

2

Now substituting 2 into 1

Calculate the magnitude of the electric field strength at the surface of a nucleus U (Z=92 M=238) . Assume that the radius of this nucleus is 7.4 × 10–15 m.............................................................................................................................................................................................................................................................................................................................................................................................................

Magnitude of electric field strength =.........................................

State the direction of this electric field.....................................................................................................................................

State one similarity and one difference between the electric field and the gravitational field produced by the nucleus.Similarity ....................................................................................................................

. ...................................................................................................................................

Difference .......................................................................................................................................................................................................................................................

The diagram shows a positively charged oil drop held at rest between two parallel conducting plates A and B.

2 .5 0 cmO il d ro p

A

B

The oil drop has a mass 9.79 x 10–15 kg. The potential difference between the plates is 5000 V and plate B is at a potential of 0 V. Is plate A positive or negative?………………………………………………………………………………………………

Calculate the electric field strength between the plates.Electric field strength =…………………………………

Calculate the magnitude of the charge Q on the oil drop.

Charge =……………………………………

How many electrons would have to be removed from a neutral oil drop for it to acquire this charge?………………………………………………………………………………………………

(3)