PHY 2049: Physics II

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quiz

The electric field at a distance of 10 cm from an isolated point particle with a charge of 2×10−9 C is:

A. 1.8N/C B. 180N/C C. 18N/C D. 1800N/C E. none of these

Quiz 2

An isolated charged point particle produces an electric field with magnitude E at a point 2m away from the charge. A point at which the field magnitude is E/4 is:

A. 1m away from the particle B. 0.5m away from the particle C. 2m away from the particle D. 4m away from the particle E. 8m away from the particle

PHY 2049: Physics II

Last week Coulomb’s law, Electric Field and

Gauss’ theoremToday Electric Potential Energy and

Electric Potentials Numerous cases

Potential Energy and Potential

Force => work => change in K=> change in Potential energy

ΔU = Uf – Ui = -W = - ΔK Work done is path

independent.

K+U = constant.

U = k q1 q2/r : interaction energy of two charges. Sign matters

PHY 2049: Physics II

Electric Potential V = U/q = -W/q Units of Joules/coulomb = volt 1 eV = e x 1V = 1.6 x 10-19 J

Also V = kq/r Vf –Vi = -∫E.ds In case of multiple charges, add as a

number

PHY 2049: Physics II

Can you tell the sign of the charge by looking at its behavior over a surface of potentials. Is the speed at the end bigger or smaller than in the beginning.

1. Neg., lower2. ?3. Pos., higher4. Neg., higher5. Pos., higher

PHY 2049: Physics II

Vf-Vi = -∫ k q/r2 dr Choose Vi = V

(∞)=0 V(r) = kq/r

V = kpcosθ/r2

E = -∂V/∂s = - Uniformly charged disk

V = ??

A

B

C

p

Consider the electric dipole shown in the figure

We will determine the electric potential created at point P

by the two charges of the dipole using superpos

V

Example : Potential due to an electric dipole

( ) ( )( ) ( )

( ) ( ) ( ) ( )

ition.

Point P is at a distance from the center O of the dipole.

Line OP makes an angle with the dipole axis

1

4 4

We assume that where is

o o

r

r rq q qV V V

r r r r

r d d

( ) ( )

2( ) ( ) 2 2

the charge separation

From triangle ABC we have: cos

cos 1 cosAlso:

4 4

where the electric dipole momento o

r r d

q d pr r r V

r r

p qd

(24 - 6)2

1 cos

4 o

pV

r

dqO A

Potential created by a line of charge of

length L and uniform linear charge density λ at point P.

Consider the charge element at point A, a

distance from O. From triangle OAP we

dq dx

x

Example :

2 2

2 2

2 20

2 2

2

2

0

2

2

2

2

have:

Here is the distance OP

The potential created by at P is:

1 1

4 4

4

ln4

l

l

n ln4

n

o o

L

o

L

o

o

r d x d

dV dq

dq dxdV

r d x

dxV

d x

V x d x

V L L x

dxx

d

x dd x

(24 - 8)

PHY 2049: Physics II

.Pr

dq

Consider the charge distribution shown in the figure

In order to determine the electric potential created

by the distribution at point P we use the pr

V

Potential due to a continuous charge distribution :

inciple of

superposition as follows:

We divide the distribution into elements of charge

For a volume charge distribution

For a surface charge distribution

For a linear charge distributi

dq

dq dV

dq dA

1.

on dq d

1 We determine the potential created by at P

4

1 We sum all the contributions in the form of the integral:

4

The integral is taken over the whole charge distrib

o

o

dqdV dq dV

r

dqV

r

2.

3.

Note 1 : ution

The integral involves only scalar quantities Note 2 :

1

4 o

dqV

r

(24 - 7)

( )F

( )F

2

2 2

Many molecules such as H O have a permanent electric

dipole moment. These are known as "polar" molecules.

Others, such as O , N , etc the electric dipole moment

is zero. These ar

Induced dipole moment

e known as "nonpolar" molecules

One such molecule is shown in fig.a. The electric dipole

moment is zero because the center of the positive

charge coincides with the center of the negative charge.

In f

p

ig.b we show what happens when an electric field

is applied on a nonpolar molecule. The electric forces

on the positive and nagative charges are equal in magnitude

but opposite in direction

E

As a result the centers of the positve and negative charges move in opposite

directions and do not coincide. Thus a non-zero electric dipole moment

appears. This is known as "induced" electric dipol

p

e moment and the molecule

is said to be "polarized". When the electric field is removed disappearsp

(24 - 9)

A collection of points that have the same

potential is known as an equipotential

surface. Four such surfaces are shown in

the figure. The work done by as it moves

a charge be

E

q

Equipotential surfaces

tween two points that have a

potential difference is given by:

V

W q V

2 1

2 1

For path I : 0 because 0

For path II: 0 because 0

For path III:

For path IV:

When a charge is moved on an equipotential surface 0

The work don

I

II

III

IV

W V

W V

W q V q V V

W q V q V V

V

Note :

e by the electric field is zero: 0W

W q V

(24 - 10)

S

AB

V

q

F

E

r

Consider the equipotential surface at

potential . A charge is moved

by an electric field from point A

to point B along a path

V q

E

The electric field E is perpendicular

to the equipotential surfaces

.

Points A and B and the path lie on S

r

Lets assume that the electric field forms an angle with the path .

The work done by the electric field is: cos cos

We also know that 0. Thus: cos 0

0, 0,

E r

W F r F r qE r

W qE r

q E r

0 Thus: cos 0

The correct picture is shown in the figure belo

9

w

0

(24 - 11)

S

V

E

Examples of equipotential surfaces and the corresponding electric field lines

Uniform electric field Isolated point charge Electric dipole

(24 - 12)

Assume that is constant constant4 4

Thus the equiptential surfaces are spheres with their center at the point charge

and radius 4

o o

o

q qV V r

r V

qr

Equipotential surfaces for a point charge q :

V

A

B

V V+dV

Now we will tackle the reverse problem i.e. determine if we know .

Consider two equipotential surfaces that corrspond to the values and

E V

E V

V V

Calculating the electric field from the potential

separated by a distance as shown in the figure. Consider an arbitrary direction

represented by the vector . We will allow the electric field

to move a charge from the equipotenbtial surfo

dV

ds

ds

q

ace to the surface V V dV

The work done by the electric field is given by:

(eqs.1)

also cos cos (eqs.2)

If we compare these two equations we have:

cos cos

From triangle PAB we see that

o

o

o o

W q dV

W Fds Eq ds

dVEq ds q dV E

ds

cos is the

component of along the direction s.

Thus:

s

s

E

E E

VE

s

s

VE

s

(24 - 13)

A

B

V V+dV

We have proved that: s

VE

s

s

VE

s

The component of in any direction is the negative of the rate

at which the electric potential changes with distance in this direction

E

If we take s ro be the - , -, and -axes we get:

If we know the function ( , , )

we can determine the components of

and thus the vector itself

x

y

z

x y z

V x y z

E

VE

xV

Ey

VE

z

E

ˆˆ ˆV V VE i j k

x y z

(24 - 14)

Potential energy of a system of point charges

We define as the work required to assemble the

system of charges one by one, bringing each charge

from infinity to its final position

Using the above de

U

U

finition we will prove that for

a system of three point charges is given by:U

2 3 1 31 2

12 23 134 4 4o o o

q q q qq qU

r r r x

y

O

q1

q2

q3

r12

r23

r13

(24 - 15)

, 1

each pair of charges is counted only once

For a system of point charges the potential energy is given by:

Her1

e is the separation between and 4

T

i

ij

ni j

i jo iji j

i j

n q U

r q qq q

Ur

Note :

he summation condition is imposed so that, as in the

case of three point charges, each pair of charges is counted only once

i j

1

1

2

2 2

1 1 22

12 12

3

3 3

1 2

13 23

1 3 23

13

Bring in

0

(no other charges around)

Bring in

(2)

(2)4 4

Bring in

(3)

1(3)

4

1

4

o o

o

o

q

W

q

W q V

q q qV W

r r

q

W q V

q qV

r r

q q qW

r

Step2 :

Step 1 :

e

St p3 :

3

23

1 2 3

2 3 1 31 2

12 23 134 4 4o o o

q q q qq

q

r

W W W W

qW

r r r

(24 - 16)

O

x

y

q1

Step 1

x

y

O

r12q1

q2

Step

2

Step 3

x

y

O

r12

r13

r23

q1

q2

q3

path

A

B

conductor

0E

We shall prove that all the points on a conductor

(either on the surface or inside) have the same

potential

Potential of an isolated conductor

Consider two points A and B on or inside an conductor. The potential difference

between these two points is give by the equation:

We already know that the electrostatic field

B A

B

B A

A

V V

V V E dS

JJJJJJJJJJJJJJ

inside a conductor is zero

Thus the integral above vanishes and for any two points

on or inside the conductor.B A

E

V V

A conductor is an equipotential surface

(24 - 17)

We already know that the surface of a conductor

is an equipotential surface. We also know that

the electric field lines are perpendicular to the

equipote

Isolated conductor in an external electric field

ntial surfaces.

From these two statements it follows that the electric field vector is

perpendicular to the conductor surface, as shown in the figure.

All the charges of the conductor reside on the surface and a

E

rrange

themselves in such as way so that the net electric field inside the

conductor

The electric field just out side the conductor is:

0.

in

outo

E

E

(24 - 18)

Electric field and potential

in and around a charged

conductor. A summary

All the charges reside on the conductor surface.

The electric field inside the conductor is zero 0

The electric field just outside the conductor is:

The electric field

in

outo

E

E

1.

2.

3.

4. just outside the conductor is perpendicular

to the conductor surface

All the points on the surface and inside the conductor have the same potential

The conductor is an eequipotential surf

5.

ace

0inE

ˆouto

E n

E

E

n̂

n̂(24 - 19)

R

1For ,

4

1For ,

4

o

o

qr R V

R

qr R V

r

2

2

1For ,

4

1For ,

4

o

o

qr R E

R

qr R E

r

R

q

Electric field and electric potential

for a spherical conductor of radius

and charge

Outside the spherical conductor the electric field

and the electric potential are identical to that of a point

charge equal to the net conductor charge and placed

at the center of the s h re

p e

Note :

(24 - 20)