slides unit 1. electrostatics of point chargesjmas.webs.upv.es/ffi/unit 1/slides unit 1....

Introduction: electric charge.

Electrostatic forces: Coulomb’s law.

The electric field. Electric field lines.

Electric flux. Gauss’s law.

Work of the electric field.

Electrostatic potential energy

Electric potential on a point

Equipotential surfaces

Unit 1: Electrostatics of point charges

Two kinds:

positive and negative

* Tipler, chapter 21, section 21.1

Introduction. Electric charge

The positive charge is located in the protons and thenegative in the electrons. The quantity of any charge mustbe a multiple of these and means the electric charge isquantized. The lowest electric charge that can be isolated ise = 1.60×10-19 C. (electric charge of the proton and electron).

In an isolated object the net electric charge is constant (lawof conservation of charge).

Electric charge unit: Coulomb (C) [Q]=IT

Neutral atom: Equal number of protons (+) as electrons (-).


Only the electrons can be removed from the atom(+ charge remains) or be added to an atom (-charge remains).

Introduction. Conductors and insulators

229

0

/100,94

1CNmk ×==

επ

dq1q2

2212

0 /1085,8 NmC−×=ε

Coulomb’s law quantifies electric forces between pointcharges in vacuum.

Experimental law similar to Newton’s gravity law.

In escalar form:

q1 and q2 of the same sign.

F


Electric forces: Coulomb’s law

F

2

21

04

1

d

qqF

⋅=

πε

dq1q2

If the sign of charges is the same, then the force isattractive:

q1 and q2 with different sign.

F


Electric forces: Coulomb’s law

F2

21

04

1

d

qqF

⋅=

πε

Usually, these forces must be written in vector form, according the considered referency system.

i,1F

q1

qiq2 i,2

F

q3

==j

i,j2

i,j

j

0

i

j

i,ji ur

q

4

q ��

επFF

i,3F

In a system of charges, the net forceon a charge is the vector sum of theindividual forces exerted on it by allthe other charges in the system.

iF


Electric forces: Principle of superposition

The electric field is a useful concept to model the effect of an electric charge on the surrounding space.


The electric field

The electric field at a point in space is defined as theelectric force acting on the positive unit of chargeplaced at this point.

The force produced by a charge q1 at the point wherea charge of 1 C is located (electric field E) is:

+

+q1

+1 C

E=F(1 C)

The electric field

1

1 2C

qk ur

= =E F

� ��

ru�

So, if we put a charge inside an electric field, theeffect is a force acting over the charge (F=qE):

E

q(>0)

EF q=

q(<0)

EF q=

The electric field

E

E

E+

The electric field at apoint only depends on thecharge creating the field(q1) and the distance tothat point.

[E]=M L T-3 I-1 The unit is N/C or V/m

q1

The electric field

Electric field is a central force field.

-E

E

If the charge creating the field is negative, theelectric field points towards the charge.

The electric field

The electric field created by a system of pointcharges is the vector sum of the field created byeach of the charges:

==i

i2

i

i

0i

i ur

q

4

1 ��

επEE

1E

q1

q2

2E

q3

3E

E

The electric field

E

-+


The lines parallel to the field vector at each pointin the space are called “Electric field lines”.

E

E

Electric field lines

+ +

They are lines running from the positive charges (orinfinite) to the negative charges (or infinite).

Electric field lines

P

E

SE dd ⋅=φ

Sd


Let us take a point P with an electricfield E. If we take a little surface(infinitesimal) dS around P, we candefine the elemental electric fluxthrough dS as (escalar quantity)

If we consider a bigger (non infinitesimal) surface (S), then the flux is not infinitesimal:

Electric Flux

==SS

dd SEφφ

Nm2/C

q

r

E


Let’s take an spherical surfacewith a point charge q at itscentre.

The electric field is pointing outside the sphere(q>0) (inside if q<0)

Gauss’s law

The electric field at anypoint on the sphericalsurface (modulus) is

2E

r

qk=

q

r

E

SE dd ⋅=φ

Sd


If we consider an infinitesimal surface (dS) around the pointon the surface of the sphere, the electric flux through dS

will be

And the electric flux (Nm2/C) on the whole surface of thesphere:

Gauss’s law

0

22 εφφ

qdS

r

qkdS

r

qkdd

SSSS

===== SE

24 rπ

00

1

εεφ

Qqd

volumeEnclosed

i

surfaceClosed

==⋅= SE

=

volumeEnclosed

iqQ

This result can be applied to any surface (not onlyspheres) and is generally valid (Gauss’s law):

The net outward flux through any closed surfaceequals the net charge inside the surface divided byε0

Gauss’s law

>0 or <0 according φ >0 or <0

* Tipler, capítulo 23, sección 23.3

Gauss’s law can be applied to any closed surface,but the calculus is easier if the surface (S) satisfiestwo features:

a) The modulus of the electric field has the samevalue at all points on the surface (is constant).

b) The electric field vector has the same direction asthe surface vector at any point on the surface.

In this way:

α=α=α⋅⋅=⋅ coscoscos ESdSEdSEd

SSS

SE

Using Gauss’s law to calculate E

These two features can only be true if the problemshows symmetrical charge distribution.

As the Gauss (closed) surface must be “created” byus, we will usually have to think about:

Spherical surfaces

Plane surfaces

Cylindrical surfaces

Using Gauss’s law to compute E

Work done to move a second charge q a trip willbe: (charge q over a distance dl)

Let’s take an electric field created by a pointcharge Q.

Work of the electric field

q

� r

P

�

dl

�

E

dr

φ

� ur

Q

� �

F=qE

ld�

2

0

2

0 r4

qQdrdr

r4

QqdrFcosdlFldFdW

πεπεϕ ==⋅=⋅⋅=⋅=

��

If Q and q have the same sign (repulsive force) andrA<rB (A closer to Q than B) then

Work is done spontaneously by the forces in the electric field.

The work done by the electric force tocarry q along any line from A to B willbe:


A

� rB

B

� rA

L

� E

Q

q

� d l

0>LABW

BA

r

r

r

r

B

A

LAB

r

qQ

r

qQ

r

qQdr

r

qQldFW

B

A

B

A000

20 4444 πε

−πε

=πε

−=πε

== ��

If Q and q have the same sign (rejectingforce) but rA>rB (B closer to Q than A)then

work is done against the forces of the electricfield due to an external force


B

� rA

A

� rB

L

� E

Q

q � d l

0<LABW

If Q and q have opposite sign (attracting force)

If rA< rB (A closer to Q than B) then

If rA> rB (B closer to Q than A) then

0<LABW

0>LABW

As general rule:

If the work is positive, it means that the work is donespontaneously by the forces of electric field:

Work done by the forces of electric field

If the work is negative, it means that the work isdone against the forces of electric field:

Work done against the forces of electric fieldby an external force


B

� rA

A

� rB

L

� E

Q

q � d l

0<LABW

0>LABW

WLAB only depends on q, Q, rA and rB. So, if

we choose another line L’ going from A to B,the work done by the electric field will be thesame:

and and

Work of the electric field.Electric potential energy

0=AAW

AB

L

AB

L

AB WWW == '

BAAB WW −=

BAAB UUW −=

U is the electrostatic (electric) potential energy of a charge q in field due to Q

BA

LAB

r

qQ

r

qQW

00 44 πε−

πε=

Fields having this feature are called conservativefields or fields deriving from potential.

For these fields

Is the electrostatic potential energy of a charge q ata point at distance r from charge Q, which createsthe field. C tells us that an infinite number offunctions can be taken.

U=0 is usually taken at r=∞∞∞∞ , and then

U represents the work done by the electric field tomove q from this point to infinite.

Cr

qQU +

πε=

04

r

qQU

04πε=

U=WA∞

Work of the electric field. Electric potential energy

Electric or electrostatic potential at a point is theelectrostatic potential energy that would have acharge of 1 C placed at this point:

taking V=0 at infinite

Represents the work done by the electric field tocarry 1 C from such point to infinite.

And circulation of E along any line from A to B

Electric potential

r

Q

q

UV

04πε==

∞

=

P

P ldEV��

=−

B

A

BA ldEVV��

Unit: Volt (V=J/C)


Equipotential surfaces are those surfaces whosepoints have the same electric potential.

Equipotential surfaces due to a positiveand a negative electric charge

Their equation is V=k

k being a constant

Equipotential surfaces (on a plane) dueto a set of two charges are really 3dsurfaces.

10 V

2 V

5 V

15 V

-10 V

-2 V

-5 V

-15 V

+ —


As potential is constant across an equipotentialsurface, the work done to move any charge qbetween two points (A and B) is zero:

A

B0)( =−= BAAB VVqW

VA=VB

As

The electric field must be perpendicular toequipotential surface.

Equipotential surface

==−

B

A

BA ldEVV 0��


Electric field lines are perpendicular toequipotential surfaces at every point in the space.

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