7/20/2006 Electric Fields (© F.Robilliard) 1flai/Theory/lectures/ElectricFields.pdf · 7/20/2006 Electric Fields (© F.Robilliard) 2 Charge: Atoms are held together as a system,

7/20/2006 Electric Fields (© F.Robilliard) 1

Q

R

r-axis

dA

E

r


Charge:

Atoms are held together as a system, by the fact that positive and negative charges attract each other, and so an attractive force exists holding the protons and the electrons together to form the atom.

Unlike charges attract, but like charges repel each other. This force between charges is known as the Coulomb force.

This unit is essentially a study of the Coulomb force.

- --+ ++

Atoms usually have equal numbers of electrons and protons, and consequently have zero net charge. However, it is possible to transfer loosely-held electrons from the atoms of one body, to another. The body with excess electrons will become negative, leaving an equal-magnitude, residual, positive charge due to excess protons, on the body, from which the electrons were removed.

All atoms are composed of protons, neutrons, and electrons. These fundamental sub-atomic particles have an intrinsic quantum property, which we call electric charge. There are only two types of charge: positive (+) and negative (-). The proton is positive (+), the electron, negative (-). The neutron has zero net charge. Protons and electrons, although they are different in other quantum properties, have charges of identical magnitude = e = (1.602 177 x 10-19) Coulomb..


Coulomb’s Experiment:

++

F

F

r

AB

C

Fibre

A pair of pith balls, B & C, were balanced at the ends of a thin rod, suspended by a thin fibre. Pith ball B was charged. Another pith ball, A, was also charged. The position of A was adjustable.

As A was brought closer to B, B experienced an increasing Coulomb force, F, which deflected B against the torsion forces of the fibre. The torsionalcharacteristics of the fibre were calibrated.

For various charges on A and B, and at various separations, r, between them, F was able to be measured, using the torsional calibration of the fibre.

After a series of experiments in 1785 Coulomb was able to formulate a law describing the Coulomb force, between charged bodies.


Coulomb’s Law:

rQ q

k F thus

rQ q

F

2

2

=

∝

20

0

rQ q

4 �1

F thus

4 �1

k

εεεε

εεεε

=

≡

+ +q Qr

FF

Between two point charges, q and Q, a force, F, acts on each charge, because of the other. This Coulomb force –

1. has the same magnitude on each charge2. acts in the straight line between the charges3. is repulsive for like, and attractive for unlike charges4 is proportional to the product of the charges, (q Q)5. is inversely proportional to the square of their separation, r.

k is often expressed in terms of another constant, εεεε0000, called thepermittivity of vacuum, where -

Thus –ε0 = 8.854 x 10-12 C N-1 m-2

2-29

2-29

C m N 109

C m N 108.988

alityproportion ofconstant k where

×≈×=

≡


Example:The hydrogen atom consists of one electron, distributed in an orbital region, about one proton. Find the Coulomb force holding the electron to the proton, given that the effective separation between them is a0 = 0.0529 nm, and the electronic charge = e = 1.6x10-19 C

( ) ( )( ) nN 82N 108.2

100.0529

101.6109.0

proton) &electron on charge (same ae

k rQ q

kF

829

219-9

20

2

2

=×=×

××=

==

−

−


How Does It Work?:

+ +q Qr

FF

There is a fundamental question that is implicit in Coulomb’s Law. How does one charged particle sense the other’s charge, and position, to get the correct Coulomb force.

This is the problem of “action at a distance”, and has only been properly answered by developments in Quantum Physics.

Furthermore, how fast does this information pass from one particle to the other? For example, if the position of one particle were changed, how long would it take for the force on the other particle to adjust?

We need to go down to the atomic level, and consider the Coulomb force between two charged particles, such as two electrons, or a proton and an electron.

Next, we need to introduce another sub-atomic particle, the photon. A photon is a lump of electromagnetic energy. A photon has both energy and momentum, and travels at the speed of light.

Charged particles interact by exchanging photons. A photon is transferred from one charged particle to the other. Since a photon has momentum, momentum is exchanged with the photon.


Exchange Interactions?

We currently identify only four forces in nature -

4. gravitational force due to “mass charge” (mass)

1. electromagnetic force due to electric charge

2. weak nuclear force due to “weak charge”

3. strong nuclear force due to “strong charge”.

Force is a rate of change of momentum, and hence the exchange of photons carrying momentum, between charged particles, results in a force acting on each particle. This interaction force is the Coulomb force.

The Coulomb force is due to an exchange interaction between charged particles, which is mediated by virtual photons. The speed of transmission of the force is the speed of the photons, namely the speed of light.

We need here, “virtual” photons. These are photons, whose energy and momentum are subject to quantum uncertainty in such a way that their values cannot be detected during the space and time of the interaction.

Similar types of exchange interaction are used to explain the other fundamental forces of nature, but use different mediating particles.

The Coulomb (or electric) force, plus the magnetic force, make up the electromagnetic force


Electric Field:+ +q Qr

FFThe classical way of dealing with “action at a distance”, for the Coulomb force, was tointroduce the idea of an electric field.

We say that charge q effects the space about it in such a way, that another charge, Q, inserted into that space will experience a Coulomb force.

We quantify this effect on the space, in terms of a vector that exists at every point in the space. Such a vector is called a field vector.

To define this vector operationally, we use a test charge, +δδδδq . Test charges are always, small and positive.

An electric field is generally due to a particular spatial distribution of source charges. If the test charge were large, it could disrupt the original source charge distribution, thereby altering the field that it is trying to define. Therefore, the test charge must be as small as possible. It is always taken as positive by convention.

To define the field vector at a point, we temporally shift a test charge to the point, and measure the Coulomb force, that acts upon it. We then withdraw the test charge.


Electric Vector, E - Definition:Let the electric field vector at a point in the field be E.

E is defined as the Coulomb force per unit test charge, placed at the point.To define E at a point P, we temporally place a test charge (+dq) at P.

��

��

→≡

�q�

0�qlim FEdF

+dq

E

P

where: δδδδF is the Coulomb force on test charge (+δq) at point PE is the electric field vector at P.

The limit in the definition, is a formal way to specify that the test charge is small.

SI Units: Newton/Coulomb = N/CAn equivalent unit, which is more often used, is Volt/metre = V/m


E Due to a Single Point Charge - Setup:

Applying Coulomb’s Law to our definition of field vector, E, we can determine an expression for the magnitude of E, at point, P, in the field.

Following from our definition of the electric field vector, E, we will find the field produced by a single point charge, q, at a point P, which is a radial distance, r, away from the charge.

+q P

rδδδδq

We temporarily introduce a test charge, +dq, to point P

δδδδF

E

Let r be the position vector of P, relative to q.

A field force dF will act on the test charge, due to the field of the point charge q.

Let E be the electric vector at P due to the point charge q.


E Due to a Single Point Charge:

+q P

rδδδδq

δδδδF

E

( )

2

2

r q

kE thus

Law sCoulomb' using r�q q

k �q1

�q�F

chargeunit test per force E

=

��

�

�=

≡

≡

+q P

rE

The direction of E will depend on whether q is (+) or (-)

-q P

r

E

There is a neat vector form to include direction.

-or becan q andr ofdirection in r unit vecto where

rq

kE 2

+≡

=

^

^

r

r


More-than-1 Point Charge:

The total field vector at a point is the vector sum of all the component field vectors at that point, due to all point charges present.

+q1

E1E

E2

q2 P

E = E1 + E2 (vector sum)

E2 = field component, at P, due to charge q2

This is called the principle of superposition.

When more than one point charge is present, each individual charge contributes to the overall field.

The best strategy for this type of situation is to express each field component as a vector (in ijk form), using Coulomb’s law, and then to add the component vectors.

E = total field vector at P

E1 = field component, at P, due to charge q1


Example:

( ) ( )

( )

( )

( )

( ) V/m 3j4i 102.25

3j4i10 49

10 3j4i2109

j q i q rk

irq

kjrq

k

3

3

62

9

2

22

+×=

+=

+×=

+=

+++=

+=

−

+−

−+

−+

E

��

kV/m 11.3 V/m 10 25.11

431025.2

EEE

3

223

22

=×=+×=

+= −+

A square, PABC, of side 2 m, has charges of +3 µC, and -4 µC, at corners A and C, respectively. Find the electric field vector at corner P.

P

A B

C

+

-

q+ = +3 µC

q- = -4 µC

r=2 m

r=2 mE+ = field component, at P, due to +3 µCE- = field component, at P, due to -4 µCE = (total field at P) = E+ + E-r = 2 m

E+

E-

E

θ

Magnitude:

Direction:

deg 36.9�

43

�tan

=

=


Flux Lines:Visualising an electric field in terms of millions of E vectors is not very picturesque.

A better, more intuitive, graphical representation is by flux lines.

A flux line is a line drawn in a field, with shape such that, the tangent to the line, at any point along it, is in the direction of the E-vector at that point.

Examples:

+q

isolated (+) charge

-q

isolated (-) charge

+q

A dipole [a pair of opposite charges]

-q

E

E E

E

Note: E vectors.


More Examples:

Parallel-plate Capacitor.

+ + + +

_ _ _ _

++q

_ _ _ _ _ _ _

+ + + + + + + Point charge near flat metal plate

By connecting an emf between the top and bottom plates, electrons from the top plate are transferred to the bottom plate, making the bottom plate negative. The top plate is left with an equal residual positive charge.

The positive point charge, +q, above the plate attracts electrons in the plate to the top of the plate, making the top surface negative. The bottom of the plate is left with the equal residual positive charge.

Because of mutual attraction, the positives will be on the bottom of the top plate, and the negatives, on the top of the bottom plate.


Properties of Flux Lines:Flux patterns have certain important properties. We can apply these properties to new charge situations, to predict the new flux pattern. We assume that the pattern is steady-state – any transients (short-term changes) have ended, and charges are in static equilibrium.

1. start on a positive charge, and end on a negative. If there is only one polarity present, the flux lines go to infinity.

2. never intersect each other. The E vector is unique at any point along a flux line.

3. are always perpendicular to a conducting surface. If they were not perpendicular, there would be a component of E along the surface, and surface charges would continue to move, until E was perpendicular.

Flux lines -

4. the closer the lines, the stronger the field. For a flat imaginary surface, placed in a uniform field, so that the flux lines intersect that surface perpendicularly, the number of lines that cut the surface, per unit area, is numerically equal to the magnitude of E.


Example:

E = 4 V/m1 m2

E = 2 V/m1 m2

(flux lines normal to surface)

How many flux lines come from a single charge of q?

+q

isolated (+) charge

r

Draw sphere, radius, r, with charge at its centre.

Let φ = number of flux lines cutting the surface of the sphere

Let A = area of surface of sphere = 4π r2

Since E is normal to surface, E =flux lines per unit area = φ/A

( )

0

22

0

�

q re therefo

r 4�rq

� 4�1

A E Thus

=

��

�

�==

φφφφ

φφφφ

q/ε0 flux lines come from a point charge of q.


Example:How many flux lines come from a point charge of 1 µµµµC?

lines thousand113

lines 101.13108.85

101�

q� 5

12

6

0

=

×=×

×== −

−

Flux lines are also called field lines, or lines of force.


Potential Difference - Setup:So far, we have described electric fields directly in terms of the Coulomb force, and its associated field vector E. We have used the notion of a vector field – a region of space, characterised at every point, by a vector E.

A second approach is to use a scalar at every point to characterise the field. This is a scalar field description.Because scalars are simpler than vectors, the second approach has some important advantages.Consider a positive test charge, δq in an electric field.

δδδδq

δδδδF

+x

A field force δδδδF will act on the test charge.To hold the charge in place, we will need to apply to it, a force δδδδF in the opposite direction.

-δδδδF

Say the applied force, δδδδF, then moves the charge, from A to B, a displacement, δx, in the (–x) direction.

δδδδx

Because a force moves the test charge through a displacement, work, δW has been done.

AB

δδδδx


Potential Difference:

( )( )( )

( )1................�x........E�q�W

thus

�q�F

E since�x �qE

work thedoes ,�F force, applied thesince�F.�x �W

⋅−=��

�

�

≡⋅−=

−−=

( )2.....................�q�W

�V

chargeunit per work difference Potential

≡

≡

δδδδq

δδδδF

+x

-δδδδF

δδδδqδδδδq

δδδδF

δδδδq

+x

δδδδqδδδδqδδδδx

AB

Definition:The work done per unit charge to move a test charge, from a point A, to a point B, is called the potential difference (PD), δδδδV, from A to B.

The LHS of (1) is defined as the potential difference between A and B.

SI Units:Joule/Coulomb =J/C= Volt = V


Potential Gradient:( ) ( )

( ) 3.......... dxdV

E limit, in the or, �x�V

E

so and�x E- �q�W

�V

:2 and 1 Combining

−=−=

=��

�

�≡

dV/dx is called the potential gradient. It measures the rate at which the field potential changes with displacement through the field. The stronger the field, the greater will be the potential gradient.

(3) shows that the field can be specified either in terms of the field force (E), or alternatively in terms of the work done against the field force (δV).(3) also shows that alternative units for E are Volt/metre = V/m.

Example:

E

δδδδx=1cm

δδδδV=100V

The plates of a parallel plate capacitor are separated by 1 cm. A potential difference of 100V exists between them. Find the E vector between the plates.

kV/m 10V/m 101.00.01100

�x�V

E

4 =×=

==


Zero of Potential:So far, we have defined differences in potential between two points. It would be better, if we could give each individual point its own individual potential. To achieve this, we need to give each point its potential difference from some common reference point, which is defined to have zero potential.

The point of zero potential is taken to be a point which is far from any fields –often called infinity.

Taking the earth to be our zero is not satisfactory, since the earth is often unavailable.

Potential at a Point, V:The potential at a point is defined to be the potential difference

between the zero of potential (infinity) and that point. Equivalently, the potential at a point is the work per unit charge, that must be done, to move a positive test charge, from infinity to the point.

This definition allows us to allocate to each point in an electric field, a scalar potential, thereby defining a scalar field.


Example:

V/m 1020.05100

�x�V

E 3×−=−=−=

- +

1 2 3 4 5 x(cm)

100Vearth

A B

E

A 100V cell is connected across the two plates, A & B, of a parallel-plate capacitor. The plates are separated by 5 cm. Plot the magnitude of both the electric vector, E, and the potential, V, as a function of distance, x, between the plates.

Field is uniform between the plates, therefore E = const. (Note: E is directed from B to A).

x

5 cm

|E|2 kV/m

( )�x 102�x E�V�x�V

E

3×=−=→

−=x

5 cm

V 100V

From A, where V=0 (since A is earthed), voltage increases in proportion to x, up to 100V at B

E:

V:


infinity++++δδδδq

x

Potential Due to a Single Point Charge - Setup:Here we will derive an expression for the potential, V, at a point, P, a distance, r, from a single point charge, q.

δδδδF

(Potential at P) = V = (work per unit charge that must be done, to bring a test charge, +δq, from infinity, to P).

Consider a test charge, on its journey from infinity to P, at distance x from q.

Say a Coulomb force, δδδδF, acts on the test charge at x. To hold the test charge, we need to apply an equal, but opposite force, -δδδδF, to the test charge.

−−−−δδδδF

Let’s move the test charge, a small displacement, δδδδx, closer to P.

δδδδx

The applied force, -δδδδF, does work δW on the test charge, where δW = δF .δx.(Although δF changes with x, δF is effectively constant over a small interval δx if δx is small)

+q Pr V


Potential Due to a Single Point Charge:

( )

( ) ( )

( )

rkq1

r1

kqx1

kqdxxkq

Law sCoulomb' from dxx�qq

k�q1-

directionx - in the isdx dx�x as limit, taking

dx�F�q1

workof steps theall of sum �W�q1

P oinfinity t from chargeunit per work totalV

rx

x

rx

x

2

P

infinity2

P

infinity

P

infinity

= ��

��

�

∞−= �

��

�+=⋅−=

⋅⋅=

��

�

� →−⋅=

=

=

=

∞=

=

∞=

−�

�

�

�

x

δδδδq

x

+q P

r V

rq

kV =

Summary:

If q is positive, then V is positive;if q is negative, then V is negative.

+q P

rδδδδF

infinityinfinityδδδδq

x

−−−−δδδδFδδδδx

V


Several Charges:

q+=2µC

q-= - 6µC

-+ 3m

4m

P

5m

Charges of 2µC and -6µC are located at corners of a rectangle, with sides 3m and 4m, as shown. Find the potential, V, at corner P.

( ) ( )

( ) kV -9.9V 109.9 1013.5-3.6 10 46

52

109

4q

5q

k VV

q todue Vq todue VV

3369 =×−=+= �

��

� −×=

�

��

� +=+=

+=

−

−+−+

−+

If several charges are present, each will contribute to the potential at a point, P.

Example:

If VA = potential at P due to charge at AVB = potential at P due to charge at B

Then total potential at P = scalar sum of individual component potentialsV = VA + VB

(Remember: +/- charges produce +/- potentials.)


x

y

R

+q

Continuous Distributions of Charge:We have looked at the case of finding the electric field vector, and scalar potential, due to a few isolated point charges. However, charges may be spread out over a line, surface, or volume, as a continuous charge distribution, in space. In this case, we divide the distribution up into small component lengths, areas or volumes of charge, which can be treated as point charges.

Example: Find the electric field vector, E, and the scalar potential, V, at the centre, O, of a semicircular, uniformly-charged wire, of radius, R, and total charge q.

dEdE sinθ

dE cosθ θ

ds = R dθdq

Consider a typical segment of the wire, at an angle θ to the x-axis, that subtends an angle dθ at the origin, and is of length ds= R dθ. The segment caries an elementary charge dq.

We break the wire up into many short segments.

Such a segment will produce an elementary field, dE, at the origin.which has x- and y-components.

Field Vector E


Find dE:

( ) ( )4.... d� �sin �cosRk�

-

�sin dE- cos� dE-

ji

j idE

+=

=

( )( )( )3.......

Rd��

k RRd��

k Rds�

k Rdq

kdE

:2 using CoulombBy

222 ====

x

y

R

+q

dE dE sinθ

dE cosθ θ

ds = R dθdq

Let µ be the linear charge density (charge per unit length) of the wire.

(charge on segment) = dq = µ ds = µ R dθ ...(2)

µ = (total charge)/(total wire length)q/(πR) ..............(1)

( )d� sin� cos� Rk�

- �

0� wirealong

jidEE +== ��=

Elementary field, dE, in terms of unit vectors i and j:

To get the total field, E, we sum the dE vector contributions [equation (4)] for all segments along the wire:

from (3)


Find E:

( )

[ ]

( ) ( )[ ]

( ) ( )[ ]

( )1 using R �

2kq-

R

2k�-

1- 1- 0 - 0 Rk�

-

0 cos - cos- 0sin - sin Rk�

-

cos� sin� Rk�

-

d� sin� cos� Rk�

-

2

�

0�

�

0� wirealong

j

j

ji

ji

j i

jidEE

=

=

−=

=

−=

+==

=

=��

ππππππππ

The integration is straightforward:

µ = (total charge)/(total wire length)q/(πR) ..............(1)

x

y

R

+q

dE

θ

ds = R dθdq

E

The total field at the origin is of magnitude

2R �2kq

E =

and is directed in the (-y) direction.


Find V:

( )[ ]( )5...... d�� k

Rd R�

k Rds�

k Rdq

k dV

:2 using and CoulombBy

==== θθθθ

[ ]Rqk

��Rq

k � �k �� k d�� k dV V �

0

��

0� wire thealong

=��

�

�===== ��=

=

Potential V

x

y

R

+q

θ

ds = R dθdq

A typical segment, dq, will produce an elementary potential, dV, at O, where:

Therefore, the total potential, V, at O, is given by the scalar sum of all the elementary potentials:

from (5) from (1)

dV

Total scalar potential, V, at O is R

qk V =

(which is the same as for a single point charge, q, placed at a distance R from the origin.)


Equipotential Surfaces:We have seen that vector electric fields can be represented graphically, by flux lines. Analogously, scalar potential fields can be represented by equipotential surfaces.

An equipotential surface is a surface drawn in an electric field, such that all points of the surface have the same scalar potential.

This means that charges can be moved about on an equipotentialsurface, without any work being done on them.

Since flux lines represent the direction of the field force, they must be perpendicular to equipotential surfaces. If they were not perpendicular, there would be a force component along that surface, and work would have to be done to move charges on it. The surface would therefore NOT be equipotential.

+r

q

Equipotential surface is a sphere centred on the point charge:V=kq/rV = const, if r = const.

Example: Isolated point charge.


Example:

Equipotential surfaces

Electric Dipole: (two separated, equal, but opposite point charges)

+q

A dipole [a pair of equal, opposite charges]

-q

It is generally easier to measure equipotential surfaces, than flux lines. We just measure voltages. Knowing the equipotential surfaces, the flux lines can then be drawn, using the perpendicularity property.


Electric Flux, φφφφ:We have introduced the idea of electric flux lines. We extend this now by considering more carefully, the number of flux lines that cut a given surface. This is called the “flux”, φ, through the surface. We will need this concept for the next section.

Case 1: E vector perpendicular to Surface:

For case 1, the surface is aligned so that the area vector, dA, is parallel to the Evector of the field.

Consider a small, flat element of surface, in an electric field.

The area of this tiny element of surface is represented by vector, dA, perpendicular to the surface.

Because the area element is so small, it can be regarded as flat, and the electric field near it, to be locally uniform

(Total flux cutting the surface) = dφ = E dA

This defines what we mean by flux, dφ, for this case. Intuitively, dφ can be visualisedas the number of flux lines that cut through the particular surface. (Example: dφ = 4 in the above figure). That is, when we draw the flux, we would draw it with 4 flux lines cutting the surface dA.

SI Units for φφφφ: (V/m) m2 = V m or alternatively (N/C) m2 = N C-1 m2

Definition:

This is the simplest case.

dA

locallyuniformfield, E

E

dA


Case 2: E vector NOT perpendicular to Surface :

For this Case the surface is oriented, so that its dA vector is aligned at an angle of θ, to the E vector. View this looking at the area element edge-on.

The E vector can be resolved into a component E cos θθθθ in the direction of the dAvector, plus a component E sin θθθθ in the perpendicular direction, which is along the surface itself.

The (E cos θ) component is Case 1, and contributes [(E cos θ) dA] flux through the surface dA.

The (E sin θ) component passes along the surface, never cuts it, and therefore contributes zero flux, through the surface.

(Thus the total flux through the surface dA) = dφ = E dA cos θ = E.dAdotproduct

E sin θθθθ

E cos θθθθ

dA

surface dA

flux lineEθ

dA

dA

locally uniformfield, E

θE


Resolving the Area:Rather than consider the two components of E, we could have equivalently considered the two components of the area vector dA.

dA

Eθ

dA

θ EdA cosθθθθ

dA sinθθθθdA

Eθ

dA cosθθθθdA sinθθθθ

The vector dA, like any vector, can be resolved into two components, dA cosθθθθ, and dA sinθθθθ.The physical significance of these two components of area is illustrated.

All the flux cuts the dA cosθθθθ component of area; none cuts through the dA sinθθθθ component.

Again, total flux through the (dA cosθ) surface = dφ = E dA cos θ = E.dAdotproduct


Case 3. Extended, Non-flat Surface:We will need to consider the flux through extended surfaces, which are not flat.

SE

dA

We subdivide the curved surface, S, into a large number of tiny elements of vector area. A typical one of these, dA, is shown.Because these elements of area are small, they can be treated as flat, and the field near them, locally uniform, even though the field, on the larger scale, may be non-uniform.

The flux, dφ, through a typical area element, dA, will be given by Case 2, namelydφ = E.dA (dot product)

where the directions of E and dA are local to the particular area element, and will change from element to element.

The total flux, φ, through the entire curved surface, S, will be the sum of the fluxes, dφ, through all the individual elements of area, dA, that make up the surface.

( ) �� •===Asurfaceover

dAEds' d theof surface over the sum φφφφφφφφφφφφ dotproduct

Integrals (or sums) over a surface, of some quantity, that has a value at each point on the surface, are called “surface integrals”


Case 4:Closed Surfaces:These are surfaces that completely surround an inner region. Such surfaces divide space into two regions, an inside, and an outside. There are no holes in a closed surface. To get from the inside to the outside, you must pass through the surface.

We will need closed surfaces, in particular, to quantify the total flux originating from a given charge. We surround the charge by a closed surface. The total flux cutting this surrounding surface will be the total flux originating from the enclosed charge.

When a flux line cuts a closed surface, it either comes from the inside to the outside, or from the outside to the inside. There is a sign convention to distinguish these two cases.

outside

inside outsideoutside

closed surface

outside


Flux Through Closed Surfaces – Sign Convention:Flux lines coming from the inside to the outside are positive.

φ = +6 (V m) φ = +3 (V m)

Flux lines coming from the outside to the inside are negative.

φ = -6 (V m) φ = -3 (V m)


Flux Lines That Pass Through:

-1 +1

A flux line that comes from the outside, to the inside, and back to the outside( that is, passes right through a closed surface) contributes zero net flux to the surface.

φ = -1 +1 = 0 (V m) φ = -5 +5 = 0 (V m)

It is only flux lines that either originate, or terminate, inside a closed surface, that add net flux to the surface.

This sign convention is consistent with the sign of the dot product in φ = E.dA, ifthe direction of vector dA, for a closed surface, is always taken from inside to outside. Assume that E is parallel to dA.

dAE

E.dA = E dA cos 0= E dA (+1)= + E dA

(positive flux)

dAE

E.dA = E dA cos π= E dA (-1)= - E dA

(negative flux)


Net Flux For Closed Surface :

( )( )

( )surface closed over then integratioan represents symbol thewhere

dAEdsurface closed over the s' d of sum

surface closedgh flux throunet

surface closedover

�

�� •===

=

φφφφφφφφ

φφφφ

On the understanding that the dot product (E.dA) encorporates the sign convention for flux, the total flux cutting a closed surface follows the argument of Case 3 .

Note: dφ is a scalar. Therefore all the dφ ‘s will sum over a surface, by simple scalar addition.

Remember: Integrals, or continuous sums, over surfaces are called “surface integrals”


Flux and Charge – Gauss’ Law:We discussed earlier, that the total number of flux lines, φ, coming from a point charge is proportional to the charge, q. We proved that -

( ) vacuumofty permittivi theis

system. SI in theality proportion ofconstant theis 1

where

0

00

εεεεεεεεεεεε

φφφφ

φφφφqq

=

∝

This was a consequence of Coulomb’s Law, and the definition of flux.

Combining this idea, with our more general definition of flux, φ, as developed immediately above, gives us a relationship between charge and electric field.

This is a powerful alternative formulation of Coulomb’s Law, and is named after its discoverer – Gauss.

Gauss’ Law has important advantages over Coulomb, where distributions of charge are concerned, particularly where there is a level of spatial symmetry in that charge distribution.


Gauss’ Law-Setup:

++q

r

We consider a point charge +q.

What will be the total flux, φ, coming from this charge?

We surround the point charge, +q, with a closed sphere of radius, r, centred at the point charge. Any flux coming from the charge, will have to cut this sphere.

Consider a typical element of area, dA, of the surface of the sphere. dA is a vector pointing radially outward.

Let the electric field vector, due to the charge, +q, at the area element, dA, be E.

From our foregoing definition of flux, the total flux, dφ, through this area element will be (E.dA) [dot product]

The total flux, φ through the entire spherical surface is the scalar sum of all the fluxes, dφ, through all the elementary areas, that compose the entire surface.

dAE

We will now develop a general relationship between a charge, and its electric field, via the notion of flux.


Gauss’ Law:( )

( )

( ) ( )

( ) ( )

0

222

0

2

�

q

r 4 sphere of area surface since r 4rq

� 4

1

Law sCoulomb' using dA rq

k

symmetryby surface,over const. E since dA E

dA E0 cosdA Esymmetryby surface, spherical on thepoint any at

toparallel is since dA E

productdot d

=

=��

�

�=

��

�

�=

==

��

�

�

�

==•→=

•==

�

�

�

��

ππππππππππππ

φφφφφφφφ

��

��

dAE++q

r

dAE

� =•=Φ0εεεε

qdAEThus, for this case of a point charge, q - This is Gauss’ Law

The total flux, φthrough the surface is the sum of all the fluxes, dφ, that pass through all the dA’sof that surface.


Generalisation:

Gauss’ Law gives a connection between charge, q, and electric field vector, E. The relationship is implicit. To find the field vector, E, we need to integrate the equation.

vacuum.ofty permittivi theis � surfaceGaussian he within tenclosed charge total theis �q

element areaat surface,Gaussian on the vector field theis where

�

�q

0

0

dAE

dAE� =•=φφφφ

Although we derived this result for a point charge, the result is valid for any distribution of charges (but we will not prove this). Although we used a spherical Gaussian surface, it is true for any-shaped Gaussian surface.

In deriving this result we used a spherical surface, to quantify the flux. Such a surface is called a “Gaussian Surface” (GS).

In order to facilitate the integration, we need to choose the shape of the Gaussian surface carefully. If E is parallel to, or perpendicular to dA, or if E = 0, the integral is easier to evaluate.

We will find that Gauss’ Law works best where there is a high degree of symmetry in the charge distribution, and hence in the electric field produced.

The general statement,for charges in vacuum, is -


Notes onGauss:

vacuum.ofty permittivi theis � surfaceGaussian he within tenclosed charge total theis �q

element areaat surface,Gaussian on the vector field theis where

�

�q

0

0

dAE

� =•= ��φφφφ

for any distribution of charges, with total charge ΣΣΣΣq

What Gauss is essentially saying is -

if we draw a closed surface (a Gaussian Surface, GS) of any shape, that totally encloses those chargesthen the total number of flux lines, φφφφ, from the charges, that cut the GS, equals (Σq)/ε0

dA

E

GS

Furthermore, we can compute the total number of flux lines, φφφφ, by decomposing the GS into tiny tiles, a typical tile having vector surface area, dA.

If the electric field vector is E, at a typical tile of area dA, then the flux (number of flux lines), dφ, cutting that tile is E.dA [dot product].Note: E and dA could be inclined at any angle to each other.

The total flux, φφφφ, cutting the whole GS will be the scalar sum of the individual fluxes, dφφφφ, through all the tiles that compose the GS.

�� •== ��

GS

φφφφd

ΣΣΣΣq

+ + ++ +


Applications of Gauss’ Law:1. Field Produced by a Uniformly-charged Insulating Sphere:

+ + ++ + + +

+ + +

Q

R r-axis We will use Gauss to find expressions for the electric field vector, E, both inside the sphere (r<R), and outside the sphere (r>R)

Consider an insulating sphere of radius R, with a total charge of Q, distributed uniformly throughout the volume of the sphere.

Outside the Sphere (r >R):

+ + + + + + + + +

+ + + + + + + + + + + + + + + + + + + + +

+ + + + + + + +

Q

Rr-axis

dA

E

r

Find the E-vector at a radial distance of r from the centre of the charged sphere.Because the charge is uniform and spherical, the flux direction will radiate out symmetrically from the centre of the charged sphere. The following arrows only represent the direction of the field, they are not flux lines, and do not represent field strength.

r Draw a spherical GS, of radius, r, concentric with the charged sphere.

By symmetry, at any typical vector area element, dA, of the GS, the electric field, E, will be parallel to vector dA.


Outside the Sphere:

( )

( )

220

0

2

0

2

0

0

0

rQ

k rQ

� �4

1E

�

Qr �4 E

�

�qr �4 E

�

�qdA E

�

�qdA E

�

�qdAE

==→

=→

=→

=→

=→

=•

�

�

�

+ + + + + + + + +

+ + + + + + + + + + + + + + + + + + + + +

+ + + + + + + +

Q

Rr-axis

dA

E

rInvoking Gauss’ Law, we get the relationship between E at the Gaussian surface, and the total charge, ΣΣΣΣq, contained within it.

dA. E cos0dA EdAE ThusdA vector. the toparallellocally always is vector E the

surface, theof dA, elements, allat symmetry, of Because

==•

Reasoning:Steps:

( ) integral. surface theduring const E

GS. on the points allat magnitude, same thehas vector E thecharge, theofsymmetry spherical theof Because

=→

( ) 2r 4 GS the sphere a of area surface Total ππππ=

Q GS e within thcontained charge Total =

Gauss

041

k whereπεπεπεπε

≡


Notes:

Outside the charged sphere, the field is the same as would be produced by a point charge of the same total charge, Q, located at its centre – namely, inverse square. However, as we will see, this is not true inside the sphere..

It is important to choose a GS with a shape that corresponds to the symmetryof the electric field. If we do this, the integration of Gauss is easy; if we choose an inappropriate shape, the integration will be difficult. In the previous derivation, a concentric, spherical surface was chosen because then, E is parallel to dA, at all points on that surface.

2r1

E

:square-inverse is field thesphere, theOutside

∝

We will next find the field inside the sphere.

The two spheres in this derivation should not be confused. The charged sphere is a real, physical body. It is made out of atoms. The Gaussian surface is an imaginary sphere. It is a geometrical construct, upon which the Gauss formulation, of the relation between charge and electric field, is expressed.


Inside the Sphere (r <R):+ + +

+ + + + + ++ + + + + + +

+ + + + + + + + + + + + + +

+ + + + + + + +

Q

Rr-axis

dA

E

r

The symmetries inside the sphere are analogous to those outside. Hence, again, we chose as GS, a sphere of radius r, concentric with the charged sphere.

As before, by symmetry, the E vector at an element of the GS, and the dA vector of that element, will be parallel.

Again, by symmetry, the magnitude, E, of the electric field vector, at all points on the GS, will be constant.

It follows, that the arguments used on the left hand side of Gauss’ Law are identical to the previous, outside-the-charged-sphere case. However, the arguments on the right hand side of Gauss are different. If the GS is outside the charged sphere, all the charge, Q, is enclosed. Here, only part of that charge is enclosed within the GS.

( )1............. Q

sphere theof volumetotalsphere on the charge total

sphere of eunit volumper chargesphere charged theofdensity charge �Let

334� R

=≡

≡≡


Derivation:

( )

( ) ( )

( )2.....r 3

E

�

r � 4 E

�

�qr 4 E

�

�qdA E

�

�qdA E

�

�qdAE

0

0

334

2

0

2

0

0

0

�

r �

�

ρρρρ

ππππ

=→

=→

=→

=→

=→

=•

�

�

�

dA. E cos0dA EdAE hussymmetry.Tby GS, over the dA vector, the toparallellocally is vector E The

==•

Reasoning:Steps:

symmetry.by GS, on the points allat magnitude,constant has vector E The

( ) 2r 4 GS the sphere a of area surface Total ππππ=

( )334 r

GS of volume

sphere charged of

eunit volumper harge

GS e within thcontained charge Total

ππππρρρρ=

��

�

��

�

�=

c

+ + + + + + + + +

+ + + + + + + + + + + + + + + + + + + + +

+ + + + + + + +

Q

Rr-axis

dA

E

rInvoking Gauss, as before:

Gauss

rE sphere theInside ∝


In Terms of Q:

( )

( )

( ) ( ) r

r

�

RkQ

E :31

3...... 3

4k

r 3

E :2

3

0

=→

��

�

�=

=

ρρρρππππ

ρρρρ

[ ]

[ ]r tosquare-inverse r1

rkQ

E :sphere theOutside

r toalproportionr r RkQ

E :sphere theInside

22

3

∝=

∝=

This result can be expressed in terms of the total charge Q.

Summary:The field is radially symmetric about the charged sphere.

( )1....... R �

Q � 3

34

=


Graphically:

R

+Q

2rkQ

E =r RkQ

E 3=

2RkQ

E =

r-axis

E

-R +R

Max field occurs at the surface of the sphere


Field Inside a Conducting Sphere:The sphere considered previously, was an insulator. The charges could not move, and thus remained uniformly distributed through the volume of the sphere.However, if the sphere is a conductor, the charges will move under their mutual repulsion, to the surface of the sphere, where they will spread out uniformly. There will be no charges in the interior of the sphere.

+ + + +

+ + + + + +

+ + + +

Q

Rr-axis

dA

E

r

P

( )0 E 0r 4 E

0 dA E

�

�qdA E

�

�qdAE

2

0

0

=→=→

=→

=→

=•

�

�

�

�

Invoking Gauss, and using the same reasoning on the left hand side as for the insulating sphere -

The symmetry of the field is analogous to the insulating sphere case. Similarly, a suitable GS is a sphere, radius r, concentric with the charged sphere.

Find the E-field at a point P, inside the sphere, which is a radial distance, r, from the centre of the sphere.

zero charge inside the GS- it’s all on the surface!

E-field inside the sphere is zero


Comments:The net field, inside a statically charged conducting sphere, is zero.

If the sphere were hollow, the same arguments would hold. Thus, there is no net field inside a hollow, statically charged, conducting sphere.

This result has some application for the shielding of electronic circuitry from interference by extraneous electrical noise. The components are placed inside an earthed metal case to reduce the intrusion of noise fields.

Another application is in coaxial cable, where the signal-carrying conductor is protected from external electrical noise by being placed inside a woven, earthed, metal, sheath.

Note: the field outside the conducting sphere (both solid, and hollow) would follow the same inverse-square relationship as for the previous insulating sphere case.


r

P

Field Due to a Long Straight Line of Charge:

+ + + + + + + + + + + + + + +

Let λ = charge per unit length along the line of charge in C/m = linear charge density

Find the electric field vector, E, at point P, a radial distance, r, from the line of charge.

The charge is uniformly distributed along the line.

By symmetry, the field will radiate out perpendicularly from the line of charge.

+λλλλ

This can be seen from an end view.

Chose as GS, a cylinder, radius r, coaxial with the line of charge.

end view

P

r+


dA and E:

+

end view

P

r A surface element, dA, of the curved part of the GS is shown, at left.

dA

Because of the common symmetry between the field, and the cylindrical part of the GS, dA is parallel to E, at all points on the curved surface.

The GS consists of a curved cylindrical surface, plus two flat circular ends.

E

r

P

+ + + + + + + + + + + + + + + + +λλλλ

An element, dA, of one of the flat ends of the GS is shown below.

dA

E

By symmetry, dA, will be perpendicular to E, at all points of the flat end of the GS.

Let the length of the GS = LL


Use Gauss to Find E:

( )

r � 2E

�

L Lr 2E

�

�qdAE

�

�q dA E 0 0

�

�qdAEdA EdA E

�

�qdAE :Gauss

0

0

0curved

0curved

0curvedendright endleft

0

ππππλλλλ

ππππ

=→

=→

=→

=++→

=•+•+•→

=•

�

�

��

�

Curved surface area = 2 π r L

Total charge enclosed within GS = λ L

E = constant over cylindrical surface, by symmetry

Total flux through the GS is the flux through the ends, plus the flux through the curved cylindrical surface.

At ends, E perpendicular to dA, hence E.dA = 0

On curved, E parallel to dA, hence E.dA = E.dA

E is inversely proportional to the radial distance, r, from the line of charge.


Field Due to a Flat Charged Metal plate:

+ + + + + + + + ++ + + + + + + + + +

+ + + + + + + + + ++ + + + + + + + + +

+ + + + + + + + + ++ + + + + + + + +

Let the charge per unit surface area on the top surface =the charge per unit surface area on the bottom surface = Q/A = σ

Consider a flat conducting plate with top surface area A. Bottom surface area is also A. A total charge of +2Q is placed on the plate. Since the plate is conducting, this charge will distribute uniformly over the surface of the plate. Because of mutual repulsion, each charge will try to maximise its distance from all other charges, thereby minimising the total PE of the charges. There will be no charges in the interior of the plate.

If the total area of the top and bottom surfaces is much greater than the total area of the sides, we can neglect the small fraction of the charge on the sides, and assume that half of the total charge = Q, is on the top, an the other half = Q, on the bottom surface.

Q

Q

A

A


Find E Above the Top Surface of the Plate:

+ + + + + + + + ++ + + + + + + + + +

+ + + + + + + + + ++ + + + + + + + + +

+ + + + + + + + + ++ + + + + + + + +

σσσσa

dA

E E

dA

The charge on the top surface is uniformly distributed, with surface charge density = σConsequently, the electric field is uniform & perpendicular to the surface.We choose a cylindrical GS, with its top surface area = a, axially aligned with the field. Let’s look more closely at the GS. The GS is embedded in the plate. Because there is zero net field inside a statically charged conductor, E at the bottom end of the GS = 0

E=0

By symmetry, an element of area, dA , on the top of the GS is parallel to the local field vector, E.

Similarly, dA on the curved surface, is perpendicular to E, locally.


Use Gauss to Find E:

0

0

0top

0top

0

0

�E

�

a a E

�

qdA E

�

q 0 dA E 0

:Gauss

σσσσ

σσσσ

εεεε

εεεε

=

=

�=

�=++

�=•+•+•

�=•

�

�

��

�

qdAEdAEdAE

qdAE

curvedtopbottom

Total flux = sum of fluxes through bottom, top, and curved sides of the GS.

E=0 over bottom of GS.

By symmetry, E parallel to dA over top of GS.

By symmetry, E perpendicular to dA over curved side of GS.

By symmetry, E=const. over top of GS.

Top surface area = a.Therefore charge enclosed within GS= σ a

The field, E, is perpendicular to, and uniform, over the top, and over the bottom, of the plate, and is proportional to the surface charge density, σσσσ.

Note: we have considered the central regions of the plate only – our derivation does not apply near the edges, where the field is not uniform.


The Capacitance of a Parallel-Plate Capacitor:Using the result from the previous section (the field near a charged conducting plate) we can derive an expression for the capacitance of a parallel-plate capacitor, in terms of its dimensions.

+ + + + + +

- - - - - -

+ + + + + +

- - - - - -

Consider the two metal plates of a charged capacitor, viewed edge-on.

Let A = area of surface of top platearea of surface of bottom plate

let d = separation between plateslet +q = charge on bottom surface of top platelet -q = charge on top surface of bottom plate

d

A

+q

-qv

Between the plates there will be a uniform electric field, E, and an associated potential difference, v.

The capacitance of the capacitor, C, is defined in terms of the potential difference, v, produced by given charge, q:

( )1...... vq

C ≡


Find C:( )1...... vq

C ≡

( )3.......... d E v thus

here dv

E dxdv

- E

gradient potential theof negative theis vector field The

=

=→=→

+ + + + + +

- - - - - -

+ + + + + +

- - - - - -d

A

+q

-qv

Since the field between the plates is uniform, the PD between the plates, v, can be expressed in terms of the magnitude of the electric field vector, E.

The density of charge, σ, on a plate is defined as its total charge, q, divided by its total plate area, A.

σ = q/A therefore q = σ A ............(2)

The field situation here is the same as in our derivation of the field near a charged plate. Hence the magnitude of E between the capacitor plates, will be given by -

( )4.......... �

E

0

=


Find C:( )1...... vq

C ≡

( )5.......... d�

v

0

=

dA �

d

�

A

vq

C 0

0

=

��

�

�=≡

( )3.......... d E v =

q = σ A ............(2)

( )4.......... �

E

0

=

(4) �� (3):

(2), (5) �� (1):

+ + + + + +

- - - - - -

+ + + + + +

- - - - - -d

A

+q

-qv

dA �

C 0=

Thus, to maximise capacitance, we maximise the area, A, of the plates, and minimise their separation, d.

In integrated circuits, the effective plate area, A, of capacitors must be small, but d is reduced correspondingly, by using a very thininsulating film between two thin metal layers. The metal layers are the capacitor plates.


Summary:In our studies of the electric field, we firstly discussed Coulomb’s law.

We then went on to the characterisation of the electric field, by the electric vector, scalar potential, and potential gradient. We saw how the electric field can be represented graphically using flux lines, or equipotential surfaces.

Using the idea of flux, we introduced an alternative formulation of Coulomb’s law, namely Gauss’ law, which proved to be powerful in determining the electric field due to charge distributions, which are symmetrical.

Finally we were able to use our field results to derive an expression for the capacitance of a parallel-plate capacitor, in terms of its dimensions.


+ + + + + + + + +

+ + + + + + + + + + + + + + + + + + + + +

+ + + + + + + +

Q

Rr-axis

dA

E

r

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7/20/2006 Electric Fields (© F.Robilliard) 1flai/Theory/lectures/ElectricFields.pdf · 7/20/2006 Electric Fields (© F.Robilliard) 2 Charge: Atoms are held together as a system,