Electric Field Concepts

Rules for constructing filed lines• A convenient way to visualize

the electric field due to any charge distribution is to draw a field line diagram. At any point the field line has the same direction as the electric field vector

• Field lines begin at positive charge and end at negative charge

• The number of field lines shown diverging from or converging into a point is proportional to the magnitude of the charge.

• Field lines are spherically symmetric near a point charge

• If the system has a net charge, the field lines are spherically symmetric at great distances

• Field lines never cross each other.

ELECTRIC POTENTIAL (The Volt)

To develop the concept of electric potential and

show its relationship to electric field intensity.

In moving the object from point a to b, the work can be expressed by:

b

a

dW LF

dL is differential length vector along some portion of the path between a and b

The work done by the field in moving the charge from a to b is

b

afieldE dQW LE

If an external force moves the charge against the field, the work done is negative:

b

a

dQW LE

ELECTRIC POTENTIAL (cont’d)

We can defined the electric potential difference, Vab as the

work done by an external source to move a charge from point

a to point b as:

b

aba d

Q

WV LE

Where,abab VVV

ELECTRIC POTENTIAL (cont’d)

‘a’ is the initial point while ‘b’ is the final point

If is negative, there is a loss in potential energy in

moving Q from ‘a’ to ‘b’; this implies that the work is being

done by the field. However. If is positive, there is a

gain in potential energy in the movement, an external

agent performs the work

is independent of the path taken

is measured in joules per coulomb, commonly referred to

as volts (V)

abV

abV

abV

abV

Consider the potential difference between two

points in space resulting from the field of a

point charge located at origin, where the

electric field intensity is radially directed, then

move from point a to b to have:

b

arr

b

aba dr

r

QdV aaLE

204

ELECTRIC POTENTIAL (cont’d)

Thus,

ab

br

arba

VVab

Q

r

QV

11

4

4

0

0

The absolute potential at some finite radius from a point charge fixed at the origin:

r

QV

04

ELECTRIC POTENTIAL (cont’d)

If the collection of charges becomes a continuous distribution, we could find:

r

dQV

04

Where,

r

dVV

r

dSV

r

dLV

V

S

L

0

0

0

4

4

4

Line charge

Surface charge

Volume charge

ELECTRIC POTENTIAL (cont’d)

N

NQ

QQV

rr...

rrrr

0

20

2

10

1

4

44

N

k k

kQV

104

1

rrOr generally,

The principle of superposition, where applied to

electric field also applies to potential difference.

ELECTRIC POTENTIAL (cont’d)

Based on figure, if a closed path is chosen, the integral will return zero potential:

Three different paths

to calculate work

moving from the origin

to point P against an

electric field.

0 LE d

ELECTRIC POTENTIAL (cont’d)

EXAMPLE 10Two point charges -4 μC and 5 μC are located at (2,1-,3)

and (0,4,-2) respectively. Find the potential at (1,0,1).

Let and CQ 41 CQ 52 So,

20

2

10

1

44 rrrr

QQ

V

Where, 262,4,12,4,01,0,1

62,1,13,1,21,0,1

2

1

rr

rr

SOLUTION - EXAMPLE 10

264

105

64

104

441,0,1

0

6

0

6

20

2

10

1

rrrr

QQVTherefore,

kVV 872.51,0,1

The electrostatic potential contours from a point charge form equipotential surfaces surrounding the point charge. The surfaces are always orthogonal to the field lines. The electric field can be determined by finding the maximum rate and direction of spatial change of the potential field.

ELECTRIC POTENTIAL (cont’d)

Therefore,

VEThe negative sign indicates that the field is pointing in the direction of decreasing potential.

By applying to the potential field:

rrr

Q

r

Q

rV aaE

200 44

ELECTRIC POTENTIAL (cont’d)

Three ways to calculate E:

If sufficient symmetry, employ Gauss’s Law.

Use the Coulomb’s Law approach.

Use the gradient equation.

IMPORTANT !!

Consider a disk of charge ρS, find the potential at

point (0,0,h) on the z-axis and then find E at

that point.

EXAMPLE 11

Find that,

dd

dSdQ

S

S

and 22 hr

With r

dQV

04

then,

a

S

r

ddV

0

2

004

SOLUTION TO EXAMPLE 11

Let and leads to

Integral then,

22 hu

How to calculate the integral?

ddu 2

duu 21

hah

hV

S

aS

22

0

0

22

0

2

2

SOLUTION TO EXAMPLE 11 (Cont’d)

To find E, need to know how V is changing with position.

In this case E varies along the z-axis, so simply replace h

with z in the answer for V, then proceed with the gradient

equation.

zS

zS

z

az

z

az

z

z

VV

aa

aE

220

220

12

12

2

1

2

Conductors and Insulators• A conductor is a substance that allows current to

flow through it :- they transfer charge across them.

• In metals, the current is composed of moving electrons.

• Electrolytic solutions also conduct current but by the movement of flow of ions.

• Insulators have few mobile electrons or ions and the flow of current is inhibited- They keep tight tabs on their electrons.

• As fields are increased, dielectric breakdown of insulators occurs and the current is discharged as a surge.

• The dielectric strength is the maximum field an insulator can support.

• Resistance is a measure of resistance to flow of electricity. It is defined by Ohm’s Law as follows:

(ohm’s Law)

Therefore, resistance is in the units of volts per ampere.

One volt per ampere is called an ohm (Ω).The reciprocal of resistance is conductance

Resistance and Ohm’s Law

I

VR

The amount of charge that accumulates as a

function of potential difference is called the

capacitance.

V

QC

The unit is the farad (F) or coulomb per volt.

Capacitance

Two methods for determining capacitance:

Q Method

• Assume a charge +Q on plate ‘a’ and a charge –Q on plate ‘b’.

• Solve for E using the appropriate method (Coulomb’s Law, Gauss’s Law, boundary conditions)

• Solve for the potential difference Vab

between the plates (The assumed Q will divide out)

Capacitance (Cont’d)

V Method

• Assume Vab between the plates.

• Find E , then D using Laplace’s equation.

• Find ρS, and then Q at each plate using

conductor dielectric boundary condition

(DN = ρS )

• C = Q/Vab (the assumed Vab will divide

out)

Capacitance (Cont’d)

Use Q method to find the capacitance

for the parallel plate capacitor as

shown.

Example 12

Place charge +Q on the inner surface of the top plate, and –Q charge on the upper surface of the bottom plate, where the charge density,

Use conductor dielectric boundary, to obtain:

dSQ SSQ

S from

zSQ aD from

SN D

Solution to Example 12

We could find the electric field intensity, E

zrS

QaE

0

The potential difference across the plates is:

S

Qddz

S

Q

dV

rz

d

zr

a

bab

00 0

aa

LE

Solution to Example 12

Finally, to get the capacitance:

SQd

QV

QC

r

ab0

d

SC r0

Solution to Example 12

• Bioelectrical impedance analysis (BIA) is a commonly used method for estimating body composition.

• Since the advent of the first commercially available devices in the mid-1980s the method has become popular owing to its ease of use, portability of the equipment and its relatively low cost compared to some of the other methods of body composition analysis.

• It is familiar in the consumer market as a simple instrument for estimating body fat.

• BIA actually determines the electrical impedance, or opposition to the flow of an electric current, of body tissues, which can be used to calculate an estimate of total body water (TBW).

• TBW can be used to estimate fat-free body mass and, by difference with body weight, body fat.

Bioelectrical Impedance Analysis

When constant electric current is applied between two electrodes through a biological medium and the corresponding voltage is measured between the two source poles, the resultant impedance or bioimpedance is determined by Ohm’s law. The recorded voltage is the sum of the potential difference contributions due to the electrical conductivity properties of the tissue medium. The exchange of electrons from source to sink occurs from electrons of the metal electrode (such as platinum or silver-silver chloride) to ions of the tissue medium. The electrode is the site of charge carrier exchange between electrons and ions and thus serves as a transducer of electrical energy. Impedance measurements most commonly use a two-electrode (bipolar) of four-electrode (tetrapolar) arrangement.

Bioelectrical Impedance Theory

The Maxwell equation most relevant to bioimpedance is Eq. (1) Eq. (2)

where H - magnetic field strength [A/m], D - electric flux density [coulomb/m2], J - current density [A/m2],E - electric field strength [V/m], - permittivity of vacuum [farad (F)/m], and P - electric polarization dipole moment pr. volume

[coulomb/m2].

• If the magnetic component is ignored, Equation 1 is reduced to:

Eq. (3) • Equations 1-3 are extremely robust and also valid under

nonhomogeneous, nonlinear, and anisotropic conditions. They relate the time and space derivatives at a point to the current density at that point.

Bioelectrical Impedance Theory

PED

JtDH

JtD

• Impedance and permittivity in their simplest forms are based on a basic capacitor model.

• Basic equation of bioimpedance is then (time vectors)

Bioelectrical Impedance Theory

CjGY

Typical body segment resistance values

SUMMARY (1)

The force exerted on a charge Q1 on charge Q2 in a medium of permittivity ε is given by Coulomb’s Law:

12212

2112

4a

RF

QQ

Where is a vector from charge Q1 to Q2

121212 aR R

2

121 Q

FE

Electric field intensity E1 is related to force F12 by:

The Coulomb’s Law can be rewritten as:

RQ

aR

E2

04

For a continuous charge distribution:

RR

dQaE

204

For a point charge at origin:

rr

QaE

204

SUMMARY (2)

•For an infinite length line charge ρL on the z axis

aE2

L

•For an infinite extent sheet of charge ρS

NS aE2

Electric flux density, D related to field intensity by: ED 0 r

Where εr is the relative permittivity in a linear, isotropic

and homogeneous material.

SUMMARY (3)

Electric flux passing through a surface is given by:

dS DGauss’s Law states that the net electric flux through any closed surface is equal to the total charge enclosed by that surface:

encQd SD

Point form of Gauss’s Law is

V D

SUMMARY (4)

The electric potential difference Vab between a pair of points a and b in an electric field is given by:

ab

b

a

ab VVdV LE

Where Va and Vb are the electrostatics potentials at

a and b respectively.

For a distribution of charge in the vicinity of the origin, where a zero reference voltage is taken at infinite radius:

r

dQV

4

SUMMARY (5)

E is related to V by the gradient equation:

VEWhich for Cartesian coordinates is:

zyx z

V

y

V

x

VV aaa

• The conditions for the fields at the boundary between a pair of dielectrics is given by:

S 2121 DDa21 TT EE and

SUMMARY (6)

Where ET1 and ET2 are the electric field components

tangential to the boundary, a21 is a unit vector from

medium 2 to 1 and ρS is the surface charge at the

boundary. If no surface charge is present, the components of D normal to the boundary are equal:

21 NN DD

At the boundary between a conductor and a dielectric, the conditions are:

0TE and SN D

SUMMARY (7)

Poisson’s equation is:

VV 2

Where the Laplacian of V in Cartesian coordinates is given by:

2

2

2

2

2

22

z

V

y

V

x

VV

In a charge free medium, Poisson’s equation reduces to Laplace’s equation

02 V

SUMMARY (8)

• Capacitance is a measure of charge storage capability and is given by:

V

QC

For coaxial cable:

For two concentric spheres:

ab

LC

ln

2 abV L

ab ln2

So,

baC

11

4

ba

QVab

11

4So,

SUMMARY (9)