Lecture 1-Coulomb_s Law

ECE Department, TUP Manila Engineering Electromagnetics Timothy M. Amado, ECE Summer

Engineering Electromagnetics

LECTURE 1: Coulombs Law

Instructor: Sir Tim

Summer Term

Coulombs Law


POINT CHARGES


ATOMIC MODEL


VALENCE

The NUMBER OF ELECTRONS in THEOUTERMOST SHELL determines thevalence of the atom

The outermost shell is called theVALENCE SHELL

The electrons residing in the outermostshell is called VALENCE ELECTRONS


Electrostatics

Deals with the phenomena due to theATTRACTIONS OR REPULSIONS OFELECTRIC CHARGES that are NOTMOVING.


Static Electricity


Coulombs Law

Coulombs First Law of Electrostatics The size of the force of attraction or repulsion

between two charges is DIRECTLYproportional to the value of the charge

Coulombs Second Law of Electrostatics The size of the force varies INVERSELY asthe square of the distance between the twocharges


Mathematical Form

k = 1/(4pi),(for air o = shift constant 32)

F force in newtons

Q1 and Q2 charge in coulombs

r distance bet. two charges in meters


NATURE OF CHARGES

LIKE charges REPEL

UNLIKE charges ATTRACT


ELECTRIC FIELD

Defined as the force per unit positivecharge that would be experienced by astationary point charge in a given locationin a field

The space between and around thecharged bodies where their influence is felt


ELECTRIC FIELD


ELECTRIC FIELD

k = 1/(4pi),(for air o = shift constant 32)

E electric field intensity in V/m orN/C

Q charge in coulombs r distance in meters


Problem 1

Point charges 1 mC and -2 mC are locatedat (3, 2, -1) and (-1, -1, 4) respectively.Calculate the electric force on a 10-nCcharge located at (0, 3, 1) and the electricfield intensity at that point.

Ans: F = -6.507ax 3.817ay + 7.506az mN

E = -650.7ax 381.7ay 750.6az kV/m


Problem 2

Two point charges of equal mass m, chargeQ are suspended at a common point by twothreads of negligible mass and length l.Show that at equilibrium, the inclinationangle of each thread to the vertical isgiven by

2 = 162 sin2 tan


Problem 3

A practical application of electrostaticsseparation of solids. Assuming zero initialvelocity and displacement, determine theseparation between the particles after falling80 cm. Take E = 500 kV/m and Q/m = 9C/kg for both positively and negativelycharged particles.

Ans: 73.4 cm

Coulombs Law


CHARGE DISTRIBUTION


VOLUME CHARGE DISTRIBUTION

=

where:

V is the volume charge density; and

Q is the total charge within some finite volume


FIELD OF A LINE CHARGE

=2

where:

L is the line chargedensity


FIELD OF A SHEET OF CHARGE

=2

where:

S is the surface chargedensity


Problem 1

Find the force on a point charge 50 C at (0,0, 5) m due to a circular disk with surfacecharge density of s = 20 C/m

2 and 5m, z = 0 m.

Ans: F = 16.5az N


Problem 2

The finite sheet 0 1, 0 1 on the z= 0 plane has a charge density =

(2 + 2 + 25)3

2 nC/m2. Find:(a) the total charge on the sheet(b) the electric field at (0, 0, 5)(c) the force experienced by a -1 mC charge

located at (0, 0, 5)

Ans: (a) Q = 33.15 nC(b) E = -1.5ax, -1.5ay, 11.25az V/m(c) F = 1.5ax, 1.5ay, -11.25az mN


Problem 3

Determine E at (2, 0, 2) m due to threestandard charge distributions as follows: auniform sheet at x = 0 m with s1 = (1/3)nC/m2, a uniform sheet at x = 4 m with s2= (-1/3) nC/m2, and a uniform line at x =6 m, y = 0 m with l = -2 nC/m.

Ans: E = 21ax V/m


Thank you.

Lecture 1: Coulomb's Law

Engineering Electromagnetics

Timothy M. AmadoFaculty, Electronics Engineering Department

Technological University of the Philippines Manila

[email protected]

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Lecture 1-Coulomb_s Law