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Module M2-1Electrical Engineering

T U T O R I A L 1 V E C T O R S A N D C O U L O M B ' S L A W

A U G U S T 2 0 1 3

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Topics

Review of vectors (with Matlab)

Coulomb's law

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Coulomb s law

Examples and in-class exercises

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Notation

Vector and scalar quantities3

Vector or or

Scalar

Vector is a quantity that has both magnitude and di idirection

We usually represent a vector by a bold font, or a font with an arrow or a hat on top

The Cartesian coordinate system4

The direction of the axes is determined by the right- The direction of the axes is determined by the right-hand rule

Each point in the space is represented by a triple (x, y, z)

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The vector from one point to another point5

http://www.colorado.edu/geography/gcraft/notes/coordsys/coordsys_f.html

The vector from point to point is given by

or

Vector operation in Matlab

Matab represents a vector A = 1ax + 2ay + 3az by>> A = [1 2 3]

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>> A = [1 2 3]

For a vector

The magnitude of A is given by

Matlab command for the magnitude (A)>> norm(A)

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Dot product

For vectorand vector

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and vector

Dot product is a scalar that equals

where is the angle between A and B

Matlab commandMatlab command>> dot(A,B)

Cross product

A cross product is a vector that equals

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Matlab command>> cross(A,B)

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Magnitude and direction of .

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The magnitude isThe magnitude iswhere is the angle between A and B

The direction is orthogonal to both A and B and in the direction of the thumb (right-hand rule)

Cross product (cont.)

The cross product of the two t A

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vectors A = 2ax + 1ay + 0az

and B = 1ax + 2ay + 0az is shown in blue.

C = 0ax + 0ay + 3az.

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Example: 2D Cartesian coordinate system

Let vectors A and B be given byA=[3 -2]; B=[1 3]; C=[-2 0]

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A [3 2]; B [1 3]; C [ 2 0]

Question 1: Draw these vectors in the same diagram

Solution (picture) to question 112

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Example: 2D (cont.)

Question 2: Find in Matlab and draw in paper the vector sums A+B and A+B+C

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Solution (Matlab):>> clear all % clear all variables>> A = [ 3 -2];>> B = [ 1 3];>> C = [-2 0];>> A+Bans =

4 14 1>> A+B+Cans =2 1

Solution (picture): A+B = [4 1]14

triangular law parallelogram law

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Solution (picture): A+B+C = [2 1]15

triangular law parallelogram law

Example: 2D (cont.)

Question 3: Find and draw vector differences A – B, B – A, and A – B – C

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Solution (Matlab):>> A-Bans =

2 -5>> B-Aans =

2 5-2 5>> A-B-Cans =

4 -5

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Solution (picture)17

Example: 2D (cont.)

Question 4: Find and draw scalar multiplications of vectors 2A, -3B, and C/1.5

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Solution (Matlab):>> 2*Aans =

6 -4>> -3*Bans =

3 9-3 -9>> C/1.5ans =-1.3333 0

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Solution (picture): 2A = [6 -4]19

Solution (picture): -3B = [-3 -9]20

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Solution (picture): C/1.5 = [-4/3 0]21

Example: 2D (cont.)

Question 5: Find the dot product:

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Solution:

>> dot(A, B)ans =

-3

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Example: 2D (cont.)

Question 6: Find and draw the cross product

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Solution:

>> cross([A 0], [B 0])ans =0 0 11

Append zero to the z-component,so we have a 3D vector needed forcomputing a cross product

Solution (picture)24

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In-class exercise: 3D Cartesian coordinate system

Let A=[3 -2 7]; B=[1 3 5]; C=[-2 0 3]

Question 1: Draw these vector in the same diagram

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Question 1: Draw these vector in the same diagram

Solution (picture)26

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In-class exercise: 3D (cont.)

Question 2: Find and draw vector sumsA+B, A+B+C

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Solution (Matlab):>> clear all % clear all variables>> A=[ 3 -2 7];>> B=[ 1 3 5];>> C=[-2 0 -3];>> A+Bans =

4 1 12>> A+B+Cans =

2 1 15

Solution (picture)28

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In-class exercise: 3D (cont.)

Question 3: Find and draw vector differencesA-B, B-A, A-C

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, , Solution (Matlab):

>> A-Bans =

2 -5 2>> B-Aans =

-2 5 -2-2 5 -2>> A-Cans =

5 -2 4

Solution (picture)30

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In-class exercise: 3D (cont.)

Question 4: Find and draw scalar multiplications of vectors 2 A, 1.75B, 0.5C

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Solution:>> 2*Aans =

6 -4 14>> 1.75*Bans =

1.7500 5.2500 8.7500>> 0.5*C ans =

-1.0000 0 1.5000

Solution (picture)32

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In-class exercise: 3D (cont.)

Question 5: Find the dot products , ,

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Solution (Matlab):

>> dot(A, A)ans =

62>> dot(A, B)ans =

3232>> dot(A, C)ans =

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In-class exercise: 3D (cont.)

Question 6: Find and draw the cross products, ,

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Solution:>> cross(A, B)ans =

-31 -8 11>> cross(B, A)ans =

31 8 1131 8 -11>> cross(A, C)ans =

-6 -23 -4

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Solution (picture)35

Coulomb’s law

Positive charge attracts negative charge

Same polarity charges repel one another

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Same polarity charges repel one another

Forces between two chargesa unit vector pointing fromcharge Q1 to charge Q2

Force acting on charge Q2due to charge Q1

Q = electric charge (coulomb, C)R12 = distance between two charges0 = 8.854 x 10-12 F/m

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Example: two positive charges

The first charge of 20p C is located at P1(2,3,4) m

The second charge of 40p C is located at P2(5 -3 6) m

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The second charge of 40p C is located at P2(5, 3,6) m p denotes charge of a proton (1.602 x 10-19 C)

Let F12 = force acting on the first charge due to the second charge

Let F21 = force acting on the second charge due to the first charge

Question 1: Draw a diagram depicting charges and directions of F12 and F21

Solution (picture)38

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Example: two positive charges (cont.)

Question 2: Find F12 and F21

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Solution: the Matlab code (next page) gives us these answers

Solution (Matlab)

%-----------------------%% adjustable parameters %%-----------------------%clear all % clear all variables, etc. from memoryP = 1 602e 19; % electrical charge of proton C

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P = 1.602e-19; % electrical charge of proton, CEp0 = 8.85e-12; % Air permittivity, F/mQ1 = 20*P; % 1st chargeQ2 = 40*P; % 2nd chargeP1 = [2 3 4]; % location of 1st chargeP2 = [5 -3 6]; % location of 2nd charge

%-----------------------%% compute the forces %%-----------------------%R12 = P2-P1; % a vector from point P1 to point P2R21 = P1-P2; % a vector from point P2 to point P1R = norm(R12); % norm of vector R12, also norm of vector R21a12= R12/R; % a unit vector in the direction of vector R12a21=R21/R; % a unit vector in the direction of vector R12

% Coulomb's law: F = (Q_1 Q_2)/(4 * pi * epislon_0 R^2 ) a_rF21 = Q1*Q2/(4*pi*Ep0*R^2)*a12; % Force acting on Q2 due to Q1F12 = Q1*Q2/(4*pi*Ep0*R^2)*a21; % Force acting on Q1 due to Q2

% Display resultsF12 F21

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In-class exercise: positive & negative charges

The first charge of 20p C is located at P1(2,3,4) m The second charge of 40e C is located at P2(5,-3,6) m

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The second charge of 40e C is located at P2(5, 3,6) m e denotes charge of an electron (-1.602 x 10-19 C)

Let F12 = force acting on the first charge due to the second charge

Let F21 = force acting on the second charge due to the first charge

Question 1: Draw a diagram depicting charges and directions of F12 and F21

Solution (picture)42

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In-class exercise: +,- charges (cont.)

Question 2: Find F12 and F21

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Solution: the Matlab code (next page) gives us these answers

Solution (Matlab)

%-----------------------%% adjustable parameters %%-----------------------%clear all % clear all variables, etc. from memoryP = 1 602e 19; % electrical charge of proton C

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P = 1.602e-19; % electrical charge of proton, Ce = -P; % electrical charge of electron, CEp0 = 8.85e-12; % Air permittivity, F/mQ1 = 20*P; % 1st chargeQ2 = 40*e; % 2nd chargeP1 = [2 3 4]; % location of 1st chargeP2 = [5 -3 6]; % location of 2nd charge

%-----------------------%% computer the forces %%-----------------------%R12 = P2-P1; % a vector from point P1 to point P2R21 = P1-P2; % a vector from point P2 to point P1R = norm(R12); % norm of vector R12, also norm of vector R21a12= R12/R; % a unit vector in the direction of vector R12a21=R21/R; % a unit vector in the direction of vector R12

% Coulomb's law: F = (Q_1 Q_2)/(4 * pi * epislon_0 R^2 ) a_rF21 = Q1*Q2/(4*pi*Ep0*R^2)*a12; % Force acting on Q2 due to Q1F12 = Q1*Q2/(4*pi*Ep0*R^2)*a21; % Force acting on Q1 due to Q2

% Display resultsF12 F21

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Summary45

Vector arithmetic vectors in Matlab vectors in Matlab

magnitude

sum, difference, scalar multiplication

dot product, cross product

Charges with same polarity repel

Charges with different polarity attractg p y

Coulomb’s law formula

examples (Matlab)