Coulomb’s law & Electric Field Intensity Electromagnetics I

Electromagnetics

Electrostatics Electrostatic force Proportional to the product of charges Inversely as the square of the distance In the direction of the line connects between charges Depending on the medium Charges of same polarity : repulsive and different polarity : attractive Coulomb’s law

Two charges of same polarity Q1 Q2 a21 a12 F1 F2

Two charges of different polarity Q2 Q1

Charge densities

Electric Field intensity, E (Field Strength) V/m, kV/m, V/cm etc E : force exerted on a positive charge introduced into the field per unit of positive charge

positive

Electric field and electric field lines

Electric field caused by positive and negative charges

Electric field intensity from a point charge

F=-0.15 N ax

E due to point charge - Cartesian A point charge of Q= 1 nC located at (-0.5 , -1 , 2) in free space. a. Determine E at a point 1 m distance from the point charge. b. Determine E at (0.9 , 1.2, -2.4) a.

=(0.9 - (-0.5)) ax + (1.2 - (-1)) ay + (-2.4 - (2)) az b. E di (0.9 , 1.2 , -2.4) QT=QT - OQ =(0.9 - (-0.5)) ax + (1.2 - (-1)) ay + (-2.4 - (2)) az =1.4 ax + 2.2 ay - 4.4 az

Determine F and E experienced by Q1= 20 mC located at (0,1,2) m caused by a charge of Q2=-300 mC at (2,0,0) m. (2,0,0) (0,1,2) Q2 Q1 x y z

E due to multi charges

Point charges q1=2x10-5C and q2=-4x10-5 C located at (1,3,-1) and (-3,1,-2) in free space. Determine E at (3,1,-2).

Line charge

A line charge located at Z axis extended from z=-5 m to z=5 m A line charge located at Z axis extended from z=-5 m to z=5 m. The density is 20 nC/m. Determine E at (2,0,0) in Cartesian and cylindrical. (2,0,0) dQ=rldz x y z R E -5 5

Exercise Determine E at (2,0,0) due to 2 line charges extended from z= 5 to +infinite and from z= - infinite to z=-5 with charge density of 20 nC/m

A line charge distributed in a line defined as x=2 m and y = -4 m with density of 20 nC/m. Determine E at point (-2,-1,4)

Exercise Determine E at (2,2,0) due to line charge defined as x=4 m and y=3 m with density of 10 nC/m and a point charge of + 2 mC at (0,0,0)

dy Z rs q Y Komponen yang ada hanya Ex, karena Ey dan Ez saling menghilangkan X dEy P q dE dEx

Infinite surface charge with dinsity of rs located at x= -1 and x =1 planes. Determine E . rs x=-1 rs x=1

Surface charge of 30 nC/m2 located at a plane of 2x – 3y + z = 6. Determine E at (0,0,0). (0,-2,0) (3,0,0) (0,0,6) x y z E

Exercise: Determine E at (1,1,0) due to a surface charge at z=1 with charge density of 2 nC/m2 , line charge along z axis with charge density of 5 nC/m and 10 nC point charge located at (0,0,0).

Field lines

Representation of field lines X Y B=Kx ay Non-Uniform X Y A=K ax Uniform D=k/r ar

Field lines equation Y Ey E Ey X Solve the differential equation

Example: field caused by a line charge with rl=2peo Y=cx

Gaya Lorentz Bila suatu muatan listrik Q bergerak pada medan magnet B maka muatan akan mengalami gaya yang sebanding dengan Q, kecepatan V, rapat fluks magnetik B dan sudut antara V dan B. Arah gaya akan tegak lurus baik pada V maupun B. Secara matematis dinyatakan dengan : F = Q v x B Dengan demikian bila suatu muatan bergerak di dalam ruangan dengan medan listrik dan medan magnet maka akan mengalami gaya F = Q (E + v x B)  Gaya Lorentz

adalah panjang lintasan maka integrasi INTEGRASI VEKTOR Untuk suatu skalar , dimana adalah panjang lintasan maka integrasi dari a ke b adalah Untuk sembarang kontur maka dengan diskretisasi diperoleh C N 2 1

INTEGRASI VEKTOR (0,1) y contoh Bila = A (konst) Maka c (0,1) x

INTEGRASI VEKTOR Bila dan c adalah dari (a,0) → (a,a) y (a,a) c x (a,0)

UNTUK VEKTOR y b c a x Untuk vektor maka integrasi sepanjang lintasan c adalah

UNTUK VEKTOR

c adalah vektordari A(0,1,2) dan B(1,0,2) UNTUK VEKTOR contoh c adalah vektordari A(0,1,2) dan B(1,0,2) A(0,1,2) B(1,0,2) y x dengan parameter x=t , y=I-t , z=2

INTEGRAL PERMUKAAN Integral vektor pada permukaan merupakan total flux menembus yaitu contoh Hitung Flux yang keluar dari box yang dibatasi oleh x=0,1 ; y=0,1 i=0,1 Jawab = 0