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**kurunji venkataramana gowda - KVG**- Electronics and Communication Engineering
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- Module- 1 Coulomb's Law, Electric Field Intensity and Flux Density - ( 1 - 27 )
- Module- 2 Gauss Law and Divergence - ( 28 - 38 )
- Current Density (I) - ( 39 - 65 )
- Module- 3 Poisson's and Laplace's Equation - ( 66 - 97 )
- Magnetic Flux and Magnetic Flux Density - ( 98 - 103 )
- Module- 4 Magnetic Forces - ( 104 - 125 )
- Module- 5 Time Varying Fields and Maxwell's Equations - ( 126 - 149 )

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ENGINEERING ELECTROMAGNETICS(15EC36)-III SEM EC MODULE 1 MODULE-1 COULOMB’S LAW, ELECTRIC FIELD INTENSITY AND FLUX DENSITY Experimental law of Coulomb Coulomb stated that the “force between two very small objects separated in a free space by a distance which is large compared to their size is proportional to the charge on each and inversely proportional to the square of the distance between them”. Q1 R Q2 Figure 1.1 Consider two point charges Q1 and Q2 separated by a distance R as shown in Figure 1.1. Then as per Coulomb’s law, F=k Q1 Q 2 R2 (1) Where, Q1 and Q2 : Positive or negative quantities of charge R: Separation between two charges k: Proportionality constant If the International System of Units(SI) is used , Q is measured in Coulombs(C) , R is meters(m) and the force should be newtons(N). This will be achieved if the constant of proportionality k is written as k= 1 4πε0 Where, ε0 : Permittivity of free space ε0 = 8.854 × 10−12 F⁄m Therefore Coulomb’s law is written as, F= Q1 Q 2 4πε0 R2 (2) Vector form of Coulomb’s law Let the vector r1 locate charge Q1 while vector r2 locate charge Q2 . Then the vector R12 = r2 – r1 represents the directed line segment from Q1 to Q2 as shown in Figure 1.2. Mr.JAGADEESH M, ASST . PROFESSOR 1 DEPT. 0F EC, KVGCE SULLIA

ENGINEERING ELECTROMAGNETICS(15EC36)-III SEM EC MODULE 1 Figure 1.2 The vector force F2 is the force on Q2 by Q1 . According to Coulomb’s law , this vector force is given by, F= Q1 Q 2 a12 4πε0 R212 (3) Where, a12 is a unit vector in the direction of R12 , or a12 = R12 r2 − r1 = |R12 | |r2 − r1| (4) The force expressed by Coulomb’s law is a mutual force i.e, F2 = -F1. Electric field Intensity Consider one charge Q1 fixed in position and move a second charge slowly around, we note that there exists everywhere a force on this second charge, in other words this second charge is displaying the existence of a force field. Call this second charge a test charge Qt . The force on it is given by Coulomb’s law, F= Q1 Q t a 4πε0 R21t 1t (5) Writing this force as a force per unit charge gives, Ft Q1 = a Q t 4πε0 R21t 1t (6) The quantity on the RHS of equation (6) is a function only of Q1 and the directed line segment from Q1 to the position of the test charge. This describes a vector field and is called the electric field intensity (E). “Electric field intensity is defined as the vector force on a unit positive test charge”. Mr.JAGADEESH M, ASST . PROFESSOR 2 DEPT. 0F EC, KVGCE SULLIA

ENGINEERING ELECTROMAGNETICS(15EC36)-III SEM EC MODULE 1 The electric field intensity can be measured by the unit newtons per Coulomb – the force per unit charge. The practical unit is Volts per meter(V/m). It can be represented by the letter ‘E’ . Ft Q1 =E= a Qt 4πε0 R21t 1t (7) In general electric field intensity can be expressed as, E= Q a 4πε0 R2 R (8) If charge Q locates center of a spherical co-ordinate system, then the unit vector aR becomes radial unit vector ar and R is ‘r’ . Hence, E= Q a 4πε0 r 2 r (9) Writing these expressions in rectangular co-ordinate system, for a charge Q at the origin we have, R = r = xax + yay + zaz aR = ar = xax + yay + zaz √x 2 + y 2 + z 2 (10) Figure 1.3 Therefore, E= 4πε0 (x 2 Q x y z ax + ay + az ) ( 2 2 + y + z ) √x 2 + y 2 + z 2 √x 2 + y 2 + z 2 √x 2 + y 2 + z 2 (11) For a charge Q located at source point r1 = x1ax + y1ay + z1az as showen in figure 1.4. The field at general point r = xax + yay + zaz is given by E= Q r − r′ Q(r − r′) = 2 4πε0 |r − r′| |r − r′| 4πε0 |r − r′|3 (12) Since the Coulomb’s force has linear , the electric field intensity due to two point charges Q1 at r1 and Q2 at r2 is the sum of the forces on Qt caused by Q1 and Q2 acting alone. i.e, E= Q1 Q2 a + a 1 4πε0 |r − r1 |2 4πε0 |r − r2 |2 2 Mr.JAGADEESH M, ASST . PROFESSOR (13) 3 DEPT. 0F EC, KVGCE SULLIA

ENGINEERING ELECTROMAGNETICS(15EC36)-III SEM EC MODULE 1 Figure 1.4 Where a1 and a2 are the unit vectors in the direction of r – r1 and r – r2 respect ively. The vector r, r1 , r2, r-r1, r-r2, a1 and a2 are shown in figure 1.5. Figure 1.5 If we add more charges at other positions, the field due to ‘n’ point charges is, E= Q1 Q2 a1 + a +⋯ 2 4πε0 |r − r1 | 4πε0 |r − r2 |2 2 + Qn a 4πε0 |r − rn |2 n (14) In general , 𝐧 𝐄(𝐫) = ∑ 𝐦=𝟏 𝐐𝐦 𝐚 𝟒𝛑𝛆𝟎 |𝐫 − 𝐫𝐦 |𝟐 𝐦 Mr.JAGADEESH M, ASST . PROFESSOR (𝟏𝟓) 4 DEPT. 0F EC, KVGCE SULLIA

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