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Unformatted text preview: Chapter 2. Coulomb’s Law and Electric Field Intensity
The Experimental Law of Coulomb Coulomb’s Law The magnitude of force between two very small objects separated in free space by a distance which is large compared to their size is given by Q1Q2 F= 4 πε0R 2
Q1 and Q2: positive or negative quantities of charge (coulomb) R: separation (meters) ε0: permittivity of free space = 8.854 x 1012 F/m = 1/36π x 109 F/m 1/4πε0 ≈ 9 x 109 In vector form: In If Q1 iis in r1 and Q2 iis in r2, then R12 = r2  r1 iis the directed line s s s then segment from Q1 to Q2. segment z F2 = force experienced by Q2: F2 Q1Q2 R12 F2 = a12 Q1 2 4 πε0R12 Q2 r2 F1 r1 y x a12 iis the unit vector in the direction of R12 = s r2 − r1  r2 − r1 
2 F1 = −F2 = − Q1Q2 4 πε0R12 a= 2 12 Q2Q1 4 πε0R 21 a21 Electric Field Intensity Electric
Consider a charge Qi located at Pi, and another charge Qt situated in Consider situated the vicinity of Qi. The position of Qt with respect to Qi iis Pit. At s with At any position Pit, Qt experiences a force due to Qi experiences Ft = Qi Q t 4 πε0Pit
2 ait Qt Ft The force exerted by Qi per unit The exerted charge is charge Pit Qi Ft Qi a = 2 it Q t 4 πε0Pit The right side of the equation is a vector field called the electric field The electric iintensity due to Qi. ntensity The unit of electric field intensity E is N/C, or V/m The For a charge Q at the origin and using the spherical coordinate system, the electric field E at any given point is given by Q E= a 2r 4 πε0r
Note: If Q is positive, E “radiates” from Note: from the charge. If Q is negative, E points points into the charge. into For a charge at rm, E at r is For at
z r E= Q a 2m 4 πε0  r − rm 
rm x Q y am iis a unit vector in the direction s of r  rm In there an n charges, the electric field at r is E(r ) = ∑ Qi a 2i i=1 4 πε0  r − ri  n Example: Example: A 1uC charge is at the origin and a 2uC charge is at P(0,10,0). What 1uC is the electric field at Q(6,4,5)? If a –0.5 uC charge is placed at Q, 0.5 how much force would it experience? how
ROQ
1 uC Q RPQ P 2 uC ROQ = <6,4,5> RPQ = <6,4,5>  <0,10,0> = <6,6,5> <6,4,5> R OQ = 6 2 + 4 2 + 52 = 77 = 8.775 R PQ = 6 2 + (−6) 2 + 52 = 97 = 9.849 aOQ = aPQ = R OQ R OQ R PQ R PQ = = 6a x + 4 a y + 5αz 8.77 6a x − 6a y + 5αz 9.85 = 0.684a x + 0.456a y + 0.570αz = 0.609a x − 0.609a y + 0.508αz EQO 1 × 10 −6 = (0.684a x + 0.456a y + 0.570az ) 4 πε0 (77) 2 × 10 −6 = (0.609a x − 0.609a y + 0.508az ) 4 πε0 (97) = 79.84 a x + 53.23a y + 66.53az V/m E QP = 112.91a x − 112.91a y + 94.18az V/m
EQ = E QO + E QP = = 192.75a x − 59.68a y + 160.71az V/m EQ EOQ FQ
1 uC 0.5 uC EPQ P 2 uC r rF v v Since E = , then F = QE : Q v FQ = −0.5 × 10 − 6 (192.75a x − 59.68a y + 160.71az ) = −96.38a x + 29.84a y − 80.36az uN Cathode Ray Tube Cathode The electric fields The formed by the parallel plates deflect the electron beam generated at the back of the tube. tube. Electric Field Due to a Continuous Charge Distribution Electric
Define ρv = volume charge density = lim ∆Q Define ∆v → 0 ∆v ∆v contains the charge ∆Q The total charge Q is Q = ∫ dQ = ∫ ρ v dv
vol vol Due to an incremental charge ∆Q, the incremental electric field ∆E is Due Q,
∆E = ∆Q r − r' ρ v (r' )∆v r − r' = 4 πε0  r − r'2  r − r' 4 πε0  r − r'2  r − r' ∆Q r' r  r' ∆E r r = position of the point in question r’ = llocation of ∆Q ocation To get the field at the point, add the To contribution of all ∆Qs origin r − r' ∆E = 4πε 0  r − r'2  r − r'
E(r ) = ∑ ρ v (r' )∆v r − r' 2 i=1 4 πε0  r − r'  r − r'
N ρ v (r' )∆v If ∆v iis shrunk so it will approach zero (N will approach infinity), the s summation becomes an integral: summation E(r ) = ∫ ρ v (r' )dv r − r' 2 vol 4 πε0  r − r'  r − r' Field of a Line Charge Field
Define ρL = line charge density (C/m) Define Consider a uniform line charge along the zaxis:
zaxis dQ1 = ρL dL dE2 dE1 Due to symmetry, the electric the field iis a function of ρ and in the field s aρ direction only. dQ2 = ρL dL z (0, 0, z)
dQ = ρL dz r = ρ aρ r’ = z az R = r  r’ = ρ aρ  z az
ρaρ − zaz ρL dz dE = 4 πε0 (ρ2 + z2 ) ρ2 + z2 r' R = r  r' y r dEρ x dEz dE Since E iis directed along aρ only, the Since s only, zcomponent may be ignored when component solving for the total electric field: solving dE = ρL ρdz 4 πε0 (ρ + z )
2 2 3/2 aρ dE = ρL ρdz 4 πε 0 (ρ2 + z 2 ) 3 / 2 aρ Eρ = E = ∫ ρL ρdz a 2 2 3/2 ρ − ∞ 4 πε0 (ρ + z ) ∞ Evaluate the integral using change of variable: Let z = ρ tan α > dz = ρ sec2 α dα Let ρL ρ(ρ sec 2 α)dα E= ∫ a 2 2 2 3/2 ρ − π / 2 4 πε0 (ρ + ρ tan α)
π/2 z α ρ tan α = z/ρ ρ =L 4 πε0 ρL π / 2 sec 2 α sec 2 α dαaρ = dαaρ ∫ ∫ 3 2 3/2 4 πε0 ρ − π / 2 sec α − π / 2 ρ(1 + tan α)
π/2 ρL π / 2 ρL ρL π/2 = cos αdαaρ = aρ sin α − π / 2 = aρ ∫ 4 πε0ρ − π / 2 4 πε0ρ 2πε0ρ Therefore, the electric field due to a uniform lline charge along Therefore, uniform ine the zaxis is equal to the
E= ρL aρ 2πε0ρ Note: 1. The electric field due to an infinite line of charge is directed radially cted outward or into to the line charge. outward 2. The electric field varies inversely with the distance from the line he charge charge 3. If the line charge density is positive, the electric field “emanates” 3. from the charge. If the line charge density is negative, the electric ectric field “converges” to the line charge. Field of a Sheet of Charge Field
Define ρS = surface charge density Define
dy ρs
y Due to symmetry, the electric Due field at a point is not a function of y and z, and does not have components parallel to the yz components plane. plane. To simplify the solution, treat To the vertical strip as a uniform line charge. The sheet of charge may now be regarded as an infinite number of line charges placed beside each other. other. (x, 0, 0) Ex If the surface charge density is ρs, the line charge density of a If the “vertical” strip is ρs dy strip dy ρs
y For a line charge in the zaxis, For recall that recall ρL E= aρ 2πε0ρ Therefore:
dE x = = = ρS dy 2πε0 x + y ρS dy 2πε0 x + y
2 2 2 2 2 cos βax ax β
(x, 0, 0) x x +y
2 dEx ρS xdy ax 2 2 2πε0 ( x + y ) ρ xdy dE x = S 2 a 2x 2πε0 x + y ∞ ρS x 1 Ex = E = ax ∫ 2 dy 2 2πε0 −∞ x + y Note: ∫ du 1 u = tan−1 + C a a2 + u 2 a
y = +∞ ρS x 1 −1 y a x = ρ S π − ( − π ) a x E= tan 2πε0 x x y = −∞ 2πε0 2 2 E= ρS ax 2ε 0 At the negative xaxis (or at the “back” of the sheet of charge), axis
E=− ρS ax 2ε 0 In general, the electric field due to an infinite sheet of charge is equal to In ρS E= aN 2ε 0
Notes: 1. aN iis a unit vector perpendicular and pointing away from the s surface surface 2. The electric field intensity is constant. 3. If the surface charge density is positive, the electric field 3. surface charge “emanates” from the sheet of charge. If the line charge density is from sheet If negative, the electric field “is into” to the surface charge. to charge. Example: Example: A lline charge with charge density equal to 10 nC/m is at x = 4, z = 3. A ine sheet of charge with surface charge density equal to 1 nC/m2 is at the xyplane. What is the electric field at P(2,3,5)? plane. z P EL ES y Side view: EL z
D ES x
x EL z
D D = −2a x + 2az D = (−2) 2 + 2 2 = 2.828 − 2a x + 2a z aD = = −0.707a x + 0.707az 2.828 ES x
10 × 10 −9 EL = (−0.707a x + 0.707az ) = −44.939a x + 44.939az V/m 2πε0 (2.828) − 1 × 10 −9 ES = az = −56.472 az V/m 2ε 0 EP = EL + ES = −44.94a x − 11.53az V/m Streamlines and Sketches of Fields Streamlines
Given an expression of E, how will one draw (sketch) the field? Given Take, for example, a point charge. The arrows show the direction of the field at every point along the line, and the separation of the lines is inversely proportional to the strength of the field. strength The lines are called streamlines. The streamlines Given a two dimensional field (Ez = 0), the equation of a streamline is Given 0), obtained by solving the differential equation obtained
Ey Ex
y = dy dx Ey
∆x E
∆y Ex
x Example. The electric field intensity is given as The E = 5e2x (sin 2y ax  cos 2y ay) V/m (sin cos Find the equation of the streamline passing through the point P(0.5, 0.5, π/10, 0). Solution: Solving the differential equation
dy − 5e −2 x cos 2y = = − cot 2 y −2 x dx 5e sin 2y
Ey Ex = dy : dx − dx = tan 2 ydy − ∫ dx = ∫ tan 2ydy 1 − x = ln(sec 2 y) + C 2 − 2 x + C' = ln(sec 2y ) e −2 x + C' = K ' e −2 x = sec 2y Ke 2 x = cos 2 y To solve for K, use the fact that the streamline passes through P (0.5, π/10, 0): Ke 2 x = cos 2 y Ke2(0.5) = cos π/5 K = 0.298 Therefore, the equation of the streamline through P is 0.298 e2x = cos 2y 2y ...
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This note was uploaded on 01/13/2011 for the course EEEI 23 taught by Professor Joeljosephmarciano during the Winter '10 term at University of the Philippines Diliman.
 Winter '10
 JoelJosephMarciano

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