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Solution to the Drill problems of chapter 02 (Engineering Electromagnetics,Hayt,A.Buck 7th ed) BEE 4A,4B & 4C Following Exercise questions are IMPORTANT! 2.4, 2.5, 2.13, 2.14, 2.16 ,2.17, 2.18, 2.19, 2.22, 2.23, 2.27, 2.28,2.29,2.30,2.31 D2.1 (a). QA = −20 20µC µC located located at A(-6,4,7) ,Q ,QB = 50 5 0µC located located at B(5,8,-2) Find RAB AB = (5 R ( 5 − (−6))ˆ ax + (8 − 4)ˆ ay + (− ( −2 − 7)ˆ az = 11ˆ ax + 4ˆay − 9ˆ az AB |= (b). | R
(112 ) + 4 2 + (− (−9)2 = 14 14..76 76m m
3 ( −20 × 10 AB (c). F AB = Q A QB RAB /4π o | RAB | = (− AB ⇒ F 3 0.76ˆax + 11. 11 .184ˆ ay − 25 25..16ˆ az mN AB = 30 3 AB (d).F AB = Q A QB RAB /4π o | RAB | = (−20 × 10 AB ⇒ F 3 0.72ˆax + 11. 11 .169ˆ ay − 25 25..13ˆ az mN AB = 30
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× 50 × 10
× 50 × 10
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− 9ˆ az ))/ ))/(4π (4π × (10
9 /36 36π π)
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− 9ˆ az ))/ ))/(4π (4π × 8.85 × 10
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D2.2(a). QA = −0.3µC located located at A(25,-30,15) in cm ,Q , QB = 0. 0 .5µC located located at B(-10,8,12) at Find E at the origin O(0,0,0). at o and it will be the sum of E A ( E due Let E at the origin is denoted by E due to Q to Q A located at point A) and E B ( E due due to Q to Q B located at point B) OA | 3 E A = Q A ROA /4π o | R OA = (0 − 25))ˆ R ax + (0 − (−30))ˆ ay + (0 − 15)ˆ az = (− ( −25ˆax + 30ˆ ay − 15ˆ az )cm OA |= |R | = (−25)2 + (30)2 + (− (−15)2 = 41. 41 .83 83cm cm 6 E A = (−0.3 × 10 ) × (−25ˆax +30ˆ ay − 15ˆaz ) × 10 2 /4π × 8.85 85× × 10 12 × | 41 41..83 83× × 10 2 |3 = −368 368..55(− 55(−25ˆ ax +30ˆ ay − 15ˆaz ) B = Q B OB |3 E ROB /4π o | R OB = (0 − (−10)))ˆ R 2 ax +2(0 − 8)ˆay2 + (0 − 12)ˆaz = (10ˆax − 8ˆay − 12ˆaz )cm OB |= (10) + (− |R (−8) + (− (−12) = 17. 17 .55 55cm cm 6 E B = (0. (0.5 × 10 ) × (−25ˆ ax +30ˆ ay − 15ˆaz ) × 10 2 /4π × 8.85 85× × 10 12 × | 17 17..55 55× × 10 2 |3 = 8317. 8317.36(− 36(−25ˆ ax +30ˆ ay − 15ˆaz ) o = E A + E B = (−368 E 368..55(− 55(−25ˆax + 30 30ˆ a ˆy − 15ˆ az ))+8317. ))+8317.36(10ˆ ax − 8ˆ ay − 12ˆ az ) = (92. (92.3ˆax − 77 77..6ˆ ay − 94 94..2ˆaz )KV/m −
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at (b). Find E at the point P(15,20,50). P A and R P B and the rest of the problem is similar to It is the same as part(a) but this time we have to calculate R part(a) D2.3 (a). Σ50 ( ( 1 + (− (−1)m )/(m2 + 1)) = (1 + (−1)0 )/(02 + 1 ) + ( 1 + (− (−1)1 )/(12 + 1 ) + ( 1 + (− (−1)2 )/(22 + 1 ) + ( 1 + (− (−1)3 )/(32 + 1) + (1 + (− ( −1)4 )/(42 + 1) + (1 + (− ( −1)5 )/(52 + 1) = 2 + 0 + 2/ 2 /5 + 0 + 2/ 2 /17 + 0 = 2. 2.52 (b). Similar to the part(a) D2.4 (a). 0.1 ≤ (| x |, | y |, | z |) ≤ 0.2 , given given ranges ranges of x,y and z coco-ord ordinat inates es doesnot doesnot constitu constitute te a cubica cubicall volume so dv so dv = 0 ⇒ Q = vol ρv dv = dv = 0
(b). Differential volume in cylindrical co-ordinates is given by dv = ρdρdφdz = ρdρdφdz , , we have Q = vol ρv dv 0.1 π 4 2 2 0.1 3 π 2 2 ⇒ Q = vol (ρ z sin(0. sin(0.6)φ 6)φ)ρdρdφdz = ρdρdφdz = 0 0 2 (ρ (ρ z sin(0. sin(0.6)φ 6)φ)ρdρdφdz = ρdρdφdz = 0 ρ dρ 0 (sin(0. (sin(0.6)φ 6)φ)dφ 2 4 dz ⇒ Q =| = | ρ4 /4 |00.1 × | (−cos(0 cos(0..6φ))/ ))/0.6 |0π × | z | 42 =| (0. (0.1)4 /4 | × | (−cos(108 cos(1080 ) − (−cos(0))) cos(0)))//0.6 | × | (64 − 8)/ 8)/3 | 4 ⇒ Q =| = | (0. (0.1) /4 | × | (1. (1.31)/ 31)/0.6 | × | 56 56//3 |= 1. 1 .018 018mC mC 1
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(C). Assuming this universe to be a perfect sphere we have limits as 0 ≤ r ≤ ∞, 0 ≤ φ ≤ 2π, 0 ≤ θ ≤ π and the differential volume for spherical co-ordinates system is gien by dv = (r sin θdφ)(rdθ)(dr) = r 2 sin θdθdφdr ⇒ Q = vol ρv dv = vol (e 2r /r2 )r2 sin θdθdφdr = 0 2π dφ 0 π sin θdθ 0 e 2r dr =| φ |20π × | (−cosθ) |π0 × | e 2r /−2 |0 ⇒ Q = 2π × 2 × 1/2 = 2πC = 6.28C ∞
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is given by E = ρL a |= ρL R/2π 2 D2.5 (a). For infinite uniform line charge E ˆR /2π o | R o | R | where R is the perpendicular distance vector between the line charge and the point under consideration, in this case the point at P A is is P A (0,0,4), since we have two infinite line charges,one along x-axis and one along y-axis so the value of E 1 (E due to infinite line charge along x-axis ) and E 2 (E due to infinite line charge along y-axis ) the sum of E 2 1 = ρL R px /2π o | R px | , so E R px =( Perpendicular distance vector between point P A and the line charge along x-axis) px = 4ˆaz , now E 1 = 5 × 10 9 × 4ˆaz /(2π × 8.85 × 10 12 × 16) = 22.479ˆ ⇒R az V /m using similar arguments we can find py = 4ˆ 2 = ρ L R py /2π o | R py |2 = 5 × 10 9 × 4ˆ R az so E az /(2π × 8.85 × 10 12 × 16) = 22.479ˆ az V /m = E 1 + E 2 = 22.479ˆ ⇒ E az V /m + 22.479ˆ az V /m = 45ˆ az V /m −
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px = 3ˆay + 4ˆaz , since the point is P B (0, 3, 4) (b). Using the same arguments as in part(a) we have R py = +4ˆaz and R 1 = ρ L R px /2π o | R px | 2 = 5 × 10 9 × (3ˆay + 4ˆaz )/(2π × 8.85 × 10 12 × 25) = 10.8ˆ E ay + 14.38ˆaz 2 = ρ L R py /2πo | R py |2 = 5 × 10 9 × 4ˆ E az /(2π × 8.85 × 10 12 × 16) = 22.479ˆ az = E 1 + E 2 = 10.8ˆay + 14.38ˆ ⇒ E az + 22.479ˆ az = 10.8ˆay + 36.9ˆ az −
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D2.6 (a). P A = (2, 5, −5) , since the point P A is located below all the given surfaces or shet of charges so the will be in the -ive ˆaz direction i.e −ˆ unit normal vector ˆaN to these surfaces ,which also shows the diretion of E az at point P A will be the sum of E ’s caused by each sheet of charge E = E z = 4 + E z=1 + E z=4 , now we have E = (ρs /2 × o )ˆ ⇒ E aN 9 12 ⇒ E z= 4 = −(3 × 10 /2 × 8.85 × 10 )ˆaz z=1 = −(6 × 10 9 /2 × 8.85 × 10 12 )ˆ ⇒ E az 9 12 z=4 = −(−8 × 10 /2 × 8.85 × 10 )ˆ ⇒ E az 9 12 ⇒ E = −(3 × 10 /2 × 8.85 × 10 )ˆ az − (6 × 10 9 /2 × 8.85 × 10 12 )ˆ az + (8 × 10 9 /2 × 8.85 × 10 12 )ˆ az 9 12 ⇒ E = −(1 × 10 /2 × 8.85 × 10 )ˆ az = −56.5ˆaz V /m −
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contributed by the surface charge at (b). P B (4, 2, −3) ,location of point P B (4, 2, −3) is suggesting that the E ’s contributed by the other surfaces will be in -ˆaz direction z = −4 will be in ˆaz direction and all other E = (3 × 10 9 /2 × 8.85 × 10 12 )ˆaz − (6 × 10 9 /2 × 8.85 × 10 12)ˆaz + (8 × 10 9 /2 × 8.85 × 10 12)ˆaz ⇒ E = (5 × 10 9 /2 × 8.85 × 10 12 )ˆ ⇒ E az = 282.4ˆ az V /m −
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(c). P C (−1, −5, 2), using the similar arguments as in part(b) about the location of the given point we can the −iveˆ notice easily that the surface at z = 4 will produce E in az direction and all other surfaces will produce E in the +iveˆ az direction ⇒ E = (3 × 10 9 /2 × 8.85 × 10 12 )ˆaz + (6 × 10 9 /2 × 8.85 × 10 12)ˆaz + (8 × 10 9 /2 × 8.85 × 10 12)ˆaz = = (17 × 10 9 /2 × 8.85 × 10 12 )ˆ ⇒ E az 960.45ˆ az V /m −
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(d). P D (−2, 4, 5), using the similar arguments as in part(b) about the location of the given point we can in the +iveˆ notice easily that all the surfaces will produce E az direction 9 12 9 ⇒ E = (3 × 10 /2 × 8.85 × 10 )ˆ az + (6 × 10 /2 × 8.85 × 10 12 )ˆ az − (8 × 10 9 /2 × 8.85 × 10 12 )ˆ az = 56.5ˆ az V /m −
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= (−8x/y)ˆ y /E x ⇒ dy/dx = (4x2 /y 2 )/(−8x/y) ⇒ D2.7(a). E ax + (4x2 /y 2 )ˆ ay , P (1, 4, −2), = E we have dy/dx dy/dx = (−x/2y) ⇒ 2ydy = −xdx ⇒ 2 ydy = − xdx ⇒ 2 × y2 /2 = −x2 /2 + c2 /2 ⇒ y2 = −x2 /2 + c2 /2 ⇒ (y2 = −x2 /2 + c2 /2)P (1,4, 2) ⇒ c2 = 33 ⇒ x2 + 2y 2 = 33 −
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y /E x = x/y(5x +1) ⇒ ydy = xdx/(5x +1) ⇒ ydy = xdx/(5x +1) ⇒ y2 /2 = xdx/(5x +1)+ c2 /2 (b). dy/dx = E put (5x + 1) = t ⇒ 5dx = dt also (5x + 1) = t ⇒ x = (t − 1)/5 ⇒ xdx/(5x + 1) = ((t − 1)/5)dt/5t = (1/25) ((t − 1)/t)dt = (1/25)( dt − dt/t) = (1/25)(t − ln | t |) = (0.04t − 0.04ln | t |)(t=5x+1) = 0.04(5x + 1) − 0.04ln | (5x + 1) |⇒ xdx/(5x + 1) = 0.04(5x + 1) − 0.04ln | (5x + 1) | now we have y 2 /2 = 0.04(5x + 1) − 0.04ln | (5x + 1) | +c2 /2 ⇒ y2 /2 = 0.04(5x + 1) − 0.04ln | (5x + 1) | +c2 /2)P (1,4, 2) ⇒ c2 = 15.66 ⇒ y2 = 0.04(5x + 1) − 0.04ln | (5x + 1) | +(15.66)/2 ⇒ y2 = 15.74 + 0.4x − 0.08ln | (5x + 1) |
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THE END
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