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Solution to the Drill problems of chapter 03 (Engineering Electromagnetics,Hayt,A.Buck 7th ed) BEE 4A,4B & 4C ~ =? at point P(2,-3,6) D3.2 (a). D ~ = QR ~ P Q /(4π | R ~ P Q |3 ) ~ = o E QA = 55mC at point Q(-2,3,-6) now D √ p ~ P Q = (2 − (−2))ˆ ~ P Q |= 42 + (−6)2 + 122 = 196 = 14 R ax + (−3 − 3)ˆ ay + (6 − (−6))ˆ az = 4ˆ ax − 6ˆ ay + 12ˆ az , | R ~ = (55 × 10−3 (4ˆ ⇒D ax − 6ˆ ay + 12ˆ az ))/(4π(14)3 ) = 6.38ˆ ax − 9.57ˆ ay + 19.14ˆ az µC/m2 ~ px /(2πo | R ~ px |2 ) ~ = ρL a ~ px |) = ρL R (b). ρL = 20mC along X axis (∞ ≤ X ≤ ∞), now E ˆpx /(2πo | R ~ px = P (2, −3, 6) − (x, 0, 0) = (2 − x)ˆ R ax − 3ˆ ay + 6ˆ az since the infinite line charge is along X axis so the E filel at point P is having only Y and Z components present and the X component is cancelled due to symmetry so √ p ~ px = −3ˆ ~ px |= (−3)2 + 62 = 45, D ~ px /(2π | R ~ px |2 ) = 20 × 10−3 (−3ˆ ~ = ρL R ~ = o E R ay + 6ˆ az , | R ay + 6ˆ az )/2π × 45 2 ~ ⇒ D = −212ˆ ay + 424ˆ az µC/m ~ = (ρs /2o )ˆ (c). E aN for infinite surface charge density also z=-5 is an infinite x-y plane located at z=-5 and the ~ = o E ~ = (ρs /2)ˆ charge is spread on this plane,so D az = (120/2)µˆ az = 60µˆ az C/m2 ~ = 0.3r2 a D3.3. D ˆr nC/m2 ~ =? at point P(r=2,θ = 25o , φ = 90o ), D ~ = o E ~ ⇒E ~ = D/ ~ o = ((0.3r2 a (a). E ˆr nC/m2 )/8.85 × 10−12 F/m)r=2 = 2 2 −12 (0.3 × 2 a ˆr nC/m )/8.85 × 10 F/m = 135.5ˆ ar V /m ~ · d~s, d~s = r2 sinθdθdφ ⇒ Q = 0.3r2 × 10−9 (b).Q= ? for a sphere of radius r=3, we have Q= D 2 2 −9 4 −9 4 ⇒ 0.3r (4πr ) × 10 = 1.2πr × 10 = (1.2πr )r=3 × 10−9 = 305nC H
R 2π R π 2 0 0 r sinθdθdφ
(c). ψ = Q =? for a sphere of radius r=4 follow the same procedure as in part (b) we will get ψ = 1.2πr4 × 10−9 = (1.2πr4 )r=4 × 10−9 = 965nC D3.4. Find total electric flux leaving the cubical surface formed by the six planes at x,y,z=±5 (a). since both the given charges are enclosed by the cubical volume ( as evident by their given locations) according to the gauss’s law ψ = Q1 + Q2 = 0.1µC + (1/7)µC = 0.243µC (b). ρL = πµC at (-2,3,z), clearly this line charge distribution is passing through point x=-2 and y=3 and is parallel to z axis, the total length of this charge distribution enclosed by the given cubical volume is 10 units as z = ±5 so ψ = Q = ρL × 10 = πµ × 10 = 31.4µC (c). ρs = 0.1µC on the plane y=3x, now this is a straight line equation in xy plane which passes through the origin,we need to find the length of this line which is enclosed by the given volume, and once we find it, this length is moving up and down along z axis between z = ±5 to form a plane, by putting y=5 p we get x=5/3 and hence the length of this line on the plane formed by the +ive x and +ive y axis is given by 52 + (5/3)2 = 5.270, the same length we will get on the plane formed by -ive x and -ive y axix and the sum of these two lengths is 10.540,now this straight line is moving between z = ±5 to form a plane whose area is given by 10×10.540 = 105.40, and this plane has a surface charge density ρs = 0.1µC now according to gauss’s law ψ = Qenclosed ⇒ ψ = ρs × (area of the plane)⇒ ψ = 0.1µC × (105.40) = 10.54µC
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D3.5.(a).at r=0.5cm It is very much clear that a sphere of this much radius will enclose only the point charge ~ = (Q/4πr2 )ˆ ~ = (0.25 × 10−6 /4π(0.5 × 10−2 )2 )ˆ 0.25 µC which is located at the origin now D ar ⇒ D ar = 796ˆ ar µC (b). at r = 1.5cm It is again clear that a sphere of this much radius will enclose the point charge 0.25µC which is located at the origin and also the uniform surface charge density of 2 × 10−3 C/m2 which is uniformly distributed over a sphere of radius r = 1 now we have Q1 = 0.25 µC and Q2 = ρs × area of sphere= 2×10−3 C × 4π(1 × 10−2 )2 = ~ = (Q1 + Q2 /4πr2 )ˆ ~ = ((0.25 × 10−6 + 2.513 × 10−6 )/4π(1.5 × 10−2 )2 )ˆ 2.513 × 10−6 C and now D ar ⇒ D ar 2 ~ ⇒ D = 977ˆ ar µC/m (c). It is the same as part(b) the only difference is that now we have r = 2.5 which not only encloses the charges Q1 and Q2 (same as in part(b)) but also encloses Q3 and we can calculate the value of Q3 just like we calculated Q2 but for Q3 calculation we are going to use r = 1.8cm and ρs = −0.6mC/m2 Q3 = ρs × area of sphere= -0.6×10−3 C × 4π(1.8 × 10−2 )2 = −2.5 × 10−6 C and then the rest of the problem is same as part(b) ~ = 0 at r = 3.5cm (d). find ρs at r = 3 to make D now the point is that the uniform surface charge density established at r = 3 should give us a charge Q4 equal to the ~ = (Q1 +Q2 +Q3 +Q4 /4πr2 )ˆ sum of Q1 , Q2 and Q3 but opposite in sign, so that when we calculate D ar it should give us −6 a 0 value, now Q1 +Q2 +Q3 = 0.25×10 C and Q4 = ρs × (area of sphere)r=3 ⇒ −0.25×10−6 C = ρs ×4π(3×10−2 )2 ⇒ ρs = 0.25 × 10−6 C/4π(3 × 10−2 )2 = −28.3µC/m2 D3.6(a). Since we are going to find the total electric Hflux passing through the given rectangular surface in the a ˆz R3R2 4 2 4 2 3 ~ ~ ~ direction at z = 2, so we get ds = dxdyˆ az and now ψ = D · ds ⇒ ψ = 1 0 (8xyz a ˆx + 4x z a ˆy + 16x yz a ˆz ) · dxdyˆ az R R R R ⇒ ψ = 13 02 16x2 yz 3 dxdy = 16z 3 13 ydy 02 x2 dx = (16z 3 × 4 × 8/3)z=2 = 1365pC ~ = o E ~ ⇒E ~ = D/ ~ o = ((8xyz 4 a (b). D ˆx + 4x2 z 4 a ˆy + 16x2 yz 3 a ˆz ) × 10−12 /8.85 × 10−12 )x=2,y=−1,z=3 ~ = 146.4ˆ ⇒E ax + 146.4ˆ ay − 195.2ˆ az V /m (c). Find Q, ∆v = 10−12 m3 ~ x /∂x = ∂(8xyz 4 )/∂x = (8yz 4 )x=2,y=−1,z=3 = −648, ∂ D ~ y /∂y = ∂(4x2 z 4 )/∂y = 0, ∂ D ~ z /∂z = ∂(16x2 yz 3 )/∂z now ∂ D ~ x /∂x + ∂ D ~ y /∂y + ∂ D ~ z /∂z) × ∆v = = (48x2 yz 2 )x=2,y=−1,z=3 = −1728, now charge enclosed in volume ∆v = (∂ D −12 −12 −24 (−648 − 1728) × 10 × 10 = −2.38 × 10 C ~ atPA (2, 3, −1), D ~ = (2xyz − y 2 )ˆ D3.7. (a). Find divD, ax + (x2 z − 2xy)ˆ ay + x2 yˆ az ~ = (∂ D ~ x /∂x + ∂ D ~ y /∂y + ∂ D ~ z /∂z) = (2yz − 2x)x=2,y=3,z=−1 = −10 div D ~ atPB (ρ = 2, φ = 110o , z = −1), D ~ = 2ρz 2 sin2 φˆ (b). Find divD, aρ + ρz 2 sin 2φaφ + 2ρ2 z sin2 φˆ az ~ ~ ~ ~ div D = (1/ρ)∂(ρDρ )/∂ρ + (1/ρ)∂ Dφ /∂φ + ∂ Dz /∂z ~ ρ )/∂ρ = (1/ρ)∂(2ρ2 z 2 sin2 φ)/∂ρ = (1/ρ)(4ρz 2 sin2 φ) = (4z 2 sin2 φ)ρ=2,φ=110o ,z=−1 = 3.532 ⇒(1/ρ)∂(ρD ~ φ /∂φ = (1/ρ)∂(ρz 2 sin 2φ)/∂φ = (1/ρ)(2ρz 2 cos 2φ) = (2z 2 cos 2φ)ρ=2,φ=110o ,z=−1 = −1.532 ⇒(1/ρ)∂ D ~ z /∂z = ∂(2ρ2 z sin2 φ)/∂z = (2ρ2 sin2 φ)ρ=2,φ=110o ,z=−1 = 7.064 ⇒ ∂D ~ = (1/ρ)∂(ρD ~ ρ )/∂ρ + (1/ρ)∂ D ~ φ /∂φ + ∂ D ~ z /∂z = 3.532 − 1.532 + 7.064 = 9.06 ⇒ div D ~ in spherical coordinates as given (c). Part (c) is similar to part (b) but we have to use the formula for div D in the book.
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D3.8. Find volume charge density ρv ~ = (4xy/z)ˆ (a).D ax + (2x2 /z)ˆ ay − (2x2 y/z 2 )ˆ az ~ x /∂x + ∂ D ~ y /∂y + ∂ D ~ z /∂z) ρv = (∂ D ρv = 4y/z + 0 + 4x2 y/z 3 = (4y/z 3 )(x2 + z 2 ) (b). Use the equation of ρv for cylindrical co ordinates,rest is the same as part(a) (c). Use the equation of ρv for spherical co ordinates,rest is the same as part(a) ~ = (6ρ sin(φ/2))ˆ D3.9. D aρ + (1.5ρ cos(φ/2))ˆ aφ H R ~ ~ D · ds = vol ∆· dv sH ~ = H ((6ρ sin(φ/2))ˆ ~ · ds ⇒ D aρ + (1.5ρ cos(φ/2))ˆ aφ ) · (ρdφdzˆ aρ − dρdzˆ aφ ) Hs
s
H
⇒ s (6ρ sin(φ/2)ρdφdz − s (1.5ρ cos(φ/2)dρdz R R R R ⇒ 0π 05 (6ρ sin(φ/2)ρdφdz − 02 05 (1.5ρ cos(φ/2)dρdz (this second surface lies at φ = 0o ) R R R R ⇒ 6ρ2 0π (sin(φ/2)dφ 05 dz − (1.5 cos(φ/2)) 02 ρdρ 05 dz ⇒24 × | −2 cos(φ/2) |π0 × | z |50 −(1.5 cos(φ/2))× | ρ2 /2 |20 × | z |50 ⇒ 24 × 2 × 5 − ((1.5 cos(φ/2)) × 2 × 10)φ=0o = 240 − 15 = 225 ~ φ /∂φ + ∂ D ~ z /∂z ~ ~ ρ )/∂ρ + (1/ρ)∂ D now ∆ · D=(1/ρ)∂(ρ D ~ ρ )/∂ρ = (1/ρ)∂(ρ(6ρ sin(φ/2))/∂ρ = (1/ρ)∂(6ρ2 sin(φ/2)/∂ρ = 12 sin φ/2 ⇒ (1/ρ)∂(ρD ~ φ )/∂φ = (1/ρ)∂((1.5ρ cos(φ/2))/∂φ = (−1.5/2) sin φ/2 ⇒ (1/ρ)∂(D ~ )/∂ρ + (1/ρ)∂ D ~ φ /∂φ + ∂ D ~ z /∂z = 12 sin φ/2 + (−1.5/2) sin φ/2 = 11.25 sin φ/2 ⇒ ∆ · D=(1/ρ)∂(ρD R R π R 2 Rρ5 Rπ R2 R5 vol ∆ · D dv= 0 0 0 (11.25 sin φ/2)ρdφdρdz = 11.25 0 (sin φ/2)dφ 0 ρdρ 0 dz ⇒ 11.25× | −2cos(φ/2) |π0 × | ρ2 /2 |20 × | z |50 = 11.25 × 2 × 2 × 5 = 225
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