# R 6 6 12 3 20 10 50 10 11 4 9 4 π48 x y z a a a

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R = 6 6 12 3 ( 20 10 )(50 10 ) (11 4 9 ) 4 π(8.854 10 )(14.7648) x y z a a a = ( 2.7923 × 10 3 )(11 a x + 4 a y 9 a z ) = 30.72 a x + 11.169 a y 25.13 a z mN D2.2. A charge of –0.3 μC is located at A (25, –30, 15) (in cm), and a second charge of 0.5 μC is at B ( 10, 8, 12) cm. Find E at: ( a ) the origin; ( b ) P (15, 20, 50) cm. ( a ) Solving first the vector and magnitude from charge A to the origin O , R AO = (0 25) a x + (0 ( 30)) a y + (0 15) a z cm = 25 a x + 30 a y 15 a z cm = 0.25 a x + 0.3 a y 0.15 a z m | R AO | = 2 2 2 ( 0.25) 0.3 ( 0.15)   m = 0.4183 m E A = 0 2 4 π ϵ A AO AO Q R a = 2 A AO AO AO Q R R R = 0 3 4 π ϵ A AO AO Q R R 0 4 π ϵ
5 = 6 3 0 0.3 10 ( 0.25 0.3 0.15 ) 4 π ϵ (0.4183) x y z a a a V/m = 9.21 11.05 5.53 x y z a a a kV/m For the charge B to the origin, R BO = (0 ( 10)) a x + (0 8) a y + (0 12) a z cm = 10 a x 8 a y 12 a z cm = 0.1 a x 0.08 a y 0.12 a z m | R BO | = 2 2 2 0.1 ( 0.08) ( 0.12)     m = 0.1755 m E B = 0 3 4 π ϵ B BO BO Q R R = 3 0 6 0.5 10 (0.1 0.08 0.12 ) 4 π ϵ (0.1755) x y z a a a V/m = 83.13 66.51 99.76 x y z a a a kV/m Combining E A and E B , E = 92.3 a x 77.6 a y 94.2 a z kV/m ( b ) Same concept in ( a ) R AP = (15 25) a x + (20 ( 30)) a y + (50 15) a z cm = 0.1 a x + 0.5 a y + 0.35 a z m | R AP | = 0.6185 m E A = 1.14 5.7 3.99 x y z a a a kV/m R BP = (15 ( 10)) a x + (20 8) a y + (50 12) a z cm = 0.25 a x + 0.12 a y + 0.38 a z m | R BP | = 0.4704 m E B = 10.79 5.18 16.41 x y z a a a kV/m E = 11.9 a x 0.52 a y 12.4 a z kV/m D2.3. Evaluate the sums: ( a ) 5 2 0 1 ( 1) ; 1 m m m   ( b ) 4 2 1.5 1 (0.1) 1 . (4 ) m m m ( a ) 0 1 2 3 4 5 2 2 2 2 2 2 1 ( 1) 1 ( 1) 1 ( 1) 1 ( 1) 1 ( 1) 1 ( 1) 0 1 1 1 2 1 3 1 4 1 5 1             = 2.52 ( b ) 1 2 3 4 2 1.5 2 1.5 2 1.5 2 1.5 (0.1) 1 (0.1) 1 (0.1) 1 (0.1) 1 (4 1 ) (4 2 ) (4 3 ) (4 4 ) = 0.176 D2.4. Calculate the total charge within each of the indicated volumes: ( a ) 0.1 ≤ | x |, | y |, | z | ≤ 0.2: ρ v = 1/( x 3 y 3 z 3 ); ( b ) 0 ≤ ρ ≤ 0.1, 0 ≤ ϕ ≤ π, 2 ≤ z ≤ 4; ρ v = ρ 2 z 2 sin 0.6 ϕ ; ( c ) universe: ρ v = e 2 r / r 2 . ( a ) The differential volume of the rectangular coordinates is dv = dx dy dz . Using the formula, Q = v vol ρ dv = 0.2 0.2 0.2 3 3 3 0.1 0.1 0.1 1 dx dy dz x y z    = 0.2 0.2 0.2 3 3 3 0.1 0.1 0.1 x dx y dy z dz = 0.2 0.2 0.2 2 2 2 0.1 0.1 0.1 2 2 2 x y z = 112.5 C Notice that the charge Q is a large number. Neglecting the rate of change of the cubical volume, we have xyz = (0.1) 3 and xyz = (0.2) 3 which are very small; then the volume Δ v is nearest to 0. Yielding, Q = ρ v Δ v = [1/( x 3 y 3 z 3 )] 0 = 0 ( b ) The differential volume of the cylindrical coordinates is dv = ρ dρ dϕ dz . Using the formula, Q = 0.1 π 4 2 2 0 0 2 ( sin0.6 ) ρ z ρ dρ d dz    = 0.1 π 4 3 2 0 0 2 sin0.6 ρ dρ d z dz = 0.1 4 π 4 3 0 0 2 cos0.6 4 0.6 3 ρ z = 4 3 3 0.1 cos0.6(180 ) cos0.6(0 ) 4 2 0 4 0.6 0.6 3 3 = 1.018 mC ( c ) The differential volume of the spherical coordinates is dv = r 2 sin θ dθ dϕ dr . Assuming this universe to be a perfect sphere, we have limits as 0 ≤
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