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Homework+Ch+21

# Homework+Ch+21 - 21.34 y x Q1 Q2 E1 E2 E3 E4 Q3 Q4 E = E1...

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21.34 Q1 Q2 Q3 Q4 E1 E3 E2 E4 y x ~ E = ~ E 1 + ~ E 2 + ~ E 3 + ~ E 4 Since Q2 = Q3 = Q3, and we are looking the origin which is equal distance to the four charges, by symmetry E 2 and E 3 will cancel each other out. ~ E = ~ E 1 + ~ E 4 = kQ 1 r 2 - kQ 4 r 2 = k Q 1 - Q 2 ( 2 s 2 ) 2 = 2 k ( Q 1 - Q 2 ) s 2 E = (2)(9 × 10 9 )(38 . 6 - 27 . 0) × 10 - 6 0 . 525 2 = 7 . 58 × 10 5 N/C The direction of the field is towards Q1. 21.38 (a) ~ E = ~ E Q 1 + ~ E Q 2 ~ E Q 1 = kQ l 2 ( - ˆy ) 1
~ E Q 2 = kQ l 2 ( - cos 30 ˆx - sin 30 ˆy ) = - 3 kQ 2 l 2 ˆx - kQ 2 l 2 ˆy E = - 3 kQ 2 l 2 ˆx - 3 kQ 2 l 2 ˆy We are essentially done. To find a magnitude for the field: E = q E 2 x + E 2 y = 3 kQ l 2 And its direction is: θ = arctan( E y E x ) = arctan( 3) = 240 (b) ~ E = ~ E Q 1 + ~ E Q 2 ~ E Q 1 = kQ l 2 ( - ˆy ) ~ E Q 2 = kQ l 2 (cos 30 ˆx + sin 30 ˆy ) = 3 kQ 2 l 2 ˆx + kQ 2 l 2 ˆy E = 3 kQ 2 l 2 ˆx - kQ 2 l 2 ˆy Magnitude: E = q E 2 x + E 2 y = kQ l 2 Direction: θ = arctan( E y E x ) = arctan( - 1 3 ) = 330 2

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21.52 (a) E = 1 4 π 0 Z b a dQ r 2 = λ 4 π 0 Z b a dy r 2 (cos θ ˆx - sin θ ˆy ) = λ 4 π 0 Z b a x sec 2 θdθ r 2 (cos θ ˆx - sin θ ˆy ) = λ 4 π 0 Z b a x sec 2 θdθ ( x/ cos θ ) 2 (cos θ ˆx - sin θ ˆy ) = λ 4 π 0 x Z b a (cos θ ˆx - sin θ ˆy ) = λ 4 π 0 x (sin θ ˆx + cos θ ˆy ) | b a
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