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Electric force Submitted by: I.D. 061110185 The problem: Given 4 charges arranged at the corners of a square with sides of a length 2a (as on the picture) find the force acting on the charge Q at some point on the x-axis. 2 sign(x) ŷ. Show that for |x| a F~ = 3Qqa πε0 · x4 The solution: Coulomb law F~ = kq1 q2 k = ~r |r|3 (1) c2 1 ε0 = 8.85 · 10−12 4πε0 N m2 (2) We use the superposition principle and find the forces acting from each one of the charges. We define ~ri as a vector pointing from the charge qi to the charge Q. Then ~r1 = (x − a)~i − aĵ ~r2 = (x + a)~i − aĵ (3) ~r3 = (x + a)~i + aĵ ~r4 = (x − a)~i + aĵ (5) (4) (6) where the the charges qi are counted from the top right one in the counterclockwise direction. Therefore, the force is F~ = F~(x,0,0) = F~1 + F~2 + F~3 + F~4 = (7) ! (x + a)~i − aĵ (x + a)~i + aĵ (x − a)~i + aĵ (x − a)~i − aĵ + − + 3/2 3/2 3/2 2 2 2 2 2 2 ((x − a) + a ) ((x + a) + a ) ((x + a) + a ) ((x − a)2 + a2 )3/2 1 1 ĵ (8) = 2akqQ − + ((x + a)2 + a2 )3/2 ((x − a)2 + a2 )3/2 = kqQ − For |x| a 1 ((x ± a)2 + a2 )3/2 = ' 1 1 = 2a 3 (x2 ± 2ax + 2a2 )3/2 |x| (1 ± x + 3a 1 1 ' 1∓ 3/2 |x|3 x |x|3 (1 ± 2a x) 1 2a2 3/2 ) x2 (9) (10) Then 3Qqa2 sign(x) · ŷ F~ = πε0 x4 (11) which is the the force acting by a quadrupole. 2 Electric force The problem: An infinite wire charged uniformly with the charge density λ is bent as shown on the picture. What is the force acting on a charge q which is put into the center of the half-circle? The solution: We divide the problems to 2 parts: (1) finding the force acting from the straight line and (2) finding the force acting from the half circle. The infinidecimal charge is dq = λdl. For the straight line: ~r0 = xî + Rĵ kQdq kQλxdx = − 3 x = − 3 (x2 + R2 ) 2 (x2 + R2 ) 2 kλRdx = − 3 (x2 + R2 ) 2 Z∞ kQλxdx 2λQk = −2 3 = − R (x2 + R2 ) 2 ~r = 0 , dFx dFy Fx (1) (2) (3) (4) 0 where the last integral could be solved (for example) by the substitution x = cos α. The ycomponents of the force are canceled since there are 2 wires, and the factor of 2 in the last equation also comes from this fact. For the half circle: ~r0 = R cos θî + R sin θĵ kQdq kQλR2 dθ dFx = R cos θ = cos θ R3 R3 2λQk Fx = R (5) (6) (7) Therefore, X F~ = 0 (8) 1 Electric field - semicircle Submitted by: I.D. 061110185 The problem: A semicircle of the radius R, 0 < θ < π, is charged with the charge Q. 1. Calculate the electric field at the center if the charge distribution is uniform. 2. Let the charge density be λ = λ0 sin θ. Find λ0 and and the electric field at the center. The solution: 1) λ - homogeneous kdq sin θ R2 dq = λdl = λRdθ dEy = − (1) (2) The electric field ~ = E Zπ dEy = − 2kλ ŷ R (3) 0 The charge density is Z Z Q Q = dq = λRdθ =λRπ ⇒ λ = Rπ (4) So that ~ = − 2kQ ŷ E πR2 (5) 2) λ = λ0 sin θ Z Q = λ0 = Z dq = λ0 R sin θdθ = 2λ0 R (6) Q 2R (7) dq kλ0 dEy = −k 2 sin θ = − 2 sin2 θdθ R R Z kλ0 π Ey = dEy = − 2 R 2 kQπ ~ = − E ŷ 4R2 (8) (9) (10) 1