Electric field problem using Gauss' law: Point charge moving near a line charge

In summary, the summary for this conversation is that a point charge with a mass of 0.1 kg and a charge of -2 µC is launched from coordinate (2,0) into the plane of the paper. The launch velocity required for the charge to reach coordinate (-2,0) is 1.20 m/s. The problem cannot be solved using the equation v^2 = U^2 + 2as because the charge does not travel in a straight line with constant linear acceleration. The trajectory of the charge would need to be determined using other equations or methods.
  • #1
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Homework Statement
Consider a long line of charge with linear charge density λ = 4 µC/m and a point charge q= −2 µC with mass m = 0.1 kg at coordinate (2,0) at t = 0. The point charge is launched straight into the plane of the paper. What is the launch velocity so that the point charge will reach coordinate (−2,0)?
Relevant Equations
Electric field due to long line of charge, E = λ/(2pi*r*ε0)
Electric field due to point charge, E = Q/(4*pi*ε0*r^2)
V^2 = U^2 +2as
F=qE
F = qE
ma = (2*10^-6) * (λ / (2pi*r*ε0) )
ma = (2*10^-6) * (4*10^-6 / (2pi*4*ε0) ) => I am not certain what to put for r ( But I sub in 4 because dist is 4)
a = ( (2*10^-6) * (4*10^-6 / (2pi*4*ε0) ) )/ 0.1
a = 0.35950

v^2 = U^2 + 2 a s
v = 0
u^2 = -2 a s => Can't sqrt negative so i just removed
u = sqrt( 2 * 0.35950 * 4)
u = 1.6959Answer: 1.20m/s
 

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  • #2
wcjy said:
Homework Statement:: Consider a long line of charge with linear charge density λ = 4 µC/m and a point charge q= −2 µC with mass m = 0.1 kg at coordinate (2,0) at t = 0. The point charge is launched straight into the plane of the paper. What is the launch velocity so that the point charge will reach coordinate (−2,0)?
Relevant Equations:: Electric field due to long line of charge, E = λ/(2pi*r*ε0)
Electric field due to point charge, E = Q/(4*pi*ε0*r^2)
V^2 = U^2 +2as
F=qE

F = qE
ma = (2*10^-6) * (λ / (2pi*r*ε0) )
ma = (2*10^-6) * (4*10^-6 / (2pi*4*ε0) ) => I am not certain what to put for r ( But I sub in 4 because dist is 4)
a = ( (2*10^-6) * (4*10^-6 / (2pi*4*ε0) ) )/ 0.1
a = 0.35950

v^2 = U^2 + 2 a s
v = 0
u^2 = -2 a s => Can't sqrt negative so i just removed
u = sqrt( 2 * 0.35950 * 4)
u = 1.6959Answer: 1.20m/s
The problem states that the charge is launned straight into the plane of the paper. It does mot travel along a straight line with comstant linear acceleration, you can not use the equation v^2 = U^2 + 2 a s. What is the trajectory of the charge?
 
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  • #3
ehild said:
The problem states that the charge is launned straight into the plane of the paper. It does mot travel along a straight line with comstant linear acceleration, you can not use the equation v^2 = U^2 + 2 a s. What is the trajectory of the charge?

solved it already thx
 
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