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

[wcjy]

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

 

[ehild]

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?

 

[wcjy]

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