Maximum Horizontal Force of Relativistic Point Charge

AI Thread Summary
The discussion focuses on determining the angle θ that maximizes the horizontal force on a stationary charge due to a moving charge. The electric field equation for a stationary charge is transformed using Lorentz transformations to account for the moving reference frame. An attempt to derive the angle led to an incorrect expression for sin θ, which was identified as a mistake. The correct expression for sin θ is given as sqrt(2 / (3 - β^2)), with both limiting cases of β = 1 and β = 0 yielding the same result. The conversation emphasizes the importance of careful calculations in deriving the correct solution.
QuantumDuck23
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Homework Statement


A charge q1 is at rest at the origin, and a charge q2 moves with speed βc in the x-direction, along the line z = b. For what angle θ shown in the figure will the horizontal component of the force on q1 be maximum? What is θ in the β ≈ 1 and β ≈ 0 limits? (see image)

Homework Equations


Equation for the electric field of a stationary point charge: Q/(4*pi*ε*R^2)
Lorentz transformations

The Attempt at a Solution


Starting out with the equation for the electric field of a stationary point charge, I used the Lorentz transformations to transform the electric field expression to the reference frame moving with βc and multiplied that expression by z / ((γx)^2 + z^2)^(1/2) (equivalent to cos θ) to account for the horizontal component of the electric field. I took the derivative of this electric field expression to obtain the x-value for which the horizontal electric field was at a maximum, which was at x = b/(sqrt(2) * γ), and I rewrote this value in terms of sin θ (since the problem is asking for it in terms of θ). My final answer was sin θ = sqrt(2γ/(2γ + 1)), but that does not match up with the actual solution, which is sin θ = sqrt(2 / (3 - β^2), although for the limiting cases of β = 1 and β = 0 they are identical.
 

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QuantumDuck23 said:
which was at x = b/(sqrt(2) * γ)
OK. Note that γ is not inside the square root.

and I rewrote this value in terms of sin θ (since the problem is asking for it in terms of θ). My final answer was sin θ = sqrt(2γ/(2γ + 1))
This isn't correct. I think you might have made a careless error here. Check your work.
 
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TSny said:
OK. Note that γ is not inside the square root.

This isn't correct. I think you might have made a careless error here. Check your work.

Found the error. Thank you.
 
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