A charged wire exerts a force on a proton.

[joe41442]

Homework Statement

An electrically charged wire on the z axis exerts a force on a proton, which moves on the x axis. The initial conditions for the proton position and velocity are:
x(0) = x0 and v(0) = 0.

The force on the proton is Fx(x) = C /x where C = 3.2×10−15 Nm.

(A) Determine the potential energy function, U(x). Choose x0 to be the reference point; i.e., U(x0) = 0.
(B) Calculate the velocity v at the time when the proton passes the point x = 2 x0.
(C) Calculate the time t when the proton passes the point x = 2 x0. Assume x0 = 1 m.

Homework Equations

E= (1/2)m*v^2 + U(x)
U(x)= -∫F(x)dx
dx/dt= ±√[(2/m)(E-U(x))]

The Attempt at a Solution

In part A I integrated F(x) and got U(x)= -C*ln(x)
When I got to part B I tried to find the energy assuming that since U(x0)=0 then KE(v0)=max=E but KE(v0)=0 meaning that energy would have to be zero and the velocity is an imaginary number so I'm lost on what to do. If someone could just point me in the right direction I think I could figure it out from there.

 

[cepheid]

Welcome to PF,

I'm not sure how you get that the velocity must be an imaginary number.

It's perfectly fine for the particle to start out with zero total energy. It just means that as the force does work on the particle, its kinetic energy increases, and its potential energy decreases by a corresponding amount. In other words, at any time after the start, the kinetic energy is positive, since it increases from 0, and the potential energy is negative, since it decreases from 0. This means that the kinetic and potential energies are always equal in magnitude, and of opposite sign (such that their sum is always zero).

To solve part b, find the change in potential energy between position x₀ and position 2x₀. The negative of this is the change in kinetic energy.

 

[joe41442]

Thanks a lot man! I went back and realized when plugging in for U in dx/dt= ±√[(2/m)(E-U(x))] I kept forgetting to account for the fact that U is negative at x0. Stupid sign mistake!

 

[joe41442]

Do you have any ideas on how to solve part C? I'm trying to solve it by integration but I don't know how to integrate 1/√lnx, I've checked wolframalpha and that only showed how to integrate it unsing an imaginary error function. Is there some other method that I could use?

 

[asteriatic]

I would like to provide some guidance on how to approach this problem.

Firstly, it is important to understand the given initial conditions and the force acting on the proton. The force, Fx(x) = C/x, is an inverse square law force, which means it decreases with distance from the charged wire. This force has a singularity at x=0, which is why the potential energy function, U(x), is negative infinity at this point.

In order to determine the potential energy function, we can integrate the given force equation with respect to x. However, since the force is not defined at x=0, we will have to choose a reference point, x0, to evaluate the potential energy at. This will result in a constant offset in the potential energy function. By choosing x0 as the reference point, we can set U(x0) = 0, which means that the potential energy at this point is zero.

Therefore, the potential energy function, U(x), can be calculated as U(x) = -C*ln(x/x0).

Moving on to part B, we need to calculate the velocity of the proton at the point x=2x0. At this point, the potential energy, U(x=2x0), is -C*ln(2). Since the total energy of the proton is conserved, we can equate the initial kinetic energy (which is zero) to the final potential energy and solve for the velocity, v. This will give us the velocity at the point x=2x0.

Finally, to calculate the time, t, when the proton passes the point x=2x0, we can use the equation dx/dt = ±√[(2/m)(E-U(x))], where E is the total energy of the proton and U(x) is the potential energy function. We already know the values of E and U(x) at this point, so we can solve for t.

I hope this helps in your problem-solving process. Remember to always carefully analyze the given conditions and equations, and approach the problem step by step. Good luck!