Tackling a Tough Physics Problem: Proton's Turning Point

[spatel600]

The surface charge density on an infinite charged plane is - 2.30 *10^-6 A proton is shot straight away from the plane at 2.20*10^6m/s .



How far does the proton travel before reaching its turning point?

I've tried this problem a few times and got 3 different answers...any suggestions?



Thanks so much for your help.

 

[chroot]

Find the E field due to the charged plane. Use the Lorentz force law to find the force on the proton due to this field. Use Newton's second law to find the acceleration due to the force. Use kinematics to find the zero of velocity.

- Warren

 

[spatel600]

We haven't covered Lorentz force law.

Any more tips?

Thanks a bunch.

 

[chroot]

The Lorentz force law is:

$$\mathbf{F} = q \mathbf{E} + \mathbf{v} \times \mathbf{B}$$

When the magnetic field is zero, it reduces to just

$$\mathbf{F} = q \mathbf{E}$$

Given the electric field strength, all you need to do to find the force on a particle is multiply by the particle's charge.

- Warren

 

[spatel600]

But what is the E field...its not given...


Any input?


Thanks.

 

[spatel600]

Got it...thanks!