How Does a Proton's Velocity Change in a Uniform Electric Field?

[Haniszmi]

Homework Statement

1. A uniform electric field is directed towards the east with a magnitude of 10 N/C. A proton, moving with a speed of 10^5 m /s, enters the field traveling parallel to it. The proton covers a distance of 4 cm while in the field. Find the proton’s final velocity as it leaves the field.

Homework Equations

F=(k)(q1)(q2)/r, F=mv^2/r, Ac=v^2/r

The Attempt at a Solution

I don't understand how to attempt this problem. Confused about the 4cm, not sure if where to use it. How does the proton's velocity change when it enters an electric field?

 

[LowlyPion]

1 Newton / coulomb = 1 volts / meter

And W = q*ΔV

If you know how far it went in the field (.04 m) * 10 v/m that's how much work was exerted on increasing its kinetic energy isn't it since it was || to its direction of motion?

 

[nlightner]

I can understand your confusion with this problem. It involves understanding the behavior of charged particles in an electric field. To solve this problem, we can use the equation F=qE, where F is the force experienced by the proton, q is its charge, and E is the electric field strength. In this case, the force experienced by the proton is equal to its mass (m) multiplied by its acceleration (a). We can also use the equation F=ma to relate these two equations.

Since the proton is moving parallel to the electric field, it will experience a force in the same direction as the field. This means that the proton will accelerate in the direction of the field and its final velocity will be greater than its initial velocity.

To find the final velocity, we can use the equation F=ma and substitute qE for F. This gives us ma=qE, or a=qE/m. We can then use the equation v^2=u^2+2as, where u is the initial velocity, a is the acceleration, and s is the distance covered. In this case, u=10^5 m/s, a=qE/m, and s=4 cm. We can convert 4 cm to meters by dividing by 100, giving us s=0.04 m.

Substituting these values into the equation, we get v^2=(10^5)^2+2(qE/m)(0.04). We know that the electric field strength is 10 N/C and the mass of a proton is approximately 1.67x10^-27 kg. Substituting these values, we get v^2=(10^5)^2+2(1.6x10^-19 C)(10 N/C)/(1.67x10^-27 kg)(0.04). Solving for v, we get v=2.83x10^5 m/s. This is the final velocity of the proton as it leaves the electric field.

I hope this explanation helps you to understand how to approach and solve problems involving electric fields. It is important to carefully consider the given information and use the appropriate equations to find the solution. Remember to always double check your units and convert them as needed. Good luck with your future studies in electric fields!