Proton Distance Help: Find After .5s

[matt72lsu]

Homework Statement

A proton is released from rest at t=o in a uniform electric field whose magnitude is 2.5e-8 N/C. How far has the proton traveled after .5 s?

Homework Equations

x = vot + 1/2at^2

The Attempt at a Solution

x = 1/2(9.81)(.5)^2 = 1.2 but it says the answer is .3m. I'm prety sure I'm using the correct equation so I don't know what is wrong.

 

[pgardn]

Homework Statement

A proton is released from rest at t=o in a uniform electric field whose magnitude is 2.5e-8 N/C. How far has the proton traveled after .5 s?

Homework Equations

x = vot + 1/2at^2

The Attempt at a Solution

x = 1/2(9.81)(.5)^2 = 1.2 but it says the answer is .3m. I'm prety sure I'm using the correct equation so I don't know what is wrong.

Fe = q*E = ma Solve for acceleration, a, first. Then do your kinematics. The proton is accelerating in an electric field, I am assuming you don't have to worry about g... Have you not been studying E fields in your class? Remember what you are studying in class and then apply to the "old" ways.

 

[matt72lsu]

thanks!

 

[pgardn]

thanks!

happy day

 

[rdl]

I would first check the units of the given electric field magnitude, which is in N/C. This unit can also be written as V/m, indicating that it is a measure of electric field strength. The equation you used, x = vot + 1/2at^2, is the equation for calculating displacement in a constant acceleration scenario, such as free fall. However, in this case, the proton is experiencing a constant force in the form of an electric field, which results in a constant acceleration. Therefore, the correct equation to use is x = 1/2at^2, where a is the acceleration due to the electric field, which can be calculated using F=ma.

In this case, the proton has a mass of 1.67e-27 kg, and the given electric field has a magnitude of 2.5e-8 N/C. Plugging these values into F=ma, we get an acceleration of 1.5e-19 m/s^2.

Now, using the equation x = 1/2at^2, we can calculate the distance traveled by the proton after 0.5 seconds as x = 1/2(1.5e-19)(0.5)^2 = 1.9e-20 meters. This is equivalent to 0.19 nanometers, or 0.3m as stated in the answer. Therefore, the correct answer is indeed 0.3m, and the discrepancy in your calculation may be due to using the incorrect equation or not considering the units of the given electric field magnitude.