Proton release inside a parallel capacitor Find velocity

[conov3]

Homework Statement

The electric field strength is 47100 N/C inside a parallel-plate capacitor with 8.60 mm spacing. A proton is released from rest at the positive plate. What is the speed of the proton (in m/s) when it reaches the negative plate?

Homework Equations

Vf=√(-2U/m) I think?
ΔU=-.5eEd

The Attempt at a Solution

I tried using that equation to find ΔU=-3.768*10^-15
then plugged it in seem to have the wrong final answer though

 

[tiny-tim]

welcome to pf!

hi conov3! welcome to pf!

ΔU=-.5eEd

why .5 ?

 

[conov3]

Thank you! I figured since I am in a college level physics, I might as well try to get some help if I need it! ha

Using my Physics book.. it says ΔU=Uf-Ui= (Uo+ (-e)Ed)-(Uo+(-e)E(d/2)
Then under that it says =-1/2eEd

I am confused about it also and am really unsure.

 

[tiny-tim]

Using my Physics book.. it says ΔU=Uf-Ui= (Uo+ (-e)Ed)-(Uo+(-e)E(d/2)

i don't understand that … it doesn't seem to fit the question

 

[asteriatic]

.

I would approach this problem by first identifying the relevant equations and principles that apply to this situation. In this case, we are dealing with an electric field and a charged particle, so we can apply the principles of electrostatics and motion of charged particles in electric fields.

The first step would be to calculate the electric potential difference (ΔV) between the two plates. This can be done using the equation ΔV = Ed, where E is the electric field strength and d is the distance between the plates. In this case, we have E = 47100 N/C and d = 8.60 mm = 0.0086 m. Therefore, ΔV = (47100 N/C)(0.0086 m) = 405.06 V.

Next, we can use the equation ΔV = -ΔU/q, where ΔU is the change in potential energy and q is the charge of the particle. In this case, the proton has a charge of +1.60 x 10^-19 C. Therefore, we can rearrange the equation to solve for ΔU, which gives us ΔU = -qΔV = (-1.60 x 10^-19 C)(405.06 V) = -6.48 x 10^-17 J.

Now, we can use the equation ΔU = -0.5mv^2 to calculate the final velocity (v) of the proton. Rearranging the equation gives us v = √(-2ΔU/m), where m is the mass of the proton. Substituting the values, we get v = √((-2)(-6.48 x 10^-17 J)/(1.67 x 10^-27 kg)) = 1.98 x 10^6 m/s.

Therefore, the speed of the proton when it reaches the negative plate is 1.98 x 10^6 m/s.