Question

A proton is released from rest in a uniform electric field of magnitude \(1.08 \times 10^5 \frac{N}{C}\) . Find the speed of the proton after it has traveled \(a. ~ 1.00 ~ cm~,~ b. ~ 10.0~cm\)

Answer 1

Ummm, energy methods?

Answer 2

you can use the equation
F = q*E

Answer 3

I know that in finding the speed of a proton, given that it starts at \(V_i = 0 ~m/s\) would be \[v_f^2 = v_i^2 +2a\Delta x\]

Answer 4

How many C is a proton? Multiply that to get N, which is a measure of force. Multiply that by each distance to get the work. Then use kinetic energy. \[
Fd = \frac 12 mv^2
\]

Answer 5

substitute that for force

Answer 6

Oh, because the electric field is \(\vec E =\dfrac{\vec F}{q}\)?

Answer 7

yes

Answer 8

I was just using dimensional analysis to guess that \(q\mathbf E =\mathbf F\).

Answer 9

Ohh..

Answer 10

Because \(\frac NC \cdot C = N\) and we wanted \(N\).

Answer 11

Ohh, I wasn't familiar with units.

Answer 12

Now it makes a little more sense.

Answer 13

then... \[\sum F = m\vec a = \vec F q \] \[\vec a = \frac{\vec Fq}{m}\]

Answer 14

And i'm guessing the units would cancel out somehow giving us... m/s^2?

Answer 15

Oh, mistake.

Answer 16

\[\vec a = \frac{\vec E q}{m} = \frac{\frac{N}{C} \cdot C}{kg} = \frac{\dfrac{kg\cdot m}{s^2}}{kg}=\frac{m}{s^2}\]

Answer 17

Hmm..

Answer 18

\[ v_f^2 = v_i^2 +2 a \Delta x~,~ \Delta x = 0.0100 ~m\]\[v_f =\sqrt{v_i^2 +2\left(\frac{\vec Eq}{m}\right)(0.0100~m)}\]

Answer 19

Right? @wio

Answer 20

Yeah

Answer 21

Alright, and... um..

\(\vec E\) is given. But what is q?...

Answer 22

The charge of a proton...

Answer 23

Oh, haha the electrical charge.... ~,~ \(1.08 \times 10^{-19} C\)... Duh.

Answer 24

Thanks!

Answer 25

Hmm, yeah. Be glad the field was uniform

Answer 26

What if it wasn't? What would you do then?

Answer 27

The same thing but calculate each part? My guess.

Answer 28

Then you'd have to know what you're doing, lol

Answer 29

If field varied with respect to your distance from it, like a point charge, then you'd have to integrate to find the work done.

Answer 30

I see.