Question

I need someone to check/ tell me whats wrong with my work.
Q: a: An electron and a proton are each released from rest in a constant electric field E=150 N/C x. In what direction does each charge accelerate? Explain. You should ignore the effect of the electron and proton on each other.
b: Determine the distance each charge moves in the first 1 microseconds and the velocity of each charge at the end of the 1 microseconds.

Answer 1

@imqwerty, I'd explain but I have to go.

Answer 2

a: proton +x direction. electron -x direction

Answer 3

(I still working on the problem so don't spoil anything for me)
b: electron:
\[F_{q}=qE\]
\[F_{q}=(-1.60*10^{-19} C)(\frac{N}{C})\]
\[F_{q}=-2.9*10^{-17}N\]

Answer 4

i think this is similar to th elast semester...except the force is now from a field and not gravity

Answer 5

F=mass*accel

Answer 6

\[F=m*a\]
\[a=\frac{-2.9*10^{-17}N}{9.11*10^{-31}kg}\rightarrow \frac{-2.9*10^{-17}m}{9.11*10^{-31}s^{2}}\]

Answer 7

\[V=0+\frac{-2.4*10^{-7}m}{9.11*10^{-31}s^{2}}*1*10^{-6}s\]
\[V=-2.63*10^{7}\frac{m}{s}\]
here this is better still wrong though

Answer 8

i like diagrams

Answer 9

would you like me to draw a FBD or something?

Answer 10

\(\large E_1 = k_e \dfrac{|q_1|}{r_1^2} = k_e \dfrac{|q_1|}{x_1^2+y_1^2}\)

Answer 11

ye can you draw it for me?

Answer 12

whoa wait you lost me.

Answer 13

see if we can provide the proper components the proton and electron from field, E, at (0, y) we can determine the charge and direction ye?

Answer 14

right.

Answer 15

so we are trying to come up with a diagram we can use as a reference and I want to make sure that we both are starting properly or if you already have a diagram, I'd like to look at it.

Answer 16

ok it took me a bit to see what you were getting, but i'm there now

Answer 17

I just mocked up by reflex, I am not even sure if this is how we should start yet. There are things to consider

Answer 18

For positive charges like protons, the field lines are radially outward; and for negative point charges like electron the field lines are directed radially inward

Answer 19

we can derive the velocity from motion of a charged particle in a uniform electric field
\[\large \bar F_e = q\bar a \]

Answer 20

yes I have this draw on my paper