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Physics 15 Online
OpenStudy (anonymous):

The diagram shows an electron, with charge e, mass m, and velocity v, entering a uniform electric field of strength E. http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Physics%20%289702%29/9702_s09_qp_1.pdf Scroll down to Question No.) 29 The direction of the field and the electron’s motion are both horizontal and to the right. Which expression gives the distance x hrough which the electron travels before it stops momentarily? the options and diagram is given in the link above the answer is D but I want to know why is it so?

OpenStudy (anonymous):

Well.. thats simple 1D kinematics whats the acceleration (retardation in this case) experienced by the electron?

OpenStudy (anonymous):

there is no number given

OpenStudy (anonymous):

tell me the expression!

OpenStudy (anonymous):

uhm that's the thing, I don't why they are using kinetic energy in this

OpenStudy (anonymous):

forget about kinetic energy.. :P just answer my question.. what is the acceleration ? (in terms of E, e, and m)

OpenStudy (anonymous):

I looked through my book: the only thing that is coming in my mind is E=F/Q or V/d I don't know how you relate that with F=ma though :/

OpenStudy (anonymous):

you already wrote two equations E = F/q and F = ma using these equations can't u relate them? :D

OpenStudy (anonymous):

*hint eliminate F * shhhhshhhhhhh..

OpenStudy (anonymous):

uhm.....F= E/Q aaand F= Ma soooo......

OpenStudy (anonymous):

sooooooo... ?

OpenStudy (anonymous):

wait.. F is NOT E/q E = F/q

OpenStudy (anonymous):

wait I'm looking for distance right?

OpenStudy (anonymous):

no u are looking for acceleration.. !!! plz focus here.. ll walking you through this.. but i can't help you if you don't follow me step by step :O

OpenStudy (anonymous):

I'm trying >___<

OpenStudy (anonymous):

so elimate F from the two equations here E = F/Q => F = EQ and F = ma so what is a ???

OpenStudy (anonymous):

so acceleration is Force/ Mass right?

OpenStudy (anonymous):

If that's what we are looking for right?

OpenStudy (anonymous):

i want a in terms of (E, m and Q).. using those two equations..

OpenStudy (anonymous):

so If I make F as the subject in the E= F/Q formula the F is E x Q over mass?

OpenStudy (anonymous):

or may be I just screwed it up

OpenStudy (anonymous):

T_T

OpenStudy (anonymous):

WHAT?!.. JUST EQUATE THE TWO EQUATIONS.. :O .. u are telling me you can't do that?!? :O

OpenStudy (anonymous):

F= ma F = EQ THEREFORE ma = EQ NOW WHAT IS 'a' ????

OpenStudy (anonymous):

if you don't answer this right.. ll kil you!

OpenStudy (anonymous):

EQ/m!!!!

OpenStudy (anonymous):

Okay! Okay!

OpenStudy (anonymous):

=_=

OpenStudy (anonymous):

THERE.. was that so hard? :P now Q = e therefore a = Ee/m now the electron is going to the right.. but which direction does it experience the force.. hence which direction is the acceleration? left or right?

OpenStudy (anonymous):

it's going along in the direction of the electric field so it's right?

OpenStudy (anonymous):

uh-oh don't tell me I got it wrong agian??? o_O

OpenStudy (anonymous):

if that is the case.. then it should speed up .. and keep going.. why would it ever stop and turn back?

OpenStudy (anonymous):

so there is a force opposing it to make it stop

OpenStudy (anonymous):

yes.. so acceleration is directed towards ...... ?

OpenStudy (anonymous):

left

OpenStudy (anonymous):

ok so we can call it negative now u know initial velocity = v u know acceleration = -eE/m u know final velocity = 0 (cause it stops) can u calculate the distance? x ?

OpenStudy (anonymous):

do we use one of the equation of motion like this one? v^2 = u^2 + 2as? (where u is the initial velocity and s is the displacement)

OpenStudy (anonymous):

yea.. precisely!

OpenStudy (anonymous):

*applauds* :P

OpenStudy (anonymous):

=_='''' thank you miss genius

OpenStudy (anonymous):

o.O .. u now made the biggest mistake of your life in your final reply i dunno who you are, but m gonna find you , m gonna hunt you.. and i WILL KILL YOU!

OpenStudy (anonymous):

what?? lol hahaha I complemented you >xD

OpenStudy (anonymous):

i meant the 'miss' part ITS MR. now did you get the answer?

OpenStudy (anonymous):

ohhh Ooops sorry sir ^.^'

OpenStudy (anonymous):

uhm......all my friends and me are dumb in maths (hence physics) so I didn't get it when you said "equate two equations" it's a utter shame (I know lol) but who do you equate two equations?

OpenStudy (anonymous):

*what do you mean by saying equate

OpenStudy (anonymous):

meaning eliminating a variable liek if x = 2y and y = 3z then i can elimnate y x = 2(3z) => x = 6y

OpenStudy (anonymous):

btw.. did you get the answer? how many times should i ask :(

OpenStudy (anonymous):

I got it don't worry, don't worry :D

OpenStudy (anonymous):

x = 6z * sorry :D

OpenStudy (anonymous):

now i wanted to tell you how to do it using energy.. u wanna know? or you are fine with acceleration method? :P

OpenStudy (anonymous):

Oooohh I see

OpenStudy (anonymous):

oh I don't mind please go ahead :)

OpenStudy (anonymous):

well.. u use the concept of conservation of energy.. |dw:1396706815519:dw| what i have drawn is equipotential line when the electron moves down the field ( its moving towards the lower potential region) since E = V/d V = Ed and since i already took intiially as V = 0, the line to the right x distance away is at a potential -Ex so far so good?

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