Question

An electron has an initial velocity of 6.00 ✕ 106 m/s in a uniform 1.00 ✕ 105 N/C strength electric field. The field accelerates the electron in the direction opposite to its initial velocity.
(b) How far does the electron travel before coming to rest?
m
(c) How long does it take the electron to come to rest?
s
(d) What is the electron's speed when it returns to its starting point?
m/s

Answer 1

Not sure though,

\(\Delta V=Ed\\V_f-V_i=Ed\\\Large d=\frac{V_f-V_i}{E}\)
where your \(V_f=0 m/s\), \(V_i=6.00 ✕ 10^6 m/s\), \(E= 1.00 ✕ 10^5 N/C\)
substitute these values to solve for \(b\)

Answer 2

For letter C.
\(t=d/\Delta v\), where d is your answer from letter b and \(\Delta V=V_f-V_i\)

Answer 3

You know F =qE =ma So, a = qE/m

Then use kinematic equations

Vf^2 = Vi^2 +-2ax

Set Vf equal to zero and solve for x

Vf = -at + Vi

When Vf is zero the electron is at rest, set Vf = 0 then solve for t

The object returns to its starting point when

S= 0 = -1/2at^2 + vt Solve for t then substitute into Vf = - at +Vi

Hope this helps :)

Answer 4

You can answer the first part by considering how the kinetic and potential energies of the electron change - if you know the expressions for kinetic energy and change in potential energy in a uniform electric field. (it's quite simple)

Answer 5

simple i wish

Answer 6

i can do the math if i have he prper formulas

Answer 7

do you remember the formula for gravitational potential energy in a uniform field, like near the surface of the earth when the accn due to gravity is g ?

Answer 8

The formula is very similar in a uniform electric field.

Answer 9

F
G
=
G
mM
r
2

Answer 10

im working on part b

Answer 11

using d= vf-vi/e but its not correct answer

Answer 12

ok - never mind the gravity thing
how do you work out the acceleration of the electron from the numbers given in the question ?

Answer 13

velocity/distance

Answer 14

v/e ?

Answer 15

what is the force on a charge q in an electric field E ?

Answer 16

coulombs law

Answer 17

no, coulombs law gives the force between 2 charges, but we are looking at a single charge in a field E

Answer 18

you have to know what force a charge q feels in a field E to answer this question

Answer 19

its a much simpler expression than coulombs law

Answer 20

e=f/q

Answer 21

well yes, i would have said f = qE, same thing

Answer 22

f=qe

Answer 23

ok, now if a mass m experiences a force f, what is its acceleration, in terms of f and m ?

Answer 24

idk

Answer 25

newtons second law tells you

Answer 26

a=f/m o yea

Answer 27

right !

Answer 28

and in our problem, we know that the force f is qE, like we just discussed above

Answer 29

so, you can put those pieces together and find that a =qE/m for an electron of charge q and mass m in an electric field E

Answer 30

Now you're in business, because that a can be used in your formulae for motion in a straight line under constant acceleration

Answer 31

\[v ^{2}=u ^{2}+2as\] and so on

Answer 32

but how do i get distance
dng im supid

Answer 33

well what do we know ? we know the velocity at the start of the motion and at the end of the distance covered

Answer 34

u is given in the question, and we want v = 0, because that's what 'coming to rest' means

Answer 35

i got that

Answer 36

ok and you can calculate a from the numbers given in the question

Answer 37

so that only leaves s to be worked out from the equation i just put up

Answer 38

whats the s represent?

Answer 39

the distance !

Answer 40

im used to r or d
o

Answer 41

sorry

Answer 42

: ) it's ok

Answer 43

so i use a=f/m to find find aaccelleration mbieng the mass of an electron?

Answer 44

yes

Answer 45

and re work the v^2=u^2+as to find the difference

Answer 46

v= velocity u = uniform field and a =cceloration and s bieng distance

Answer 47

u is initial velocity v is final velocity

Answer 48

aww so i dont use the 1e5 of the field no where?

Answer 49

you need that to calculate f

Answer 50

f being the force, f = Eq, E being the field

Answer 51

whch is udes to find a

Answer 52

yep

Answer 53

o starting to click thanks a bunch

Answer 54

welcome

Answer 55

q bieng the 1.6e-19

Answer 56

yes, that's right, the charge on the electron

Answer 57

e being the 1E5

Answer 58

usually written E for electric field, but yes

Answer 59

tanks again your a genuis and thanks forhelping me understand instead of here your answer

Answer 60

you're welcome, i hope it starts to seem clearer for you - btw i'm nowhere on the genius scale - the needle on the genius detector wouldn't even flicker

Answer 61

i seeing the relationships any way lol

Answer 62

good

Answer 63

well u know the lectic aspect very well

Answer 64

so inorder to rearrange the v2=u2+as it would be 0=6e6^2+1.76e16*?

Answer 65

find s

Answer 66

6e6^2/1.76216 ?

Answer 67

still here?

Answer 68

ok, tell me what you have for a, the electron acceleration, the numerical value

Answer 69

1.756e16

Answer 70

that is correct
so the other number you need is u squared , what do you have for that ?

Answer 71

3.6e13

Answer 72

then divide

Answer 73

also correct
so now , ignoring the minus sign, we have s = u squared/2a
you've got the right numbers so far, so if you plug them in you should get the right value for s

Answer 74

don't forget the factor of 2 from the equation

Answer 75

haha crap i didnt write that down earlier dang it

Answer 76

the 2 tat is

Answer 77

thats what got me du once again you da boom

Answer 78

it should come out at about 1e-3 meters, about a millimetre - does that look about right to you ?

Answer 79

i get 3.16e29

Answer 80

what !!!!!!!!!!! you must have miscalculated lol

Answer 81

3.6e13/(2 x 1.76e16)

Answer 82

3.16e13/2(1.765e16)

Answer 83

its roughly 1e-3

Answer 84

opps parent in wrong order

Answer 85

did it wrong on the the ol t-83

Answer 86

yes i got that

Answer 87

the next part you can get easily cos you know u, v and a, there's another formula that will give you the time from those three

Answer 88

and for the last part there's a trick you can use - you don't need to calculate anything

Answer 89

for partc t=d/v

Answer 90

that would only work for constant speed
how about using v=u+at

Answer 91

and the trick for last part

Answer 92

i'll give you a hint - it's about the total energy of the electron

Answer 93

e=mc^2 ?

Answer 94

its like the first problem except we find final instead of intal

Answer 95

the total energy is conserved

Answer 96

right isnt lost

Answer 97

so when the electron comes back to its starting point, but moving in the opposite direction, at that point it has the same kinetic energy it had when it set off

Answer 98

so same as vi