two horizontal plates erated by 4.00cm , potetinal differece 120V horizaonal beem of eletrons with speed 6.50*10^6 , the electron enter 1.00 above the negative plate , the horizontal distance tht the electrom travels before striking the postive plate?

- anonymous

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- schrodinger

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- IrishBoy123

draw it first and you will get a lot of help

- anonymous

|dw:1437369186841:dw|

- anonymous

@dan815 @e.mccormick @Luigi0210

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## More answers

- anonymous

What would be the electric field between the two parallel plates?

- anonymous

i did fnet=Fe+Fg

- anonymous

is that how u start?

- anonymous

Yep, you can use the forces to do the problem.
Then, F=ma. Then find where the electron intersects on of the plates.

- anonymous

they didnt geive the accelration

- anonymous

If you know the forces, you can find the acceleration.

- anonymous

do we have to find it ?

- anonymous

to find the distane?

- anonymous

what formila are we goona conclude to ? in the end ?

- anonymous

Yeah, we'll need to.

- anonymous

Usually, in these types of problems, we'll ignore gravity.
Find the electric field.
F = qE=ma ==> a = qE/m
We can then use:
\[y(t) = \frac{1}{2}at^2 + v_ot + y_o\]
to find when the electron meets one of the plates

- anonymous

ok so for acc i get 2.85*10^17

- anonymous

not sure if correct too mnay calcs maybe screwd

- anonymous

Then, when you have t for that, you can find x
There's no acceleration in the x direction, so you can just use: x(t) = v*t
The v used here is the one given in the problem.

- anonymous

not getting the answer

- anonymous

first i found time , then found d but wrong ansswr1

- anonymous

I get a different acceleration. Be careful with units.

- anonymous

whatu get/

- anonymous

what do u get as an answerr?

- anonymous

5.276x10^14 is the acceleration I'm getting

- anonymous

m/s^2

- anonymous

even if i use ur acc i am getting wrong answr

- anonymous

the answer is 6.94

- anonymous

Show me what youre doing

- anonymous

first speed didved by acc

- anonymous

then i get the time , so i times by speed

- anonymous

to find d

- anonymous

No, can't do it quite that way.
\[E = \frac{V}{d}\]gives the E field between the plates. V is the voltage between the plates, d is the separation between them (be careful of units! We want this in meters!). As shown above, we can find the acceleration using
\[a = \frac{qE}{m} = \frac{V}{d}\frac{q}{m}\]
You can look up the value for q/m online. It's very commonly used.
After that, we'll find the time of flight. The electron starts 1cm above the bottom, negative plate. It will travel upwards to the positive plate, which sits at a height of 4cm.
\[y(t) = \frac{1}{2}at^2 +v_ot + y_o\]
From the information given, we see that vo is zero (no initial velocity in the y direction), and y_o = .01m (1cm). We want to know what t will be when y(t) = 4. That is:
\[4 = \frac{1}{2}at^2+1\]
Only one step after you find t. What do you get for t?

- anonymous

Bleh. 4 should be .04, and 1 should be .01 in the last equation. Units.

- anonymous

the answer is in cm

- anonymous

You can convert to whatever units you'd like.

- anonymous

ok yea got it thanks , could u help me with soe theory questions

- anonymous

I'll try :)

- anonymous

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- anonymous

which way will the electron accelerate?

- anonymous

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- anonymous

You can think about this two ways. We can use F = qE = ma, or we can think about which way electric fields point between charges.
Which way does the field point between a negative and positive charge?

- anonymous

2?

- anonymous

If we say that the direction the field is pointing in is the positive direction, then
F = qE will be negative, because q is negative (charge of the electron).
This means it would move in the direction opposite of the electric field.

- anonymous

so 4

- anonymous

ohh i was think why the answrr is 4

- anonymous

beacuse i was 100% sure the direction would be 2 and then i looked at the answer it was 4 i ws soo confused

- anonymous

its an electron so would make sence since they are nagative

- anonymous

Exactly :)

- anonymous

so if it was a proton the answr would be 2?

- anonymous

Yep!

- anonymous

their is one more

- anonymous

their are 2 netual metal objcts x and y on a wooden table

- anonymous

a negativly charged rod is brought near object y

- anonymous

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- anonymous

so the answer is x received a negative charge by induction and y received a positive charge by induction

- anonymous

i dont get what they mean

- anonymous

Right.
So, the negative charges in the rod will repel other negative charges. Since the objects are metal, and thus conductive, their electrons can move relatively freely. This means that the electrons in the metal will move away from the rod.
Since the rod is near object y, they'll move away to object x. This will make object x have a net negative charge. The lack of electrons on object y will give it a net positive charge.
This displacement of charge is called induction.

- anonymous

i am reading from my notes about induction and it says it causes the chrages to become opposite

- anonymous

so what ur saying is

- anonymous

in the other way

- anonymous

if the rod was postive then y would be negative and also x ?

- anonymous

i mean x will be negative

- anonymous

If the rod were positive, the y would become negative and x would be positive.

- anonymous

oh yea

- anonymous

some of the other answrrs are by conductions whats the differenr

- anonymous

if the boxes were not neutral then it would be conductions right?

- anonymous

I don't understand your questions.
If the objects weren't conductors, then the charges wouldn't be able to move freely. If they were perfect insulators, they couldn't move at all, and thus no charge would be induced.

- anonymous

like the correct answer is induction the wrong answer is conduction

- anonymous

Oh, the phenomenon of the rod inducing a charge on the boxes is called induction.

- anonymous

what would make it conduction

- anonymous

if the boxes werent neutral?

- anonymous

like lets say box x and y were postive then the answer would be conduction right

- anonymous

Conduction is just the metal carrying the charge across a potential. Like a current moving down a wire.

- anonymous

oh ok

- anonymous

their is one more

- anonymous

a posively charger sphere near a metal rod . both are insulated stands and the rod is grounded

- anonymous

|dw:1437374305785:dw|

- anonymous

the disturbution of the charge on the rod is ?

- anonymous

This is somewhat similar to what we just did. What will happen when the sphere approaches the rod?

- anonymous

umm x will become neagative?

- anonymous

That's what I'd say.
Charge would be drawn up through the ground wire toward the sphere.

- anonymous

how do u knw tht?

- anonymous

beacuse they are grounded ? like can't move right?

- anonymous

Grounded just means that it can exchange charges with a very large, external source.
So, the positive charge of the sphere will cause electrons to accumulate on the rod, mostly near the sphere.

- anonymous

to show that an objct is negativly charged we need to ?

- anonymous

it should be positve right?

- anonymous

but the answer is negative?

- anonymous

The positive sphere will cause the rod to become negatively charged.

- anonymous

ok thats it for tonight

- anonymous

do u come here often ?

- anonymous

around this time ?

- anonymous

occasionally. I'm kind of erratic about when I'm on

- anonymous

ok i am going to sleep learned so much today need to settle in the brain

- anonymous

thanks alot!

- anonymous

hope to see u again (kinda hard to meet people here that are good at physics lol)

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