Homework Help: Electric fields

1. Jul 4, 2009

leena19

1. The problem statement, all variables and given/known data

ABCD are 2 parallel plates,between which a potential difference of 10,000V is maintained.An electron starting from rest,from point P,moves towards a small hole at Q.
After emerging from Q,the electron passes through a region independent of any electric fields and then enters the space between 2 plates EF and GH ,between which a vertical electric field 1000NC-1 is maintained.
The electron then strikes the lower plate at a point X.
If the distance d is 0.01m,find the distance GX

2. Relevant equations

Equations of motion
F=ma
F=Eq
E=V/x (I think?)

3. The attempt at a solution
For plates AB,CD,

F=ma
F/m=a
Eq/m=a
Vq/xm = a q= charge of an electron
m=mass of an electron

v2 = u2 + 2as
v2 = 0 + 2ax
v2 = 2*Vq/xm
v2 = m2*10,000q/m
v=$$\sqrt{m2*1000q/m}$$

for EF,GH
$$\downarrow$$ s= 1/2gt2 s=d=0.01
0.01=1/2gt2
t=$$\sqrt{0.02/g}$$

$$\rightarrow$$ s=ut s=GX=y, u= v
y =$$\sqrt{m2*10,000q/m}$$*$$\sqrt{0.02/g}$$
But I don't know q,m and I haven't used the electric field intensity (1000NC-1)anywhere in my calculations.
What am I doing wrong?

Thank you.

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2. Jul 4, 2009

Staff: Mentor

OK, but you lost a factor of 10 under the square root. That's the electron's speed after exiting the hole.

The acceleration is not g. (That's the acceleration due to gravity--not relevant here.) Find the acceleration on the electron due to the electric field.

The charge and mass of an electron are constants that you can look up.
You'll need that to find the acceleration. (Just like you did in the first part.)

3. Jul 4, 2009

leena19

Thank you so much for replying,sir
Oh yes,what a careless mistake,
so after correcting it I get,
v=$$\sqrt{2*10,000q/m}$$

Another terrible mistake.I just thought of it as a projectile,I forgot to take into account the acceleration due to the electric field force in EFGH,but now I understand.

Correction:
Then,using F=ma ,to find the acceleration of the electron through EF,GH,
a=F/m
Eq/m=a
1000q/m=a

$$\downarrow$$ s = 1/2at2
0.01= 1/2*(1000q/m )*t2
0.02*m/1000q=t2
$$\sqrt{0.02*m/1000q}$$=t

$$\rightarrow$$s=ut
GX=($$\sqrt{2*10,000q/m}$$) $$\sqrt{0.02*m/1000q}$$
GX=$$\sqrt{2*10}$$ * $$\sqrt{0.02}$$
GX=$$\sqrt{2*10}$$ * $$\sqrt{2*10^{-2}}$$
GX=2*10-1 * $$\sqrt{10}$$
GX=0.63m

Is this correct,now?I haven't got the answers to check.

THANK YOU

Last edited: Jul 4, 2009
4. Jul 13, 2009

leena19

Can someone check this for me, please?The answer's supposed to be 0.02m,which is nowhere close to what I get.
I really don't know where I'm going wrong,

5. Jul 13, 2009

Staff: Mentor

Your work looks OK to me. (Sorry for not responding earlier.) What book are you using?

6. Jul 13, 2009

merryjman

When I use a distance d of 0.1 m, I get almost exactly 2 m as my answer for the horizontal displacement. In other words, when I use a distance which is wrong by a factor of 10, I get an answer which is wrong by a factor of 100. Indicates to me that maybe there's a unit mismatch somewhere? Centimeters or millimeters, maybe?

7. Jul 13, 2009

Staff: Mentor

Good spot! If d = 0.01 mm (instead of 0.01 m), the answer would match the book's answer.

8. Jul 14, 2009

leena19

Thank you,merryjman.
Thank you Doc Al.Thank you both so much!
our teacher dictated the question for us,so I may have got the units wrong while taking it down & I don't know what book it's from to check,so
I apologise for any trouble caused.