Calculating Distance Between Plates EF & GH with Varying Electric Fields

[leena19]

Homework Statement

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

Homework Equations

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

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.

 

[Doc Al]

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

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

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

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.

But I don't know q,m

The charge and mass of an electron are constants that you can look up.

and I haven't used the electric field intensity (1000NC^-1)anywhere in my calculations.

You'll need that to find the acceleration. (Just like you did in the first part.)

 

[leena19]

Thank you so much for replying,sir

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

Oh yes,what a careless mistake,
so after correcting it I get,
v=\sqrt{2*10,000q/m}

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.

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


\rightarrows=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

 

[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,

Thanks in advance

 

[Doc Al]

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

 

[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?

 

[Doc Al]

Indicates to me that maybe there's a unit mismatch somewhere? Centimeters or millimeters, maybe?

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

 

[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.