Could Someone Look Over My Work? (particle accelerator design problem)

[UF6]

Homework Statement

1. 
   You want to construct a simple particle accelerator and hit on the following design: you will take two oppositely charged plates with small holes drilled into them and set them up parallel to each other. You will create a vacuum between the two plates and fire a beam of electrons through the small hole in the negatively charged plate, aimed so that the beam will emerge through the hole in the positively charged plate. You want the electrons in the beam to gain 10.0 eV of energy, so you need to work out what magnitude of voltage difference to set up between the two plates.
   Correct, computer gets: 1.0E+01 V
   
   Your electron beam consists of electrons that are initially moving at around 2.7×105 m/s. You want to work out how fast the electrons are moving after passing through your accelerator.
   
   Correct, computer gets: 1.9e+06 m/s
   
   You decide to investigate the physical details of your electron accelerator a bit more closely. You start by determining the strength of the electric field between the plates. You've set the plates up so that their separation is 3 μm.
   
   Correct, computer gets: 3.3E+06 V/m
   
   The plates you chose had areas of 9 mm^2. You want to know how much electrical energy is stored by the electric field between the plates.

Homework Equations

C=(ε₀)A/d

Capacitor Energy=1/2C*v²

The Attempt at a Solution

This is fairly easy problem, which is a great reason why it's also the one that's holding me back from the entire homework set. I think I made a conversion error I just don't know where.

-(8.854E-12)(9E-6/3E-6)
=2.655E-11 C

1/2(2.655E-11)(9)²
=1.075275E-9 J

 

[RUber]

This is fairly easy problem, which is a great reason why it's also the one that's holding me back from the entire homework set. I think I made a conversion error I just don't know where.

-(8.854E-12)(9E-6/3E-6)
=2.655E-11 C

1/2(2.655E-11)(9)2
=1.075275E-9 J

Where did the (9)^2 come from? I assume that is supposed to be velocity?

 

[UF6]

Where did the (9)^2 come from? I assume that is supposed to be velocity?

It would be the charge in voltage on the area on the capacitors.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capeng2.html#c1

 

[RUber]

edit: I was wrong.

 

[mfb]

You already know the voltage: 10 V.

@RUber: that calculation does not make sense at all.

 

[UF6]

You already know the voltage: 10 V.

@RUber: that calculation does not make sense at all.

Would you convert 10 eV into 1E19 V or leave it as so?

The answer given for either case as the final answer is 1.3275E27 J or 1.3275E-9 J. I just find the first to be rather larger.

 

[mfb]

There is nothing to convert.

so you need to work out what magnitude of voltage difference to set up between the two plates.
Correct, computer gets: 1.0E+01 V

"Converting 10 eV into 1E19 V" does not make sense. It's like trying to convert 1 meter to 5 kilogram of apples.

 

[UF6]

There is nothing to convert.
"Converting 10 eV into 1E19 V" does not make sense. It's like trying to convert 1 meter to 5 kilogram of apples.

Thanks by the way. I understand how that works now a bit better after reading about electron volts to volts after I read your response.