Electric Field Strength Between Two Parallel Plates Problem

  • #1

Homework Statement


If the electric field strength between two parallel plates 1cm apart is 3.1kN/C, how much energy will be used by the electric field to move an electron (placed half way between the plates) 3mm nearer to the positively charged plate?



Homework Equations


Work Done = 1/2V^2C
F=KxQ1Q2/r^2
E=KxQ/r^2
E=V/D


The Attempt at a Solution


I know that the answer to this question is 1.49x10^-18 J. I feel that I may be going about this question the complete wrong way but this is what I did:

I worked out the potential difference between the plates to be 31V by using the equation V=ED
V=(3.1x10^3)x(0.2x10^-2)-----> I got this value for D by taking away 3mm from the 5mm half way point between the plates.

I then subbed in the values that I know into the equation Work Done = 1/2V^2C
Work Done = 1/2 x 6.2^2 x (1.6x10^-19)
This gave me 3.08x10^-18.

I feel like I'm going the wrong way about this question. Hope someone can help and hope I've worded this well enough for you all to understand haha.
 

Answers and Replies

  • #2
Doc Al
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I feel like I'm going the wrong way about this question.
Yes you are. What's the most basic definition of work? What's the force acting on the electron?
 
  • #3
Yes you are. What's the most basic definition of work? What's the force acting on the electron?
Work Done = Force x Distance? The force produced by the electric field?
Should I use F = KxQ/r^2 to find the force and then substitute my answer into the Work Done = FxD equation?
 
  • #4
Doc Al
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Work Done = Force x Distance? The force produced by the electric field?
Right and right.

Should I use F = KxQ/r^2 to find the force and then substitute my answer into the Work Done = FxD equation?
No, it's easier than that. You are given the field strength! (It's a uniform field between those plates, not the field from a point charge.)
 
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  • #5
No, it's easier than that. You are given the field strength! (It's a uniform field between those plates, not the field from a point charge.)
Ah yes! I see where you're coming from! So if I have the the field strength of 3.1x10^3 and use the equation, Force = Electric Field Strength x Charge, to work out the force this gives me:
3.1x10^3 x 1.6x10^-19 = 4.96x10^-16

Then apply this to the work done equation:
4.96x10^-16 x 0.3x10^-2 = 1.49x10^-19

Thank you for your help, I really appreciate it! :biggrin:
 

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