What is the Work Required to Increase Separation of a Parallel Plate Capacitor?

In summary, the problem requires calculating the work done to pull apart an isolated parallel-plate capacitor with a constant charge and variable separation between the plates. The formula for calculating potential energy and capacitance are used, along with the given values for charge and initial separation, to find the final potential energy and capacitance at the new separation distance. The work done is then calculated by taking the difference between the two potential energies. The correct formula for capacitance is C=ε*A/d, and the correct formula for electric field is E=Q/d^2.
  • #36
damn and i was tryin to figure out where i went wrong... kool sounds good i understand it then
 
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  • #37
sw1mm3r said:
damn and i was tryin to figure out where i went wrong... kool sounds good i understand it then
I will rewrite my post.
C1 = εοA/d1
In the second case
C2 = εοA/d2

C1/C2 = (εοA/d1)/(εοA/d2)
= d2/d1
Substitute the values of d1, d2 and C1, and find the value of C2.
Using the formula for energy, find the energies in C1 and C2 and find the difference.
 
  • #38
That's perfect. Sorry again if my false recollection that one of the d's was 1.5mm instead of 1.2mm was what was throwing you off. Oh, that's you rl.bhat, thought is was sw1mme3r. Maybe time to quit while I'm not too far behind.
 
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