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Breakdown Potential in a cylinder

  • Thread starter cd80187
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  • #1
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You are asked to construct a capacitor having a capacitance near 1 nF and a breakdown potential in excess of 9000 V. You think of using the sides of a tall plastic drinking glass as a dielectric (with a dielectric constant 5.0 and dielectric strength 10 kV/mm), lining the inside and outside curved surfaces with aluminum foil to act as the plates. The glass is 18 cm tall with an inner radius of 3.64 cm and an outer radius of 3.98 cm. (a) What are the capacitance and (b) breakdown potential of this capacitor?



I have already solved part A. I used the equation C = 2 x pi x permitivity constant x (Length/ ln b/a) where b is the outer radius and a is the inner radius. However, now that I know the capacitance, I have no clue where to even start for the breakdown potential. I am also unsure of what formulas to use as well, so even just where to get started would be great.
 

Answers and Replies

  • #2
ranger
Gold Member
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If your dielectric has a dielectric strength of 10 kV/mm and you are using a glass that is 18cm long. Set up a proportion and solve for the breakdown voltage. Be sure to convert cm to mm.
 
  • #3
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What formula should I be using for this then? I tried just using a simple proportion, but it didn't work out
 
  • #4
ranger
Gold Member
1,676
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How did you set up the proportion?

btw, the dielectric thickness is 18cm right?
 
  • #5
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I set it up as 10 kV/mm = x/180mm. And the length of the cylinder is 18cm, the thickness is .34 cm.
 

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