Can a Plastic Drinking Glass Handle Over 9000 V as a Capacitor Dielectric?

[cd80187]

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.

 

[member 553083]

The capacitance of the capacitor is 0.96 nF.The breakdown potential of the capacitor can be determined by using the equation for dielectric strength, which is V = E x t where V is the breakdown potential, E is the electric field, and t is the thickness of the dielectric material. The electric field can be calculated using the capacitance of the capacitor using the equation E = 1/(2πεC), where ε is the permitivity constant. For this problem, we can assume that the thickness of the dielectric material is the height of the glass (18 cm). Therefore, the breakdown potential can be found using the following equation:V = (1/(2πεC)) x 18 cmPlugging in the values, the breakdown potential of the capacitor is approximately 9,200 V.