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.

 

[Nate810]

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.

 

[m2003]

I would first commend you on your creative thinking in using a plastic drinking glass as the dielectric for your capacitor. However, in order to determine the breakdown potential of the capacitor, we need to consider the electric field strength within the dielectric material.

The breakdown potential, also known as the breakdown voltage, is the maximum voltage that can be applied to the capacitor before the dielectric material breaks down and allows current to flow through it. This can lead to permanent damage to the capacitor and potentially cause a safety hazard.

To calculate the breakdown potential, we can use the formula V = Ed, where V is the breakdown potential, E is the electric field strength, and d is the thickness of the dielectric material. In this case, the thickness of the dielectric is equal to the height of the glass, or 18 cm.

To calculate the electric field strength, we can use the formula E = V/d, where V is the voltage applied to the capacitor and d is the distance between the plates. In this case, the distance between the plates is equal to the difference between the outer radius and inner radius, or 3.98 cm - 3.64 cm = 0.34 cm.

Now, we can calculate the electric field strength using the given breakdown potential of 9000 V and the distance between the plates of 0.34 cm.

E = V/d = 9000 V / 0.34 cm = 26,471 V/cm

Finally, we can calculate the breakdown potential using the electric field strength and the thickness of the dielectric.

V = Ed = 26,471 V/cm x 18 cm = 476,478 V

Therefore, the breakdown potential of the capacitor is approximately 476,478 V. It is important to note that this is an estimated value and may vary depending on the actual electric field strength within the dielectric material.

In conclusion, the capacitance of the cylindrical capacitor is approximately 1 nF and the breakdown potential is approximately 476,478 V. However, it is important to thoroughly test and verify the breakdown potential of the capacitor before using it in any applications.