Capacitance and Max Potential Difference

In summary, the capacitance of a Teflon-filled parallel-plate capacitor with a plate area of 1.90 cm2 and insulation thickness of 0.0600 mm is calculated using the equation C = epsilon_0 * A * K / d, where K is the dielectric constant of Teflon (2.1). The maximum potential difference that can be applied to the capacitor is determined by using the equation V = Ed, where E is the electric field needed to conduct Teflon and d is the distance between the parallel plates. The dielectric breakdown of Teflon is 60 kV/mm, so the maximum potential difference for this capacitor is 3.6 kV.
  • #1
sonastylol
17
0

Homework Statement


(a) Determine the capacitance of a Teflon-filled parallel-plate capacitor having a plate area of 1.90 cm2 and insulation thickness of 0.0600 mm.

(b) Determine the maximum potential difference that can be applied to the capacitor.
kV

Homework Equations


I THOUGHT C = [tex]\epsilon[/tex]0*A all divided by d

The Attempt at a Solution



I did 8.85x10^-12 * 1.90 / .06x10^-3
it came up wrong.

I have to answer in pF btw.

I also have no idea how to answer part b since I can't get part a correct. Can anyone give me the equation for part b as well?Thank you so much PhysicsForums!
 
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  • #2
sonastylol said:

Homework Statement


(a) Determine the capacitance of a Teflon-filled parallel-plate capacitor having a plate area of 1.90 cm2 and insulation thickness of 0.0600 mm.

(b) Determine the maximum potential difference that can be applied to the capacitor.
kV

Homework Equations


I THOUGHT C = [tex]\epsilon[/tex]0*A all divided by d

This is only valid if there is vacuum between the plates. (or air with a very small error).
 
  • #3
C = epsilon_0 * A * K / d where K is the dielectric constant of Teflon (2.1)
(so just multiply the answer you got by 2.1)

the dielectric breakdown of Teflon according to the website below is 60 kV/mm so I think your answer to b) is:
V = Ed --> (60 kV/mm) * (0.06 mm) = 3.6 kV. Could you please check this answer for me? I don't want to use up my last guess.

got my info from:
http://hypertextbook.com/physics/electricity/dielectrics/
 
  • #4
I was right. V = Ed ; where E is the electric field needed to conduct teflon and d is the distance between the parallel plates. the electric field needed to break down teflon is 60 kV/mm so (60 kV/mm) * (0.06 mm) = 3.6 kV. You must raise the potential difference of the parallel plates to 3600 Volts in order to cause the teflon to conduct and the capacitor to short circuit. This is strange because part a) of the question asks you to find the capacitance and that value is not needed for part b)
 

1. What is capacitance?

Capacitance is the ability of a system to store an electric charge. It is represented by the symbol C and is measured in farads (F).

2. How is capacitance related to potential difference?

Capacitance and potential difference are inversely proportional. This means that as capacitance increases, potential difference decreases and vice versa. This relationship is represented by the equation C=Q/V, where Q is the charge stored and V is the potential difference.

3. What is the maximum potential difference in a capacitor?

The maximum potential difference in a capacitor is equal to the voltage of the source connected to it. In an ideal capacitor, this voltage is reached when the capacitor is fully charged.

4. How does the shape and size of a capacitor affect capacitance?

The shape and size of a capacitor can affect its capacitance. A larger surface area and smaller distance between the plates will result in a higher capacitance. This is because there is more space for the electric field to store charge between the plates.

5. Can capacitance be changed?

Yes, capacitance can be changed by altering the physical properties of the capacitor, such as the distance between the plates or the material used for the plates. It can also be changed by connecting capacitors in series or parallel, which affects the total capacitance in the circuit.

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