Finding Minimum Plate Separation for Mica-Filled Capacitor

[mattbeatlefreak]

Homework Statement

You want to maintain a potential difference of 5000 V between the plates of a parallel-plate capacitor.
What is the minimum value of the plate separation if the space between the plates is completely filled by a slab of mica?

Homework Equations

V=Ed (parallel plate capacitor)
Dielectric constant κ = 5 for mica
κ=V₀/V_d
E=σ/ε₀

The Attempt at a Solution

I know that the voltage of the capacitor needs to stay constant, so since the voltage decreases by a factor of five, the distance must increase by a factor of five. However, the dielectric also lowers the field. Whenever I try to approach the problem, I end up getting stuck because I do not know the charge density or any charge values.
Any help would be greatly appreciated!

 

[gneill]

Besides the dielectric constant κ, dielectrics have another property that describes their ability to resist electric breakdown, which is what occurs when the applied electric field becomes strong enough to free the bound electrons in the dielectric material. Essentially, it's when its insulating property is overcome and current can flow through the dielectric. Usually the capacitor is destroyed when this happens.

So, look up the term "dielectric strength".

 

[mattbeatlefreak]

Besides the dielectric constant κ, dielectrics have another property that describes their ability to resist electric breakdown, which is what occurs when the applied electric field becomes strong enough to free the bound electrons in the dielectric material. Essentially, it's when its insulating property is overcome and current can flow through the dielectric. Usually the capacitor is destroyed when this happens.

So, look up the term "dielectric strength".

Thank you! As a follow up question, I am supposed to find the ratio of the maximum charges that could be put on a plate of the capacitor in part B to the capacitor in part A. Part A used mica, and part B used barium titanate as the dielectric. I know q_free = V_d*C_d. If we use the same capacitors as in the previous parts, then their voltage would be the same (5000 V). This leads me to believe that the only thing that would change the charge would be the capacitance. Since the κ for mica is 5 and the κ for barium titanate is 1200, I did 1200/5, which is 240. However, the program wants only one significant digit in the answer, and 200 was incorrect. What am I doing wrong here?

 

[gneill]

Will the spacing of the plates be the same for both capacitors? (Did you take the available dielectric strengths into account as a contributing factor to the difference in capacity?)

 

[mattbeatlefreak]

Will the spacing of the plates be the same for both capacitors? (Did you take the available dielectric strengths into account as a contributing factor to the difference in capacity?)

So then, you can write the capacitance with a dielectric as C=κε₀A/d. Do you assume that the plate area is the same for both? If so then I get that the charge q on the capacitor plates is q=κV/d. I then used the dielectric constants for each, used 5000 V in both, and used their respective distances found from the previous problems. Doing this I get a ratio of 120. Is this method correct?

 

[gneill]

So then, you can write the capacitance with a dielectric as C=κε₀A/d. Do you assume that the plate area is the same for both?

I would presume so. It seems that only the dielectric and plate separation are changed.

If so then I get that the charge q on the capacitor plates is q=κV/d.

Strictly speaking it's not true, since the actual charge depends upon the area and the constant $\epsilon_o$. Replace the "equals"with "proportional to" and I'd agree. Since you're looking for a ratio then that's fine.

I then used the dielectric constants for each, used 5000 V in both, and used their respective distances found from the previous problems. Doing this I get a ratio of 120. Is this method correct?

It looks correct. I can't confirm your result because you haven't supplied all the data that you've used for the materials involved.

 

[mattbeatlefreak]

I would presume so. It seems that only the dielectric and plate separation are changed.

Strictly speaking it's not true, since the actual charge depends upon the area and the constant $\epsilon_o$. Replace the "equals"with "proportional to" and I'd agree. Since you're looking for a ratio then that's fine.

It looks correct. I can't confirm your result because you haven't supplied all the data that you've used for the materials involved.

It worked out, thanks again for all the help!