Capacitance 3 dielectrics sandwiched between two conductors.

  • #1
MostlyHarmless
345
15
Find that capacitance of the system consisting of 3 dielectrics w/ length, l=1.0m, width, w= 1.0m and depth, d= 1.0 CM. ##k_1=1.5, k_2=2, k_3=2.5## *dimensions of conducting plates not given*

Equations: Capacitance, ##C= Q/{\delta}V##

Field in the dielectrics
##E={\frac{\sigma}{k{\epsilon}_0}}##

I've found the fields in each dielectric, and the potential difference in each, but I'm getting hung up there. I've gotten down to, ##|{\delta}V|=({\frac{Q}{A{\epsilon}_0}})({\frac{1}{k_1}}+{\frac{1}{k_2}}+{\frac{1}{k_3}})##

My problem is, the A refers to the area of the conducting plate, but as i pointed out, nothing is said concerning dimensions of those plates.

Is there another method that does not care about the area that I'm missing?
 

Answers and Replies

  • #2
36,103
13,027
You can assume the plates have the same dimensions as width and length of the dielectric materials.
 
  • #3
MostlyHarmless
345
15
That's what I thought, but the diagram provided clearly shows that conductors being larger than the dielectrics.

Is the portion that is the same dimension as the dielectrics the only portion if the conductor that has any effect?
 
  • #4
36,103
13,027
That's what I thought, but the diagram provided clearly shows that conductors being larger than the dielectrics.
Can you show this diagram?
Is the portion that is the same dimension as the dielectrics the only portion if the conductor that has any effect?
No. For large k this is a good approximation (as you can neglect the air then), but your k-values are not large.
 

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