Three square metal plates A, B, and C, each 15.0cm on a side and 1.50mm thick, a

Question

Three square metal plates A, B, and C, each 15.0cm on a side and 1.50mm thick, are arranged as in the figure . The plates are separated by sheets of paper 0.40mm thick and with dielectric constant 4.3. The outer plates are connected together and connected to point b. The inner plate is connected to point a.

What is the capacitance between points a and b?

Side: 0.15 m, capacitor thickness: 1.5*10^-3 m, paper thickness: 4*10^-4 m

Area of the plates= 2.25*10^-4 m^2

C= (epsilon*k*A)/d

C=[(8.85*10^-12)(4.3)(2.25*10^-4)]/4*10^-4

C= 2.1406*10^-11 F   

But since the capacitors are in parallel, Ceq=C1+C2, so I tried 2*(2.1406*10^-11)= 4.28*10^-11 F, but it was wrong...

Please help, I don't know where I am going wrong, or how to interpret this question, I've already asked my teacher, but that got me nowhere.

Thank-you in advance.

Explanation / Answer

Side: 0.15 m, capacitor thickness: 1.5*10^-3 m, paper thickness: 4*10^-4 m

Area of the plates= 2.25*10^-2 m^2

C= (epsilon*k*A)/d

C=[(8.85*10^-12)(4.3)(2.25*10^-2)]/4*10^-4

C= 2.1406*10^-9 F   

But since the capacitors are in parallel, Ceq=C1+C2, >>>>> 2*(2.1406*10^-9)= 4.28*10^-9 F

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