An isolated parallel-plate capacitor of area A = 1 cm^2 with an air gap of lengt

Question

An isolated parallel-plate capacitor of area A = 1 cm^2 with an air gap of length s = 1 mm is charged up to a potential difference of 120 V. A second parallel-plate capacitor, initially uncharged, has the same area A and a gap of length s but is filled with plastic whose dielectric constant is 1.32. You connect a wire from the positive plate of the first capacitor to one of the plates of the second capacitor, and you connect another wire from the negative plate of the first capacitor to the other plate of the second capacitor. What is the final potential difference across the first capacitor? Recall that Q = CV and that C = 0A/s for a parallel plate capacitor.

THe answer is 68V. Could you explain (solve) this?

Explanation / Answer

capacitance=eoA/d

so C1=8.85*10^-12*10^-4/10^-3

=8.85*10^-13 F

C2=1.32*8.85*10^-12*10^-4/10^-3

=1.1682*10^-12 F

so qo=C1V

=1.062*10^-10 C

so,

let the charge flowed be q.so,

(1.062*10^-10-q)/(8.85*10^-13)=q/(1.1682*10^-12)

or q=6.04*10^-11 C

so charge on the first capacitor=1.062*10^-10-6.04*10^-11

=4.58*10^-11 C

so V=q/C

=51.75 V

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