I realized that I can seperate these into 2 capacitors in series. When convertin
ID: 2059190 • Letter: I
Question
I realized that I can seperate these into 2 capacitors in series. When converting all the units and working it out I get C total= 4.07114E-9 F but its incorrect, Please help.
Figure 25-48 shows a parallel-plate capacitor with a plate area A = 7.72 cm2 and plate separation d = 1.09 mm. The top half of the gap is filled with material of dielectric constant ?1 = 11.9; the bottom half is filled with material of dielectric constant ?2 = 14.3. What is the capacitance? I realized that I can seperate these into 2 capacitors in series. When converting all the units and working it out I get C total= 4.07114E-9 F but its incorrect, Please help.Explanation / Answer
Whenever you have a capacitor with more than one dielectric, consider there to be that many capacitors in series. In this case, consider two series capacitors. The total plates separation is 1.09 mm, so we need to divide that by two for each dielectric. Therefore each considered capacitor will hava a plate separation, d, of .545 mm
C1 = koA/d = (11.9)(8.85 x 10-12)(7.72 X 10-4)/(5.45 X 10-4) = 1.49 X 10-10 F
C2 = koA/d = (14.3)(8.85 x 10-12)(7.72 X 10-4/(5.45 X 10-4) = 1.79 X 10-10 F
The CEQ is found by 1/ 1.49 X 10-10 F + 1/1.79 X 10-10 F
Net Capacitance of the capacitor is
8.13 X 10-11 F or 81.3 pF