Two parallel-plate capacitors C 1 and C 2 are connected in series to a battery.

Question

Two parallel-plate capacitors C1 and C2 are connected in series to a battery. Initially, both capacitors have air bewteen the plates and C1=3x10-6 F and C2 =6X10-6 F. When fully charged, the charge on each capacitor is 8x10-8 C. Then, without disconnecting either capacitor, a dielectric with constant K=4 is inserted bewteen the plates of capacitor C1, completely filling the space bewteen the plates. After the dielectric has been inserted, what is the magnitude of the charge on each plate of the capacitor C1?

****Please provide ALL formulas/concepts used***

***I don't know how to setup the problem; or even what is actually going on in the problem...our solutions booklet says the correct answer is 16*10-6 C. ****

Explanation / Answer

Given: capacitance C1 = 3*10-6 F capacitance C2 = 6*10-6 F charge on each capacitors q = 8*10-8 C .......................................................................................... if the capacitors connected in series then the equivalent capacitanec Ceq = C1 C2 / C1+ C2         = (3*10-6 F )(6*10-6 ) / ( 3*10-6 F) + ( 6*10-6 F)         = 2*10-6 F V = q/C     = 8*10-8 C / 2*10-6 F     = 4*10-2 V C1 = KCair       = 4*2*10-6 F       = 8*10-6 F charge on each plate q1 = C1*V                                    = (8*10-6 F)(4*10-2 V)                                     = 32*10-8 C if the capacitors are connected in series charge on each capacitors is same.       = 4*2*10-6 F       = 8*10-6 F charge on each plate q1 = C1*V                                    = (8*10-6 F)(4*10-2 V)                                     = 32*10-8 C if the capacitors are connected in series charge on each capacitors is same.

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