Physics 21Fall, 2005Solution to HW-724P59Two identical capacitors are connected in parallel,and each acquires a chargeQ0when connected to a sourceof voltageV0. The voltage source is disconnected and thena dielectric (K=2.5) is inserted to Fll the space betweenthe plates of one of the capacitors. Determine (a) the chargenow on each capacitor, and (b) the voltage now across eachcapacitor.+Q0+Q02Q0-QQC0C0C0KC0(a)(b)±or panel (a), the two capacitors are the same andQ0=C0V0.±or panel (b), the charge on the two upper plateswill rearrange so that the voltages across the two capacitors(C0on the left andKC0on the right) are equal.Sincethe capacitors are in parallel, the total charge on the twopositive plates does not change. We can Fnd the chargeQthat ends up on the modiFed capacitor by using the equalityof voltages:2Q0−QC0=Q0Solving this equation forQgivesQQ0KK+1and2Q0−QQ01K.The voltage across both capacitors isV=Q0=Q0C02K=V02K.
This is the end of the preview. Sign up
access the rest of the document.