# HW_5_Solutions - 13. THINK Charge remains conserved when a...

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13.THINKCharge remains conserved when a fully charged capacitor is connected to anuncharged capacitor.EXPRESSThe charge initially on the charged capacitor is given byq = C1V0, whereC1= 100 pFis the capacitance andV0= 50 V is the initial potential difference. After the battery is disconnectedand the second capacitor wired in parallel to the first, the charge on the first capacitor isq1=C1V,whereV= 35 V is the new potential difference. Since charge is conserved in the process, the chargeon the second capacitor isq2=q – q1, whereC2is the capacitance of the second capacitor.ANALYZESubstitutingC1V0forqandC1Vforq1, we obtainq2=C1(V0V). The potentialdifference across the second capacitor is alsoV, so the capacitance of the second capacitor is()022150V35V100pF42.86 pF43pF.35VVVqCCVV====LEARNCapacitors in parallel have the same potential difference. To verify charge conservationexplicitly,wenotethattheinitialchargeonthefirstcapacitoris10(100 pF)(50 V)5000 pC.qCV===After the connection, the charges on each capacitor are1122(100 pF)(35 V)3500 pC(42.86 pF)(35 V)1500 pC.qCVqC V======Indeed,q=q1+ q2.15. (a) First, the equivalent capacitance of the two 4.00μF capacitors connected in series is givenby 4.00μF/2 = 2.00μF. This combination is then connected in parallel with two other 2.00-μFcapacitors (one on each side), resulting in an equivalent capacitanceC= 3(2.00μF) = 6.00μF.This is now seen to be in series with another combination, which consists of the two 3.0-μFcapacitors connected in parallel (which are themselves equivalent toC'= 2(3.00μF) = 6.00μF).Thus, the equivalent capacitance of the circuit is() ()eq6 00F6 00F3 00F.6 00F6 00F..CCC.CC..μμμμμʹ′===ʹ′++(b) LetV= 20.0 V be the potential difference supplied by the battery. Thenq = CeqV= (3.00μF)(20.0 V) = 6.00×10–5C.(c) The potential difference acrossC1is given by() ()16 00F20 0V10 0V6 00F6 00F..CVV..CC..μμμ===ʹ′++(d) The charge carried byC1isq1=C1V1= (3.00μF)(10.0 V) = 3.00×10–5C.(e) The potential difference acrossC2is given byV2=V – V1= 20.0 V – 10.0 V = 10.0 V.
(f) The charge carried byC2isq2=C2V2= (2.00μF)(10.0 V) = 2.00×10–5C.(g) Since this voltage differenceV2is divided equally betweenC3and the other 4.00-μF capacitors

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Term
Spring
Professor
DODD
Tags
Electric charge, NC, Dielectric
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