Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

oooo TFW F 3:45 PM webassign.net In the following figure ca as ca 2.10 and ca-sa

ID: 1769030 • Letter: O

Question

oooo TFW F 3:45 PM webassign.net In the following figure ca as ca 2.10 and ca-sao The capacitor network is connected to an applied potential V After the charges on the capacitors have reached their final values, the charge on C2 is 60.0HC. (a) What are the charges on capacitors C1 and C3? c (b) What is the applied voltage Vab? The following figure shows a system of four capacitors, where the potential difference across ab is 50.0 V. (Let C1 13.0 HF, C2 4.0 F, C3 8.5 HF, and C 7.0 HF.) (a) Find the equivalent capacitance of this system between a and b. HF (b) How much charge is stored by this combination of capacitors? HC (c) How much charge is stored in each of C1 and C2? A parallel-plate air capacitor has a capacitance of 790 pF. The charge on each plate is 2.75 HC.

Explanation / Answer

1.

a)

Charge on C2 is

Q2=Q*(C2/C1+C2)

60 =Q*(2.1/2.1+7.05)

Q=261.4 uC

Charge on C1 is

Q1=Q-Q2=261.4-60

Q1=201.4uC

Charge on C3 is

Q3=Q=261.4 uC

b)

equivalent capacitance

1/Ceq =1/(C1+C2) +1/C3=1/(2.1+7.05) +1/5.5

Ceq=3.44 uF

Applied voltage is

Vab=Q/Ceq=261.4/3.26

Vab=76.1 Volts

2.

equivalent capacitance

1/Ceq =1/C1 +1/(C2+C3)+1/C4 =1/13 +1/(4+8.5) +1/7

Ceq=3.34 uF

b)

Total Charge

Q=CeqV=3.34*50

Q=166.8 uC

c)

Charge on C1 is

Q1=Q=166.8 uC

Charge on C2 is

Q2=Q*(C2/C2+C3) =166.8*(4/4+8.5)

Q2=53.4 uC