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In Fig., let C1 = 3.00 mF, C2 = 5.00 mF, and Vab = +52.0 V. Calculate

(a) the charge on each capacitor and

(b) the potential difference across each capacitor.

Short Answer

Expert verified

The potential difference is the same for the capacitor and equals the potential between a and b and that is 52.0V

The charge on the capacitor C1is 156.0μC.

The charge on the capacitorC1is156.0μC

Step by step solution

01

Potential difference and charge on each Capacitor.

We are givenC1andC2are connected in parallel where C1=3.00 uF and C2= 5.00 pF and the potential between a and b is Vab= 520 V

(a) In Figure 24.9a, the two capacitors are connected in parallel. As shown in the

figure the upper plates of the two capacitors are connected by conducting wires

to form an equipotential surface, and the lower plates form another. Therefore, in a

parallel connection, the potential difference for all individual capacitors is the

same and is equal to

Vab=V1=V2=52.0V

So we can get the charge for each capacitor by using the value of the given

potential and the capacitance of each capacitor as next

For the first capacitor C1, the chargeQ1is given by

Q1=C1Vab=3.00μF52.0V=156.0μC

For the second capacitor C2, the chargeQ2is given by

Q2=C2Vab=5.00μF52.0V=260.0μC

02

Conclusion

Hence, the potential difference is the same for the capacitor

and equals the potential between a and b.

Vab=V1=V2=52.0V

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Most popular questions from this chapter

In the circuit shown in Fig. E26.18,ε=36.V,R1=4.0Ω,R2=6.0Ω,R3=3.0Ω(a) What is the potential difference Vab between points a and b when the switch S is open and when S is closed? (b) For each resistor, calculate the current through the resistor with S open and with S closed. For each resistor, does the current increase or decrease when S is closed?

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