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Suppose you have a 9.00 V battery, a 2.00$\mu \mathrm{F}$ capacitor, and a 7.40$\mu \mathrm{F}$ capacitor. (a) Find the charge and energy stored if the capacitors are connected to the battery in series.(b) Do the same for a parallel connection.

a) 14.17$\mu \mathrm{C}$, $6.35 \times 10^{-5} \mathrm{J}$b) 84.6$\mu \mathrm{C}$, $3.80 \times 10^{-4} \mathrm{J}$

Physics 102 Electricity and Magnetism

Chapter 19

Electric Potential and Electric Field

Electric Charge and Electric Field

Gauss's Law

Electric Potential

Cornell University

Simon Fraser University

University of Winnipeg

Lectures

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In physics, a charge is a …

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The electric force is a ph…

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Given a $2.50-\mu \mathrm{…

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A $2.00-\mu F$ and a $4.00…

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(II) A $0.50-\mu \mathrm{F…

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Two capacitors, $C_{1}=5.0…

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05:10

(II) A 0.50-$\mu$F and a 1…

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A $3.00-\mu \mathrm{F}$ an…

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Two capacitors are connect…

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A 4.00$\mu \mathrm{F}$ and…

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What If? The two capacitor…

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A $4.00-\mu \mathrm{F}$ ca…

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Two capacitors, one 7.5$\m…

Two capacitors, $C_{1}=25.…

03:35

A $7.0-\mu \mathrm{F}$ and…

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03:16

The two capacitors of Prob…

so to find the charge and energy stored in the capacitor, we are going to use a question Q equal to Stevie, and energy is equal to have CV square. But before that we need to find what what is the capacitance of the system? So it says that in part A the capacitors are connected in Siri's. So let's find too serious capacitors. Serious capacitance is when I was three yeses equal to one divided by C one plus one divided by sea to And now let's substitute the values of C one and C two. We know that C one is toe to micro parrot on DSi to is 7.4 micro flattered. That gives us so when by two is 20.5 plus one by one divided by 7.4 is zero. So if you add these two up you get 0.635 So C s is invoice off that so C s is equal to inverse appoint 65 which gives me the value one point 574 micro fired Since we began with the values here which are micro fired, Final answer will be in micro federal as well. So That's the capacitance charge store in that case is C S V. So see them? It's capacitance is 1.574 times 10 power, negative six. The voltage applied is nine. So charge here is 14 point two Michael Michael Kula. So that's the charge also. Now you confined energy stored by using half see the squared. So that will be half 1.574 times 10 power negative six times nine squared. So this calculation will yield energy stored as 63.8 micro Jules. So this is the energy stored and charge stored in capacitor on DDE. In the second part off this question, we are given that these capacitors are now connected in the parallel connection and parallel capacitance is simple to find because it's simply the addition of the two. Capacitance is so one capacitance is to microfibers. Another. A 7.4 addition will give us nine point for micro fatter. So charge here is again the same C times we but capacitance now is 9.4 times the voltage, which is nine. So answer will be in my coke alone because I'm not writing to empower negative six the 9.4 times nine is given us 84.6. Let me try that again. Six. I'm sorry about that. Micro cooler on energy stored is half seanie squared. So energy is I have times 9.4 times, Jen Power negative. Six times nine squared. So energy store in this case is 3 81 micro juice. See that? Capacitance on dhe energy stores are significantly higher in this case because this is parallel capacitance on parallel capacitance. Parallel connection increases. Look up. Net capacitance off the circuit.

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