(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Two large area parallel plate capacitors, labeled P and N, are connected as shown in the figure below. The charge on each plate is indicated in the figure, in μC.

I. The capacity of N (on the right) is 27.5 μF. Calculate the capacity of P.

1.34×10-5 F (Correct)

The plate separation of capacitor of N (on the right) is doubled. Calculate the new value of the charge on P.

2. Relevant equations

q=CV

total initial charge=total final charge

C=[tex]\epsilon[/tex]A/d

3. The attempt at a solution

[tex]C_{Ni}[/tex]=[tex]C_{Nf}[/tex]/2

because the distance has doubled (C=[tex]\epsilon[/tex]A/d)

[tex]q_{Ni}[/tex]+[tex]q_{Pi}[/tex] = [tex]q_{Nf}[/tex]+[tex]q_{Pf}[/tex]

[tex]q_{Ni}[/tex]+[tex]q_{Pi}[/tex] = [tex]C_{Nf}[/tex]V+[tex]q_{Pf}[/tex]

[tex]q_{Ni}[/tex]+[tex]q_{Pi}[/tex] = [tex]q_{Pf}[/tex]*[tex]C_{Nf}[/tex]/([tex]2*C_{Pf}[/tex])+[tex]q_{Pf}[/tex]

using q=CV because the voltage drop will be the same between the two final capacitors

then I solved for qPf

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# Homework Help: Two Charged Capacitors in Parallel

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