A 3.00 F capacitor is charged to 475 V and a 4.00 F capacitor is charged to 545

Question

A 3.00 F capacitor is charged to 475 V and a 4.00 F capacitor is charged to 545 V .

A)These capacitors are then disconnected from their batteries, and the positive plates are now connected to each other and the negative plates are connected to each other. What will be the potential difference across each capacitor?

B)What will be the charge on each capacitor?

C)What is the voltage for each capacitor if plates of opposite sign are connected?

D)What is the charge on each capacitor if plates of opposite sign are connected?

Explanation / Answer

a) Q1= C1*V1

Q1=3*475 = 1425 uC

similerly Q2= C2*V2

Q2 = 4*545 =2180 uC

We have the total charge which is the sum of the charges which were on the individual capacitors before they were connected together. Now, after they are connected in parallel, they are sharing that total charge. They are now two capacitors in parallel with a total charge of Q =Q1+Q2 = 3605 µC.

The capacitance of the parallel pair is

Ceq = C1+C2 = 7.0 µF.

The voltage across Ceq is given by

V = Q/Ceq,

V= 515 V

So that's the voltage that the paralleled capacitors have on them. Both of the individual capacitors of that parallel pair have that voltage across them -- they must have the same voltage because they are wired in parallel.

b) Now, after they are connected in parallel, they are sharing that total charge so both have same charge which is equal to sum of individual charge

Q=Q1+Q2 =3605 uC

c) because capacitor are connected in opposite polarity so charge must be subtracted and now equivalent charge is become

Q=Q2-Q1=755uC

and correspondind to this, resultatent voltage difference is

V = Q/Ceq

V= 107.9 V

because both are connected in parallel so each has the same potential = 107.9V

d) both have same charge and equal to the difference between their individual charges

Q=Q2-Q1 = 755uC

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