A 3.00 F capacitor is charged to 490 V and a 3.95 F capacitor is charged to 530

Question

A 3.00 F capacitor is charged to 490 V and a 3.95 F capacitor is charged to 530 V .

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?

What will be the charge on each capacitor?

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

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

Explanation / Answer

Q1 = C1 . V1  = 1.470 * 10-3 C

Q2 = C2 . V2  = 2.093 * 10-3 C

As per the question they are in parellel combination then-

Part A)-

QT = CT . VT

( Q1+ Q2 ) =  ( C1 + C2 ) VT

  VT = 3.563 * 10-3 / (6.95 * * 10-6 )

  VT = 512.66 V (Voltage across both capacitors will be same because they are in parelle combination)

Part B)-

After combination charge on capacitors will be as per-

Q1 = C1 . VT = (3*10-6)(512.66) = 1.537 * 10-3 C

Q2 = C2 . VT = (3.95*10-6)(512.66) = 2.025 *10-3 C

Part D)-

In this condition they will be in series combination then-

QT will be the same for both capacitor and

VT = V1 +V2 = 490+530 = 1020 V

Then -

QT = CT . VT

QT = (3*3.95 / 6.95) * 10-6 (1020)

QT = 1.739 * 10-3 C

Part C)-

V1 = QT / C1 = 1.739* 10-3 / 3*10-6

= 579.67 V

V2 = QT / C2 = 1.739* 10-3 / 3.95*10-6

= 440.25 V

Thank you !!! Have fun

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