PLEASE ANSWER CORRECTLY AND I WILL RATE YOU HIGH 1) A 2.50-F capacitor is charge

Question

PLEASE ANSWER CORRECTLY AND I WILL RATE YOU HIGH

1) A 2.50-F capacitor is charged to 748 V and a 6.80-F capacitor is charged to 563 V . These capacitors are then disconnected from their batteries. Next the positive plates are connected to each other and the negative plates are connected to each other. What will be the potential difference across each and the charge on each? [Hint: Charge is conserved.]

PART A

Determine the potential difference across the first capacitor.

Express your answer using three significant figures and include the appropriate units.

V1 =

PART B

Determine the potential difference across the second capacitor.

Express your answer using three significant figures and include the appropriate units.

V2 =

PART C

Determine the charge across the first capacitor.

Express your answer using three significant figures and include the appropriate units.

Q1 =

PART D

Determine the charge across the second capacitor.

Express your answer using three significant figures and include the appropriate units.

Q2 =

Explanation / Answer

The charge (Q) on each capacitor is the product of its capacitance (C) and the
voltage (V) across it: Q = C times V
So at first the 2.5 µF capacitor has a charge of 2.5 times 748, or 1870 micro
coulombs (µQ).

The 6.8 µF capacitor has a charge of 6.8 times 563, or 3824.4 µC.

When you connect them in parallel, the total charge stays the same, but the total
capacitance is the sum of their individual capacitances, or 9.3 µF

and the total charge is 1870+ 3824.4= 5698.4 µC.

Some electrons flow from the 2.5 µF capacitor (with higher voltage) to the 6.8 µF capacitor till the voltages are equal.
The final voltage (Vf) can be found by dividing the total charge (5698.4 µC) by the
total capacitance, which is 9.3 µF.

So,

5698.4 / 9.3 = 612.7 volts.
So the final voltage across both capacitors is 612.7 volts.
The charge on the 2.5 µF capacitor is (2.5)(612.7) = 1531.8 µC
and the charge on the 6.8 µF capacitor is (6.8)(612.7) = 4166.6 µC.
So, the final answers are :

PART A

V1 = 612.7 volts

PART B

V2 = 612.7 volts

PART C

Q1 = 1531.8 µC

PART D

Q2 = 4166.6 µC

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