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## Homework Statement

A 2.70µF capacitor is charged to 475 V and a 4.00µF capacitor is charged to 525 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 and the charge on each? (b) What is the voltage and charge for each capacitor if plates of opposite sign are connected?

## Homework Equations

C=Q/V

## The Attempt at a Solution

For part (a) I tried the following:

I used this formula: C=Q/V ---> Q=CV

For C, I added the individual capacitances (2.70µF + 4.00µF) and got a total capacitance of 6.70µF.

C = 6.70µF

For Q, I added the individual charges (475 V + 525 V) and got a total charge of 1000 V.

Q = 1000 V

I then applied the previously stated formula.

V=Q/C

V=1000 / 6.70µF

V≈149.25 V

[Are those units correct? Did I even do the problem right?]

For part (b), the first thing that came to mind was to divide my answer for part (a) in half.

V≈149.25 V/µF / 2

V≈74.63 V/µF

Is all that correct?

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