Finding the Mistake in Calculating Charge on a Capacitor with a Shorted Wire

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The discussion centers on a problem involving two capacitors (2μF and 3μF) shorted by a wire, with a 6V drop across a 1μF capacitor. Participants agree that the charge on the 2μF capacitor should be zero due to the short circuit, leading to confusion over the provided answer options. The consensus is that the book's answer may be incorrect since the potential across the 2μF capacitor is indeed zero. The conversation highlights the importance of understanding circuit behavior in capacitor arrangements. The conclusion emphasizes the need for clarity in textbook solutions regarding capacitor charge calculations.
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Homework Statement


capacitor2.jpg


Homework Equations

The Attempt at a Solution


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Sorry for the unclear image .

There is a wire across the combination of 2μF and 3μF capacitors which means the two capacitors are shorted . The whole of 6V drops across the 1μF capacitor . In that case charge on the 2μF capacitor should be 0 . But this is not an option .

What is my mistake ?
 

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I think it's a wrong answer by the book honestly.
 
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Hydrous Caperilla said:
I think it's a wrong answer by the book honestly.

You mean none of the options are correct ?

Do you agree there would be 0 charge on 2μF capacitor ?
 
yea because the potential across the capacitor is 0
 
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Thanks !
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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