Capacitors: find he charge for the new battery

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Homework Help Overview

The problem involves a capacitor's charge when connected to different voltage sources, specifically comparing the charge stored at 6.0 V and 9.0 V. The subject area is related to electric circuits and capacitance.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between capacitance, charge, and voltage, with attempts to derive equations that connect these variables. Some participants suggest solving for capacitance first and then rearranging to find charge.

Discussion Status

There are multiple interpretations of the problem, with some participants providing guidance on relevant equations and relationships. A few participants share their calculated results, indicating a collaborative exploration of the problem.

Contextual Notes

Participants note the significance of precision in their answers, referencing the significant figures of the given data. There is an acknowledgment of potential complexity in the problem setup.

Alice7979
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Homework Statement


A capacitor stores 7.9 x 10-2 C of charge when connected to a 6.0-V battery. How much charge does the capacitor store when connected to a 9.0-V battery?

Homework Equations


PE=CQ/2

The Attempt at a Solution


QV^2=CQ^2
(7.9 x 10-2)(6)/2 = Q(9)
C= .05 C
 
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You are looking for an equation that relates capacitance, charge held on the plates, and the voltage across the plates.
1. Solve for capacitance.
2. Rearrange the equation and solve for charge.
 
I guess I made that more complicated
 
If you want to post the answer you came up with, I will see if it matches mine.
 
.1185 C
 
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Likes   Reactions: David Lewis
That's what I got. The precision of the given data is 2 significant figures, so I rounded to 0.12 C.
 
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Likes   Reactions: scottdave

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