Charge on 6.4 µF Capacitor | 47.0 V Emf Source

Thread starter phy112
Start date Feb 24, 2009
Tags

Capacitor Charge

AI Thread Summary

To find the charge on a 6.4 µF capacitor connected to a 47.0 V emf source, the formula Q = C(deltaV) is used, where Q is the charge, C is the capacitance, and deltaV is the voltage. Plugging in the values, the charge can be calculated as Q = 6.4 µF * 47.0 V. This straightforward application of the formula yields the charge on the capacitor after it has been connected for a long time. The discussion emphasizes the simplicity of the calculation involved. Understanding the relationship between capacitance, voltage, and charge is crucial for solving similar problems.

Feb 24, 2009

phy112

Homework Statement

What is the charge on the 6.4 µF capacitor if the emf source between terminals A and B is 47.0 V, and it remains connected for a long time?

Homework Equations

Q=C(deltaV)

The Attempt at a Solution

Do i just plug and chug or is there more to it?

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Feb 24, 2009

Delphi51

Homework Helper

Just an easy one!

Feb 24, 2009

phy112

thats what i thought. thanks!

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