Capacitance: 24µF, 30V - Calculate Charge

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SUMMARY

The discussion focuses on calculating the charge of a 24µF capacitor with a potential difference of 30V. The correct formula to use is the definition of capacitance, expressed as C = q/U, where C is capacitance, q is charge, and U is voltage. By rearranging the formula to solve for charge, q = C * U, the charge can be calculated as 0.72 mC (milliCoulombs). This calculation is essential for understanding capacitor behavior in electrical circuits.

PREREQUISITES
  • Understanding of capacitance and its units (µF)
  • Basic knowledge of electric potential difference (voltage)
  • Familiarity with the formula C = q/U
  • Ability to perform unit conversions (microfarads to farads)
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Students studying physics, electrical engineers, and anyone interested in understanding capacitor functionality and calculations in electronic circuits.

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Homework Statement



A 24-µF capacitor has an electric potential difference of 30 V across it. What is the charge on the capacitor?

Homework Equations



kq1q2/r2?, q/change in V?

The Attempt at a Solution


ive tried looking through my notes but i was absent don't understand this if you could give me a step by step i would appreciate it.
 
Last edited:
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Hi isuck@physics, welcome to PF.
The relevant equation given by you is wrong. Search for the relation between capacitance , potential difference and the charge in a capacitor.
 
Do you really understand the concept of capacitance?
Just put the figures into the formula of definition
[tex]\[<br /> C = \frac{q}{U}<br /> \][/tex]
 

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