Charge on 6.4 µF Capacitor | 47.0 V Emf Source

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SUMMARY

The charge on a 6.4 µF capacitor connected to a 47.0 V EMF source is calculated using the formula Q = C(ΔV). By substituting the values, the charge Q equals 6.4 µF multiplied by 47.0 V, resulting in a charge of 0.0002992 coulombs or 299.2 µC. This calculation confirms that the capacitor will hold this charge after being connected for a long time.

PREREQUISITES
  • Understanding of capacitor fundamentals
  • Familiarity with the formula Q = C(ΔV)
  • Knowledge of microfarads (µF) as a unit of capacitance
  • Basic principles of electric potential (voltage)
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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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Just an easy one!
 
thats what i thought. thanks!
 

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