Charge on 6.4 µF Capacitor | 47.0 V Emf Source

In summary, a 6.4 µF capacitor is used to store electrical charge and regulate current in a circuit. A 47.0 V EMF source affects the charge on the capacitor by providing a potential difference that pushes electrons towards it. When the EMF source is disconnected, the capacitor will maintain its charge due to its insulating properties. The charge on a 6.4 µF capacitor can be increased by increasing the EMF source or connecting multiple capacitors in series or parallel. The charge on a 6.4 µF capacitor can be calculated using the formula Q = CV, where Q is the charge in coulombs, C is the capacitance in farads, and V is the voltage in volts.
  • #1
phy112
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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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  • #2
Just an easy one!
 
  • #3
thats what i thought. thanks!
 

1. What is the purpose of a 6.4 µF capacitor?

A 6.4 µF capacitor is used to store electrical charge and regulate the flow of current in a circuit. It acts as a temporary energy storage unit that can release its stored charge when needed.

2. How does a 47.0 V EMF source affect the charge on the capacitor?

The 47.0 V EMF (electromotive force) source provides a potential difference that pushes electrons towards the capacitor, causing it to accumulate a charge. The higher the EMF, the more charge the capacitor can hold.

3. What happens to the charge on the capacitor when the EMF source is disconnected?

When the EMF source is disconnected, the capacitor will maintain its charge. This is because the capacitor acts as an insulator and prevents the flow of electrons, keeping the charge in place.

4. Can the charge on a 6.4 µF capacitor be increased?

Yes, the charge on a 6.4 µF capacitor can be increased by either increasing the EMF source or by connecting multiple capacitors in series or parallel. This will increase the total capacitance and therefore, the amount of charge that can be stored.

5. How is the charge on a 6.4 µF capacitor calculated?

The charge on a capacitor can be calculated using the formula Q = CV, where Q is the charge in coulombs, C is the capacitance in farads, and V is the voltage in volts. In this case, the charge on the 6.4 µF capacitor can be calculated by multiplying its capacitance (6.4 µF = 6.4 x 10^-6 F) with the voltage of the EMF source (47.0 V).

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