Capacitance: 24µF, 30V - Calculate Charge

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In summary, the conversation discusses a problem involving a 24-µF capacitor with an electric potential difference of 30 V. The person is seeking help in determining the charge on the capacitor and is asked to review the relevant equation and understand the concept of capacitance before attempting the problem. The suggested formula is C = q/U.
  • #1
isuck@physics
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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.
 
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  • #2
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.
 
  • #3
Do you really understand the concept of capacitance?
Just put the figures into the formula of definition
[tex]\[
C = \frac{q}{U}
\]
[/tex]
 

Related to Capacitance: 24µF, 30V - Calculate Charge

1. What is capacitance?

Capacitance is the ability of a system to store an electric charge. It is measured in farads (F).

2. What is the significance of 24µF and 30V in capacitance?

24µF (microfarads) is the value of the capacitance, or the maximum charge that can be stored. 30V (volts) is the maximum voltage that can be applied to the capacitor before it breaks down.

3. How do you calculate the charge of a capacitor with a capacitance of 24µF and a voltage of 30V?

The charge of a capacitor can be calculated by multiplying the capacitance (in farads) by the voltage (in volts). In this case, the charge would be 24µF x 30V = 720 µC (microcoulombs).

4. What is the relationship between capacitance, voltage, and charge?

The relationship between capacitance, voltage, and charge can be described by the equation Q = CV, where Q is the charge (in coulombs), C is the capacitance (in farads), and V is the voltage (in volts). This equation shows that the charge stored in a capacitor is directly proportional to the capacitance and voltage.

5. How does capacitance affect the performance of electronic circuits?

Capacitance is a crucial factor in the design and performance of electronic circuits. It can affect the time it takes for a circuit to charge and discharge, and it can also affect the stability and frequency response of the circuit. Proper consideration of capacitance is essential for the efficient operation of electronic devices.

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