Finding the cafpacitance, I thought it was in series

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In summary, the capacitance of a parallel plate capacitor is dependent on the charge on the plates. If the charges are the same, the capacitance is 16.14 pF. If the charges are changed to +226 pC and -226 pC, the capacitance becomes 4.429 pF and V becomes 14 V.
  • #1
mr_coffee
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Hello everyone...I'm runnning int problems on this problem:
he two metal objects in Fig. 25-26 have net charges of +62 pC and -62 pC, which result in a 14 V potential difference between them.
image is here: http://www.webassign.net/hrw/hrw7_25-26.gif
(a) What is the capacitance of the system?
4.429 pF
(b) If the charges are changed to +226 pC and -226 pC, what does the capacitance become?
wrong check mark pF
(c) What does the potential difference become?
V

If the charges are the same, i thought that ment it would be in series, which is Q = CVtotal
So i put in:
C = 226pC/14V = 16.14 pF
I also tried 0, both wrong. ANy ideas? Thanks
 
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  • #2
This isn't a question about series capacitors. The two metal objects form a single capacitor. Is the capacitance of a parallel plate capacitor dependent on the charge on the plates? Is the capacitance of a general capacitor dependent on the charge?
 
  • #3
The capcitance of a parallel plate capacitor is dependent on the charge on the plates because the formula is. Q total = V(Ceq). Now is the capacitence of a general capacitor depedent on the charge? I"m assuming yes, the general formula is Q = CV...
 
  • #4
The capacitance of a parallel plate capacitor is [tex] C = \epsilon_0 A /d [/tex] with no mention of charge. The capacitance of a system is dependent only on the geometry which is partly why it's a useful concept. In the equation [tex] Q = CV [/tex], any change in Q is accompanied by a change in V so that C is always the same (so long as the geometry is the same). There lies the answer to your question. Once you know the capacitance of your system, it's always the same unless you change the geometry.
 
  • #5
ohhh i c what your saying, thank you very much, i got the problem now! :)
 

1. What is capacitance and how is it measured?

Capacitance is the ability of a material to store an electric charge. It is measured in units of farads (F) and can be calculated by dividing the charge by the voltage. Capacitance is typically measured using a device called a capacitor meter.

2. How do you determine if capacitance is in series or parallel?

In a series circuit, the capacitors are connected one after the other, while in a parallel circuit, they are connected side by side. To determine if capacitance is in series, you can use the following formula: Ceq = C1 + C2 + ... + Cn where Ceq is the equivalent capacitance and C1, C2, etc. are the individual capacitances in the circuit.

3. How do you find the total capacitance in a series circuit?

In a series circuit, the total capacitance is equal to the inverse sum of the individual capacitances. This can be expressed as: 1/Ceq = 1/C1 + 1/C2 + ... + 1/Cn. To find the total capacitance, simply take the inverse of this value.

4. What is the significance of capacitance in electronic circuits?

Capacitance plays a crucial role in electronic circuits as it allows for the storage and release of electrical energy. It is used in a variety of applications, such as smoothing power supplies, filtering out unwanted signals, and tuning radio frequencies. Capacitors also help to regulate the flow of current and provide stability to circuits.

5. How does the placement of capacitors affect the overall capacitance in a circuit?

In a series circuit, the overall capacitance decreases as more capacitors are added. This is because the total capacitance is equal to the inverse sum of the individual capacitances. In a parallel circuit, the overall capacitance increases as more capacitors are added. This is because the total capacitance is equal to the sum of the individual capacitances. Therefore, the placement of capacitors in a circuit has a significant impact on the overall capacitance.

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