# Potential difference across each capacitor

• w3390
In summary, the correct solution for this problem is to use the equation C=Q/V, where Q is the total charge of 1 micro-Coulomb and V is the potential difference of 10V. This results in a total capacitance of 1*10^-5 F. Therefore, 10 1 microF capacitors connected in parallel would be needed to store the total charge of 1 mC.
w3390

## Homework Statement

How many 1 microF capacitors connected in parallel would it take to store a total charge of 1 mC if the potential difference across each capacitor is 10.0V?

## Homework Equations

C=Q/V
in parallel --> Ceq=C1+C2+C3...

## The Attempt at a Solution

I attemped this problem by first using the equation C=Q/V, using 1mC as my Q and 10V as my V. This gave me a total capacitance of 1*10^-4 F. I then divided this number by the capacitance of a single capacitor, 1*10^-6. This result showed that it would take 100 capacitors. I thought this answer seemed reasonable, but the next sentence in the problem tells me to diagram the parallel combination. I don't think they want me to draw out 100 capacitors in parallel. Can anyone double check my answer and tell me if it is correct?

If you can write an equation like "Ceq=C1+C2+C3... skip a few +C100", what is your hesitancy in making a drawing like that?

Hi w3390!
w3390 said:
… if the potential difference across each capacitor is 10.0V?

erm each capacitor!

LowlyPion said:
If you can write an equation like "Ceq=C1+C2+C3... skip a few +C100", what is your hesitancy in making a drawing like that?

If I do trust my math, I will be drawing out 100 capacitors on parallel on my sheet of paper. Does this sound reasonable to you?

w3390 said:
If I do trust my math, I will be drawing out 100 capacitors on parallel on my sheet of paper. Does this sound reasonable to you?

Sounds tedious when an ellipsis will do.

Does this mean that the potential difference across the entire combination will be 1000V by multiplying the number of capacitors by the drop across each one?

Never mind, I figured it out. Thanks.

Last edited:
Yes, but it's not 1000V … call the number of capacitors n, and see how many times n comes into the equation.

I'm getting 1x10^-5 F for total capacitance. So 10 of the 1microF capacitors in parallel. Assuming you meant "total charge of 1mC" to mean micro-Coulomb and not milli-Coulomb. Otherwise it's like 10,000 capacitors.

## 1. What is potential difference across each capacitor?

Potential difference across each capacitor refers to the difference in electrical potential between the two plates of a capacitor. This difference in potential creates an electric field which allows the capacitor to store electrical energy.

## 2. How is potential difference across each capacitor measured?

Potential difference across each capacitor is typically measured using a voltmeter. The voltmeter is connected in parallel to the capacitor, and the reading on the voltmeter indicates the potential difference between the two plates of the capacitor.

## 3. What factors affect potential difference across each capacitor?

The potential difference across each capacitor is affected by the capacitance of the capacitor, the amount of charge stored on the plates, and the distance between the plates. It is also affected by the voltage applied to the capacitor and the dielectric material between the plates.

## 4. Why is potential difference across each capacitor important?

Potential difference across each capacitor is important in understanding the behavior of capacitors in electrical circuits. It determines the amount of charge that can be stored on the plates and the energy that can be stored in the capacitor. It also plays a role in determining the capacitance and the overall function of the capacitor in a circuit.

## 5. Can potential difference across each capacitor be negative?

Yes, the potential difference across each capacitor can be negative. This occurs when the polarity of the voltage source is reversed, causing the electric potential on one plate to be higher than the other. In this case, the capacitor will discharge instead of charging, and the potential difference will be negative.

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