Potential difference across each capacitor

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w3390
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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?
 
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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?

Trust your math.
 


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?

Trust your math.

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:


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.:bugeye: