Capacitors and Power (two problems)

  • Thread starter JWSiow
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  • #1
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Homework Statement


When a battery is attached to a capacitor, why do the two plates acquire charges of the same magnitude? Will this be true if the two conductors are different in size or shape? Explain.

Homework Equations


Q=CV and C=epsilon x A/d

The Attempt at a Solution


I wasn't sure if my answer was correct. I said, assuming the two plates are the same size and parallel, they will have the same capacitance, hence carry the same charge (from C=QV, where V is also the same for both plates). Also, the current flowing through the plates will be of equal, individual charges. As C=epsilon x A/d, this will not be true between conductors of different size and shape. This is because the capacitance will change, and they will be able to store different amounts of charges, hence may have different magnitudes.


Homework Statement


When a lightbulb needs to be replaced, you have a choice between a 100-W bulb and a 75-W bulb. Which would draw more current? Which has the highest resistance?

Homework Equations


V=IR, P=IV=I2R=V2/R

The Attempt at a Solution


Wasn't very sure with this one. But, using the equations above, and assuming voltage is the same for both bulbs, the 100W bulb would draw more current, and since R=I/V, it would also have a larger resistance.
 

Answers and Replies

  • #2
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for question 1: You are talking about single capacitor in your problem. So there is only one capacitance not two in your circuit even though you change size/ shape of conductors.
The equation
[tex] C=\frac{\epsilon A}{d} [/tex]

holds for single capacitor. I don't know how you are considering it as two capacitors!

Question 2: You are right in saying 100W bulb draws more current. But how you said that it has large resistance? It is contradicting because more resistance allows less current!!!!
 
Last edited:
  • #3
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woops.. Thanks! I started calling the two plates capacitors :| and for Question 2, I just rearranged the equation in my head wrong (thought R=I/V instead of R=V/I)
 
  • #4
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For your first problem, you assume that both plates of the capacitor are initially uncharged. You will also assume that each terminal of the battery has zero initial charge. Using conservation of charge show that the charge on the plates of the capacitor are equal and opposite.
 
  • #5
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Okay, thanks!
 

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