A capacitor is completely charged with 660 nC by a voltagesource that had 300 V.
ID: 1669119 • Letter: A
Question
A capacitor is completely charged with 660 nC by a voltagesource that had 300 V. Now the plates of the charged capacitor arepulled apart with the voltage source alreadydisconnected.1. The charge on the plates increases.
2. The voltage drop between the platesincreases.
3. The capacitance increases.
4. The energy stored in the capacitor remains thesame.
5. None of the above. The initial air gap of thecapacitor above was 8 mm. What is the stored energy if the air gapis now 10 mm?
A capacitor is completely charged with 660 nC by a voltagesource that had 300 V. Now the plates of the charged capacitor arepulled apart with the voltage source alreadydisconnected.
1. The charge on the plates increases.
2. The voltage drop between the platesincreases.
3. The capacitance increases.
4. The energy stored in the capacitor remains thesame.
5. None of the above. The initial air gap of thecapacitor above was 8 mm. What is the stored energy if the air gapis now 10 mm?
Explanation / Answer
We want to know how the capacitor changes as we increase thedistance between the plates. First, you know the initialcapacitance of the plates: C = Q/Vwhere C is the capacitance (what you're solvingfor) and Q is the charge stored in the capacitor (660 nC) and V isthe potential drop (the voltage of the source, 300 V) For a parallel plate capacitor, we know an equation that relatesdistance to capacitance: C = Ao/d The area A doesn't change and o is aconstant, so you can see how much C changes if youincrease d. You know d is initially 8 mm, and itincreases to 10 mm. So the new d is 10/8, or 1.25 timeswhat it used to be. This will help you determine if (3) is true orfalse. To check if (1) or (2) are true, you have an equation for therelationship between capacitance C, charge Q, andpotential (voltage drop) V. C = Q/VYou'll need to think about this a bit more. Thecapacitor was charged by a battery, so it has some chargeQ. Is there any reason why the charge will change if youincrease the distance between the plates? As for the voltage drop, voltage drop is the same thing aselectrical potential. Electrical potential changes when theelectric field changes. Does the electric field between the plateschange if you pull them apart? Finally, to check (4), we know the work that a capacitor can do(the energy stored in the capacitor) is given by: W = 1/2CV2At this point you should know howthe capacitance changed (or didn't change) and how the voltage dropchanged (or didn't change), so you can find out how workchanges. Hope that makes sense. Let me know if you need additional helpsolving the problem.
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