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# Capacitor question Watch

1. Question:

I'm facing a problem in part (b). Half of the charge is equal to 0.06 C - so ½QV(½ times 0.06 times 24 = 0.72 J) should give the correct value of energy dissipated in the bulb. But the mark scheme gives 1.08 J as the answer. Where am I going wrong?
2. (Original post by Zishi)
Question:

I'm facing a problem in part (b). Half of the charge is equal to 0.06 C - so ½QV(½ times 0.06 times 24 = 0.72 J) should give the correct value of energy dissipated in the bulb. But the mark scheme gives 1.08 J as the answer. Where am I going wrong?
How to do this...
Initial energy on capacitor is 0.5CV2 where V is 24V
After half the charge leaves, the voltage will be 12V (V=Q/C)
New energy in capacitor is 0.5CV2 where V=12V
Subtract the 2nd from the 1st for the energy in the bulb
3. (Original post by Stonebridge)
How to do this...
Initial energy on capacitor is 0.5CV2 where V is 24V
After half the charge leaves, the voltage will be 12V (V=Q/C)
New energy in capacitor is 0.5CV2 where V=12V
Subtract the 2nd from the 1st for the energy in the bulb
Alright. So the voltage drops to half because the charge drops to half (according to Q=CV)?
4. (Original post by Zishi)
Alright. So the voltage drops to half because the charge drops to half (according to Q=CV)?
Absolutely. The pd on a capacitor is directly proportional the the amount of charge on it.
5. (Original post by Stonebridge)
Absolutely. The pd on a capacitor is directly proportional the the amount of charge on it.
Thank you so much. (PRSOM)

Updated: April 19, 2012
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