Here is a link to the question: http://prntscr.com/amgop5
I had a bit of trouble with this question, so what I deduced from increasing the charge by 50% is that the energy E must also increase in turn, as seen in this equation E = 1/2QV, so am I right in saying that the voltage V is constant and should remain unaffected by the increase of the charge? However I do not understand how the value of energy is is 2.25 E? The correct answer is C by the way. Thanks in advance!
Capacitance - Unit 4 Physics! Help Please. Watch
- Thread Starter
- 31-03-2016 18:41
- Official Rep
- 02-04-2016 20:03
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Just quoting in Puddles the Monkey so she can move the thread if needed
Spoiler:Show(Original post by Puddles the Monkey)
- 02-04-2016 21:34
Q = CV so p.d. is proportional to Q.
If charge is increased by a half that is a factor of 1.5 that V will increase by
E = ½ QV = ½ CV2 = ½ Q2/CE is proportional to Q2
If charge is increased by a half that is a factor of 1.5 so energy will increase by 1.52 = 2.25