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GR0177 #10
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Electromagnetism}Capacitors

The problem gives and and V=300 V.

The capacitors are connected in series, and thus . (Recall that series and parallel capacitor formulae are opposite to that of resistors.)

Energy is , as in choice (A).  Alternate Solutions
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walczyk
2011-04-06 23:16:56
off the subject but how do you find the voltages over the two capacitors?
 busy32011-11-08 14:14:54 V1=Q1/C1 , V2=Q2/C2, we know the C's and Q1=Q2 (in series). We can eliminate Q from the equations and get a ratio. We also know V1+V2 =300 h0dgey84bc
2007-09-21 05:14:03
if C=2E-3 F, and V^2=9E4, then surely 0.5*C*V^2 is

=1E-3*9E4
=90J

not 0.09J, where does the 1E-6 come from in your answer, confused
 h0dgey84bc2007-09-21 05:15:35 sorry i read micro as milli, my apologies      LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces . type this... to get...$\int_0^\infty\partial\Rightarrow\ddot{x},\dot{x}\sqrt{z}\langle my \rangle\left( abacadabra \right)_{me}\vec{E}\frac{a}{b}\$