anonymous
  • anonymous
The capacitance of a variable radio capacitor can be changed from 50PF to 95PF by turning the dial from 0 to 180 degrees. With the dial set at 180 degrees , the capacitor is connected to a 400V battery. After charging the capacitor is disconnected from the battery and the dial is turned to 0 degrees. a - What is the charge on the capacitor b- What is the p.d across the capacitor when the dial reads 0 degrees. c - WHat is the energy of the capacitor in this position?
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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radar
  • radar
a. Q=CV=(95 10^-12)(400) = 38 10^-9 coulomb b.760 volts c. Use approriate equation (I don't have it memorized)
anonymous
  • anonymous
@radar how did you arrive at your answer pls explain @experimentX @RaphaelFilgueiras @Algebraic! any info
anonymous
  • anonymous
which part?

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anonymous
  • anonymous
c) E=1/2 C V²
anonymous
  • anonymous
@woleraymond are you there?
anonymous
  • anonymous
yep im here
anonymous
  • anonymous
what you didnt understood?
anonymous
  • anonymous
how did @radar arrive at 760 volts for question b..also what impact does the turning have on the capacitor
anonymous
  • anonymous
the charge will be the same,when you turn the dial to zero degrees ,you will change the capacitance for 50 PF
anonymous
  • anonymous
then q=c.v-> 38E-9=50E-12.v->v=760
anonymous
  • anonymous
now energy will be 1/2.c.v²=1/2. 50E-12.(760²)
anonymous
  • anonymous
14.4 microjoules
anonymous
  • anonymous
interesting....

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