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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
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a. Q=CV=(95 10^-12)(400) = 38 10^-9 coulomb b.760 volts c. Use approriate equation (I don't have it memorized)
@radar how did you arrive at your answer pls explain @experimentX @RaphaelFilgueiras @Algebraic! any info
which part?

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Other answers:

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

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