Question

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?

Answer 1

a. Q=CV=(95 10^-12)(400) = 38 10^-9 coulomb
b.760 volts
c. Use approriate equation (I don't have it memorized)

Answer 2

@radar how did you arrive at your answer pls explain

@experimentX @RaphaelFilgueiras @Algebraic! any info

Answer 3

which part?

Answer 4

c) E=1/2 C V²

Answer 5

@woleraymond are you there?

Answer 6

yep im here

Answer 7

what you didnt understood?

Answer 8

how did @radar arrive at 760 volts for question b..also what impact does the turning have on the capacitor

Answer 9

the charge will be the same,when you turn the dial to zero degrees ,you will change the capacitance for 50 PF

Answer 10

then q=c.v-> 38E-9=50E-12.v->v=760

Answer 11

now energy will be 1/2.c.v²=1/2. 50E-12.(760²)

Answer 12

14.4 microjoules

Answer 13

interesting....