A capacitor consists of two closely spaced metal conductors of large area, separ

Question

A capacitor consists of two closely spaced metal conductors of large area, separated by a thin insulating film. One of the many practical uses of a capacitor is to store up some energy, to be released in a sudden burst. A decent (not too expensive) capacitor in such a camera might be 200 uF, and could release a total electrical energy of 11.6 J in a flash.

A. What voltage difference would you need across this capacitor? (I got this question right. The answer is 340 V).

I NEED HELP FOR PARTS B AND C TO THIS QUESTION:

B. How much charge is stored on the positive plate of the capacitor?

C. If the two plates of the capacitor in the previous question have their separation increased by a factor of 3 while the charge on the plates remains constant, by what factor is the energy stored in the capacitor increased?

Explanation / Answer

C = 200uF
Energy = 11.6J

A)
Energy = 0.5 * C* V^2
11.6 * 2 = 200 * 10^-6 V^2
V = 340 Volt

B)
We Know, Q = C* V
Q = 200 uF * 340 v
Q = 68 mF
Charge stored on the positive plate of Capacitor, Q = 68mF

(C)
Capacitance = (A*Eo) /d
Where d = seperation between plates
If new distance = 3*d
Then Capacitance will become small by a factor of 3

New Capacitance = C/3

Q = C *V
Q = C/3 * V
V = 3Q / C
This means potential is increase by a Factor of 3.

Initial Energy = 0.5 * C * V^2 = 0.5 * C * (Q/C)^2 = 0.5 Q^2/C

Final Energy = 0.5 * C/3 * (3Q/C)^2
Final Energy = 3 * 0.5 *Q^2/C
Final Energy = 3 * Initial Energy

IT IS INCREASED BY A FACTOR OF 3.

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