wh U2-Oul-Chap 24 \\ G physics question I Cheo E myLTL-Blackboard Lee x / -? www
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wh U2-Oul-Chap 24 G physics question I Cheo E myLTL-Blackboard Lee x / -? www.webassi x, G For the circuit shown in 1 x 12 points Tiplerß 24 P056 My Nates Ask Your A 4 ??Lapacitor dric a?dpacitor o unknown cap titance dre both charged o 2.75 kV After charging he woudpacitors are disconnected rom he voltage source. The ca duitors are then connected to each other positive plate to negative plate and negative plate to positive plate. The final voltage across the terminals of the 14-?F capacitor is 1.00 kV. (a) What is the capacitanoe of the second capacitor? (b) How much energy Is dissipated when the connection was made? eBook Submit Answer Save Progrss Practice Another Version 5. -12 points Tipler8 21.079.scln. My Notes Ask Your Tes A para el combination of two Identical 4.30 ?F parallel piate capacitors is connected o the charge on each of the capacitors. capacitor with doubled separation capacitor with the orlglnal separatlon 75 V battery. The battery is then removed and the separation between the plates ot one of he capacitors is doubled. Find HC HC eBook Subiriil Assignuient Save Assignrnenl Progresss Ilrme My Assignments Cxtension Reques! 3:43 PM 4/6/2018Explanation / Answer
(a)
Q = CV,
and we know the same charge is present on the capacitor(s) before and after they are connected because the battery has been disconnected. So if C is the unknown, we have:
Q = (2.75*103 V)C = (1*103V)(14 * 10-6 F + C)
2750 *C = 14*10-3 + 1000*C
1750*C = 14*10-3
C = 8*10-6 F = 8 uF
(b)
Ctotal = 8uF +14uF = 22 uF
energy dissipated
E = 1/2*C*V2
E = 0.5*22*10-6*(2.75*103)2
E = 83.18 J
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