A 100 pF capacitor is charged to 100 volts. After the charging battery is discon

Question

A 100 pF capacitor is charged to 100 volts. After the charging battery is disconnected, the capacitor is connected in parallel with another capacitor. If the final voltage is 30 volts, what is the capacitance of the second capacitor? How much energy was lost, and what happened to it?

My Work:

So, I said C = Q/V

Therefore, Q = CV = 100 x 10-12 x 100 = 1x10-8 Columbs.

Will Q remain constant when the other capacitor is connected?

Since the two capacitors are connected in parallel. That means that the final capacitance = The total charge Q / The voltage across the two capacitors

But how do I calculate the total V? Is it just the final 30 V?

Also, how do I find the energy lost?

Explanation / Answer

charge on the first capacitor is Q1 = C1*V1 = 100*10^-12*100 = 100*10^-10 C

energy stored is 0.5*C*V^2 = 0.5*100*10^-12*100*100 = 5*10^-7 J

let's say C2 be the second capacitor

then in parallel combination

C1+C2 = (100*10^-10) + C2

now charge on C1 is Q1= C1*V = 100*10^-12*30 = 30*10^-10 C

then charge on C2 is Q2 = C2*V = (100*10^-10)-(30*10^-10) = 70*10^-10

then C2 = 70*10^-10/30 = 2.333*10^-10 F = 233.3*10^-12 F = 233.3 pF

now energy stored is 0.5*(C1+C2)*V^2 = (0.5)(100+233.3)*10^-12*30^2 = 1.5*10^-7 J

lost in energy is (5-1.5)*10^-7 = 3.5*10^-7 J

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