A 15 pF capacitor is charged to 1 kV and then removed from the battery and conne

Question

                            A 15 pF capacitor is charged to 1 kV and
                            then removed from the battery and connected
                            in parallel to an uncharged 65 pF capacitor.
                            What is the new charge on the second                         

                            ca-pacitor?

                            Find the change in energy.                         

                            Answer in units of ?J                         

Explanation / Answer

Charge is conserved. Q = CV
Initial charge is 1kV x 15pF = 15 nC
new C = 15+65 = 80 pF
new V = Q/C =15 nC / 80 pF = 0.1875 kV
Charge on second Cap Q = CV = 0.1875 kV x 65 pF = 12.1875nC ?
Charge on first cap Q = 0.1875 kV x15 pF = 2.185 nC
they add up to 15 nC, which they should.

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