a 40 pF capacitor is charged to 5 kV and is then removed from the battery?

and connected in parallel to an uncharged 70 pF capacitor. What is the new charge on the second capacitor

2 Answers

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  • 8 years ago
    Best Answer

    Charge is conserved. Q = CV

    Initial charge is 5kV x 40pF = 200 nC

    new C = 40+70 = 110 pF

    new V = Q/C = 200 nC / 110 pF = 1.82 kV

    Charge on second Cap Q = CV = 1.82 kV x 70 pF = 127.3 nC ⬅

    Charge on first cap Q = 1.82 kV x 40 pF = 72.7 nC

    they add up to 200 nC, which they should.

  • Rick
    Lv 7
    8 years ago

    Actually, half the energy is lost in the wire resistance between the 2 caps. Even thought the resistance is small, it is finite.

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