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
- billrussell42Lv 78 years agoFavorite 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.
- RickLv 78 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.