Nathan's answer is wrong, because 1/2 the charge is lost in the (nonzero) resistance of the wire that hooks the caps together. Other than that he has the right idea.
discover the preliminary value on the 1st capacitor. whilst the battery is disconnected, that lots value is trapped on the plates. whilst that's linked to the different capacitor, a number of that value flows onto the plates of the different capacitor. This keeps until the two capacitors have the comparable...what? (hint, they are linked to a minimum of one yet another by making use of conductors.) you're able to get 2 equations in 2 unknowns, the quantity of value on each. sturdy success!
Decaying to what fee? via certainty the poster above reported, perfect capacitors by employing no ability decay to 0. A capacitor decays to 36% (a million/e) of its preliminary fee in time T=RC. on your case, if it decayed to (.36)(30V) = 10.8V in T=3s, the resistance is R=T/C = 150k? yet in case you think approximately that we will no longer see your graph, we don't understand what fee it decayed to in 3s.
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Nathan's answer is wrong, because 1/2 the charge is lost in the (nonzero) resistance of the wire that hooks the caps together. Other than that he has the right idea.
I am not sure what you are asking here but I will take a guess.
You want to connect these capacitors together and find out what the voltage is after they connect.
If I am wrong you may find this useful in the future
Since charge is conserved then the Q after connection is the sum of the Q's before connection.
So add the Q values and add the capacitors.
C1 = 6E-6 c
C2 = 4E-6 c
V1 = 50V
V2 = 34V
solve for Q1
= C1*V1
= (6E-6)*(50)
= 300E-6 c
solve for Q2
= C2*V2
= (4E-6)*(34)
= 136E-6 c
Since Q = C*V
V = Q/C
V = (Q1 + Q2)/(C1 + C2)
V = (300E-6 + 136E-6)/(6E-6 + 4E-6)
V = (436E-6)/(10E-6)
V = 43V after the capacitors are connected
discover the preliminary value on the 1st capacitor. whilst the battery is disconnected, that lots value is trapped on the plates. whilst that's linked to the different capacitor, a number of that value flows onto the plates of the different capacitor. This keeps until the two capacitors have the comparable...what? (hint, they are linked to a minimum of one yet another by making use of conductors.) you're able to get 2 equations in 2 unknowns, the quantity of value on each. sturdy success!
Decaying to what fee? via certainty the poster above reported, perfect capacitors by employing no ability decay to 0. A capacitor decays to 36% (a million/e) of its preliminary fee in time T=RC. on your case, if it decayed to (.36)(30V) = 10.8V in T=3s, the resistance is R=T/C = 150k? yet in case you think approximately that we will no longer see your graph, we don't understand what fee it decayed to in 3s.
jumper wire ?
u in nyp ?