Charge and Voltage on Capacitors.

AI Thread Summary
The discussion focuses on calculating the charge and voltage across capacitors in a circuit with a given capacitance of 3 μF and a battery voltage of 4.0 V. There is confusion regarding the equivalent capacitance of capacitors in series and parallel, specifically the derivation of C23 as 2 μF and the charge distribution among the capacitors. Participants clarify that series capacitors share the same charge, contradicting the initial equation presented. The need for clear initial conditions and values is emphasized to resolve misunderstandings in the calculations. Overall, the conversation highlights the importance of correctly applying capacitor rules in circuit analysis.
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Homework Statement


Determine the charge on each capacitor and the voltage across ech, assuming C=3yF(micro) and the battery voltage is V=4.0 Vhttps://fbcdn-sphotos-h-a.akamaihd.net/hphotos-ak-xpf1/v/t34.0-12/11006374_10152682348496806_273021698509210726_n.jpg?oh=128f2d03e7e0ad275cb982e11d827c2c&oe=54F5E264&__gda__=1425403951_252b026b8105e21db6642c5d28a9f753

Homework Equations


so,
C23 = 2C/3 = 2microF
Qeq = Q1 + Q23 = Ceq V <--- why is Qeq is the sum of the series capacitors, i thought they had the same charge
Q eq = 8. micorC
Q1 = Q23 = 4.00 microC
V1= Q1/C = 1.33 V
V23 = V - V1 = 2.67 V
Q2 = Q3 = CV23 = 2.0 micro C <--- not sure how did they get this value, when I plug in the found valued I don't get 2.

The Attempt at a Solution

 
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The problem statement is unclear to me. You state that C = 3 μF, does that mean the circuit is fundamentally made up of a several 3 μF capacitors? If so, how do you arrive at C23 = 2 μF?

Capacitors in series, unless there are circuit changes when capacitors are already charged, should bear the same charge as you indicate.

Is the solution that you present as Relevant equations your solution or someone else's?
 
This is an example problem from one of my lectures, and I was not sure how they arrive to the solution.
Originally the C23 capacitor, was C2 and C3 capacitors in parrallel. They combined them into Ceq by additing them two.
 
Can you list all the initial capacitor values and the initial circuit conditions?
 
https://fbcdn-sphotos-h-a.akamaihd.net/hphotos-ak-xpf1/v/t34.0-12/11042003_10152682449531806_1029796180_n.jpg?oh=fcaa90dacb0f79caf862a01ccfe11094&oe=54F5C0C6&__gda__=1425387735_717d66c414ad18a91e2dbebad58aa77e
 
Okay, it's much more clear now that I can see the whole circuit. Their derivation of the equivalent capacitance looks fine.

In your first post you showed and questioned the step:

Qeq = Q1 + Q23 = Ceq V

which I agree is incorrect. Series connected capacitors should have the same charge. So what would be your version of a solution?
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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