Capacitor negative charge question

[Jamessamuel]

Homework Statement

(See image)

2. Homework Equations
C=Q/V

The Attempt at a Solution

Part a took a while but I now understand that the electrons gather on the right plate of the 2μF capacitor and on the right plate of the 4μF capacitor. This imparts a negative charge of 133μC on each capacitor.

Part b is where I cannot justify the answer. The charge should flow to equalise the pd's, that's All I know.

Part c is fine.

 

[cnh1995]

Homework Statement

(See image)

2. Homework Equations
C=Q/V

The Attempt at a Solution

Part a took a while but I now understand that the electrons gather on the right plate of the 2μF capacitor and on the right plate of the 4μF capacitor. This imparts a negative charge of 133μC on each capacitor.

Part b is where I cannot justify the answer. The charge should flow to equalise the pd's, that's All I know.

Part c is fine.

Your idea for part b is correct..But here the voltage polarities of the capacitors will be additive (since they were charged serially). So when the current will flow, voltages of both the capacitors will reduce..when would they be equal?? Use your idea..

 

[Jamessamuel]

I believe that the voltages will reduce because energy will be lost across the resistor. As voltage is charge per unit capacitance, the capacitor with twice the size will store twice as much as the other. And so on that basis, roughky 88 micro C will end up with the largest capacitor and 44 on the smallest. In order to achieve this, 44 micro C must flow clockwise from the largest to the smallest capacitor.

 

[CWatters]

The lower voltage capacitor will drop to 0V first and discharging will continue till the second one drops to 0V.

Err how does that work? They are in series. If charge continues to flow after one has reached 0V it will become charged with the opposite polarity.

 

[CWatters]

Part A

100V is applied. Same charge flows through both capacitors until the combined voltage = 100V

Q = C₁V₁ = C₂V₂ ...(1)

V₁+V₂ = 100 ......(2)
or
V₁ = 100 - V₂

Substitute for V₁ in (1)

C₁ * (100 - V₂) = C₂V₂
Rearranging, substituting values and solving gives...

V₂ = 66.667V (voltage on the 2μF)
V₁ = 33.333V (voltage on the 4μF)
and then
Q = 133μC

Part B

The same charge (133μC) flows out of both capacitors and through the resistor because they are all in series. Both caps reach 0V at the same time.
.
Part C

What was your answer?
PS: If in any doubt why not replace the two caps with the equivalent single capacitor before you start answering Part A. It makes the calculations a lot simpler :-)

Equivalent single cap has value...

1/(1/2+1/4) = 1.333μF

Q = VC
=100V * 1.333μF
= 133.333μC (same as before)

 

[cnh1995]

So all the charge stored (133 μC) will be lost (by capacitors) in the form of current and capacitors will be completely de-energized..

 

[CWatters]

Yes. Try redrawing the circuit with the two caps replaced by one equivalent capacitor.