Capacitors help!

Original post by hhattiecc

I will be INFINITELY grateful to anyone who can help me understand why :-)

I agree with your reasoning. I think the given answer is incorrect.

Reply 2

hhattiecc

OP

9

Original post by atsruser

I agree with your reasoning. I think the given answer is incorrect.

Ok, thank you! :smile:

Reply 3

Original post by hhattiecc

Just done a past paper question on a fully charged capacitor connected to an uncharged capacitor (so the two are in parallel with 0 circuit resistance). The initially uncharged capacitor has a capacitance 1000 times greater than the charged one. It asks you to explain why the charge remaining on the initially charged capacitor after the two are connected is Q/1000, where Q is its initial charge before being connected to the circuit.

I don't really understand why.
My understanding so far is that the two capacitors are in parallel, and so the potential difference over both, V, will be the same.
V=Q/C, so Q1/C1=Q2/C2, and rearranging gives Q1/Q2=C1/C2, so the ratio of charge stored is the same as the ratio of the capacitances. In other words, Q1/Q2=1/1000 (*) (Q1 is the initially charged capacitor and Q2 is for the initially uncharged capacitor).

Charge is conserved, and so Q1 + Q2 = Q.
However, we know from (*) that Q2=1000Q1.
Substituting that in: 1001Q1=Q.
So why is Q1, the charge remaining on the initially charged capacitor, Q/1000 not Q/1001?

I will be INFINITELY grateful to anyone who can help me understand why :-)

Original post by atsruser

I agree with your reasoning. I think the given answer is incorrect.

Flawed logic, the question is correct.

Q/1001C represents the p.d. across the combined capacitance of (C + 1000C). i.e. the shared p.d. must fall from the initial p.d. across the initial charged capacitor.

The question asks for the charge remaining on the initially charged capacitor only and not the p.d. developed across the combined capacitance after charge sharing.

i.e.

if

Q_0

= initial charge and

Q_t

= final charge, then:

Q_0 = Q_t(1 + \frac{1}{1000})

Q_0 = Q_{t} + \frac{Q_t}{1000}

Charge remaining on the smaller capacitance is the difference between the initial charge and charge remaining on the larger capacitance:

Q_0 - Q_t = \frac{Q_t}{1000}

(edited 8 years ago)

Reply 4

Original post by uberteknik

Q/1001C represents the p.d. across the combined capacitance of (C + 1000C). i.e. the shared p.d. must fall from the initial p.d. across the initial charged capacitor.

The question asks for the charge remaining on the initially charged capacitor only and not the p.d. developed across the combined capacitance after charge sharing.

This is true but I'm not sure why you think it is relevant here.

if

Q_0

= initial charge and

Q_t

= final charge, then:

Q_0 = Q_t(1 + \frac{1}{1000})

How did you derive this?

Charge remaining on the smaller capacitance is the difference between the initial charge and charge remaining on the larger capacitance:

Q_0 - Q_t = \frac{Q_t}{1000}

This has confused me now. Is

Q_t

the charge on the larger capacitor, or the final charge on the smaller capacitor?

It's not impossible that I'm being dimwitted here, but as yet, I'm not convinced that I'm wrong. A bit of googling turns up:

http://physics.stackexchange.com/questions/21418/how-to-calculate-the-charge-and-the-potential-across-a-charged-capacitor-chargin

which agrees with the OP's result.

Are you using some strange definition of "charge remaining"? The result Q/1001 seems to be the only possible one for the idea that I have of "charge remaining"

Reply 5

Original post by atsruser

but as yet, I'm not convinced that I'm wrong.

Hi.

Apologies for not replying sooner, hangovers are a bum! :colone:

Let's try this a different way.

Q = CV

V = \frac{Q}{C}

.........(i)

We know that the larger capacitor is 1000x the capacity of the smaller initially charged capacitor.

i.e. in the ratio of

c

and

1000c

(NB Small (c) represents a normalised unit of capacitance.)

Equilibrium of charges will be achieved when the p.d. developed across both capacitors is the same:

V_{c} = V_{1000c}

.........(ii)

i.e. when (sub' i into ii)

\frac{Q_{c}}{c} = \frac{Q_{1000c}}{1000c}

(where

Q_c

is the charge on the smaller capacitor and

Q_{1000c}

is the charge on the larger capacitor after redistribution)

cross multiplying:

1000cQ_C = cQ_{1000c}

eliminate c

1000Q_c = Q_{1000c}

Q_c = \frac{Q_{1000c}}{1000}

(edited 8 years ago)

Reply 6

Original post by uberteknik

Q_c = \frac{Q_{1000c}}{1000}

Yes, this is true. However, it simply says that ratio of charge on the caps equals the ratio of the capacitances.

The question, according to the OP, says: "explain why the charge remaining on the initially charged capacitor after the two are connected is Q/1000, where Q is its initial charge before being connected to the circuit."

i.e in your notation, it asks us to show that

Q_c = \frac{Q}{1000}

.

This cannot agree with your result unless

Q_{1000c} = Q

which contradicts conservation of charge (since

Q=Q_c+Q_{1000c} \ne Q_{1000c}

unless

Q_c=0

)

The result that you have given is correct, but it answers the wrong question.

Reply 7

Original post by atsruser

Yes, this is true. However, it simply says that ratio of charge on the caps equals the ratio of the capacitances.

The question, according to the OP, says: "explain why the charge remaining on the initially charged capacitor after the two are connected is Q/1000, where Q is its initial charge before being connected to the circuit."

i.e in your notation, it asks us to show that

Q_c = \frac{Q}{1000}

.

This cannot agree with your result unless

Q_{1000c} = Q

which contradicts conservation of charge (since

Q=Q_c+Q_{1000c} \ne Q_{1000c}

unless

Q_c=0

)

The result that you have given is correct, but it answers the wrong question.

Be careful, we are dealing with ratios. If the smaller capacitor is 1 part, the bigger capacitor is 1000 parts, with a combined capacitance of 1001 parts.
i.e. divide the original charge by 1001 to get the unit charge, but this is distributed 1 part to 1000 parts between the two capacitors.

Put some numbers in:

let the smaller capacitor be 68uF
the larger one would be 68uF x 1000 = 0.068F

Then charge the smaller capacitor to say 600volts

Q = CV
Q = 68uF x 600V = 0.0408 coulombs

Now connect the two capacitors together:

New total capacitance is: 68uF + 0.068F = 0.068068F

Charge is conserved so the voltage on both capacitors must now be:

V = Q/C = 0.0408/0.068068 = 0.5995005994 volts.

which means the charge on each capacitor is:

(small cap) Q = CV = 68uF x 0.5994005994V = 40.75924076uC
(large cap) Q = CV = 0.068F x 0.5994005994V = 40.75924076mC

My original relationship was:

Q_0 = Q_{t} + \frac{Q_t}{1000}

Plugging in the values:

0.0408C = 40.75924076mC + 40.75924076mC/1000
0.0408C = 40.75924076mC + 40.7592406uC

i.e. total charge is conserved

The original charge is now distributed in the ratio of 1:1000, which is identical to the ratio of the capacitances 1:1000

I also said:

Q_c = \frac{Q_{1000c}}{1000}

again plugging in values:

40.75924076uC = 40.75924076mC/1000

This is commensurate with the relationship

\frac{Q1}{Q2} = \frac{C1}{C2}

which in my notation

\frac{Q_c}{Q_{1000c}} = \frac{C_c}{C_{1000c}}

As i said, the original charge comprised 1001 parts of which 1 part remains on the smaller capacitor and 1000 parts are now on the large capacitor.

(edited 8 years ago)

Reply 8

Original post by uberteknik

Be careful, we are dealing with ratios. If the smaller capacitor is 1 part, the bigger capacitor is 1000 parts, with a combined capacitance of 1001 parts.
i.e. divide the original charge by 1001 to get the unit charge, but this is distributed 1 part to 1000 parts between the two capacitors.

...

As i said, the original charge comprised 1001 parts of which 1 part remains on the smaller capacitor and 1000 parts are now on the large capacitor.

As we're going round in circles this will be my final response.

1. The question does not ask for a ratio; it asks for an absolute amount: "explain why the charge remaining on the initially charged capacitor after the two are connected is Q/1000, where Q is its initial charge before being connected to the circuit."

This asks for the absolute value of charge remaining in terms of the original total charge, Q. I think that is unarguable. The word "ratio" or "fraction" or anything similar is absent, so I can't see why you would want to give an answer in terms of a ratio. Also note that they ask for an explanation of the quantity Q/1000 which must have units of coulomb; a ratio would be unitless.

2. The charge

Q_s

on the smaller capacitor turns out to be

\frac{Q}{1001}

and that on the larger

Q_l

is

\frac{1000Q}{1001}

. And, of course, since

Q_s

remains on the capacitor, it is the answer to the question.

Note that:

a)

Q_s+Q_l = \frac{Q}{1001} + \frac{1000Q}{1001} = Q

as required by conservation of charge

b)

\frac{Q_s}{Q_l} = \frac{Q/1001}{1000Q/1001}=\frac{1}{1000}

which is the ratio that you are stressing.

I strongly suspect that there is merely a typo in the question.

Reply 9