# A Level physics question on capacitorsWatch

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#1
Hi everyone, I'm studying A2 Physics and I came across this question from an exercise book which has thrown me a bit. I'll write it out an explain where I'm stuck.

"The diagram shows an RC series circuit with the switch open - I'll edit or post a scan of this circuit from the book later.

- battery has a voltage of 12 V
- a microammeter is connected in series to it
- there is a switch in the circuit
- there is one 1.0 megaohm resistor connected in series
- there is a 22 microfarad capacitor connected in series

a) At the instant the switch is first closed, what is
i) the p.d. across the cap? (it's 0 F - I don't understand why)
ii) the p.d. across the resistor (it's 12 V, the voltage of the battery - that's to be expected I suppose. Current through the resistor decreases proportionately)
iii) the current in the circuit. I = V/R implies that the current is 12/1x10^6 = 12 micro amps. I worked this out myself an it is correct.
iv) the charge on each cap. plate - this is 0 - but I don't see how - unless the electrons haven't actually reached the cap. plates yet

b) What are the same quantities i) - iv) after the switch has been closed and a steady condition reached? - What's a steady condition?

the answers for part b) are respectively

12 V ?
0 V ?
0 A ?
26 mC - I tried to find Q using Q = CV with 2.2x10^-5 F (12 V) = 260 microcoulombs to 2 s.f. = 0.26 mC. The answer in the book is 26 mC - is the book wrong?

part c) (the final part) of this quesion I am okay with, it says: "At a certain instant after the switch has been closed the microammeter reads 9.5 microamps. Calculate, for this instant, i) the p,d, across the resistor - I used 9.5 microamps x 1 megaohm = 9.5 V, which is correct according to the answers. ii) the p.d. across the cap.? - the p.d. is 2.5 V - I can understand that the p.d. has been split up, as 9.5 V + 2.5 V = 12 V, but why does this happen? What is the physical reason? iii) the charge on each capacitor plate - I used the equation Q = CV to find Q = (2.2x10^-5F)(2.5V) = plus or minus 55 microcoulombs - I'm okay with this part of the question.

0
6 years ago
#2
(Original post by Mr Cosine)
a) At the instant the switch is first closed, what is
i) the p.d. across the cap? (it's 0 F - I don't understand why)
Charge is built up on the capacitor plates by electrons pilling up under voltage pressure supplied by the battery. Electrons must travel to the plates and this takes a finite time. At t=0, there is no net charge on the plates.

Q=CV and C is constant. Then since the charge on the capacitor at t=0 is 0,
V must also be 0.

(Original post by Mr Cosine)
ii) the p.d. across the resistor (it's 12 V, the voltage of the battery - that's to be expected I suppose. Current through the resistor decreases proportionately)
At t=0, charge will start to flow (current). Kichoffs Voltage law say's the sum of the p.d.'s dropped around the circuit must equal the supply voltage. (an instance of the energy conservation laws).

If the voltage across the capacitor is 0, then:
Vc= V s - Vr
= 12 - 0
= 12V

(Original post by Mr Cosine)
iv) the charge on each cap. plate - this is 0 - but I don't see how - unless the electrons haven't actually reached the cap. plates yet
Exactly.

(Original post by Mr Cosine)
b) What are the same quantities i) - iv) after the switch has been closed and a steady condition reached? - What's a steady condition?

the answers for part b) are respectively
In this question, steady state d.c. conditions are when no more charge can be accomodated by the capacitor.

12 V ?
Correct. The plates cannot hold any more charge and since no charge is flowing, the current through the circuit is 0. All the supply voltage pressure is now across the capacitor.

0 V ?
Correct. Vr=IrR and since Ir = 0, then Vr = 0.

0 A ?

26 mC - I tried to find Q using Q = CV with 2.2x10^-5 F (12 V) = 260 microcoulombs to 2 s.f. = 0.26 mC. The answer in the book is 26 mC - is the book wrong?

Q = CV

Q = 22x10-6 x 12

Q = 0.264x10-3 Coulombs
= 0.264 mC or 264 uC

(Original post by Mr Cosine)
ii) the p.d. across the cap.? - the p.d. is 2.5 V - I can understand that the p.d. has been split up, as 9.5 V + 2.5 V = 12 V, but why does this happen? What is the physical reason?
As stated previously, the charge on the capacitor takes a finite time to build up. This time is governed by the current allowed to flow as a result of the series resistance.

The power source can only provide a limited amount of voltage pressure to push more electrons onto the capacitor plates (electrons repel electrons).

As the charge builds up, it becomes harder for new electrons to enter the plates because the electric field between the plates is also increasing, which provides a back-pressure.

So the voltage developed across the capacitor at time (t) seconds, is a measure of that back pressure.

By Kirchoffs Voltage law, the p.d. developed across the capacitor must therefore be 12V - 2.5V = 12.5V
0
#3
Thanks for the advice uberteknik - unfortunately I can't attach a scan of the circuit.
0
6 years ago
#4
(Original post by Mr Cosine)
Thanks for the advice uberteknik - unfortunately I can't attach a scan of the circuit.

12V battery with +ve terminal connected to

switch contact. Other switch contact connected to

1 Megohms resistor connected to

22uF capacitor connected to

-ve battery terminal.

Swicth contact normally open but closed at t=0 seconds

Then remains closed for >>5CR to achieve steady state.
0
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