# A level Physics Capacitor Question

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#1
Have stumbled across a capacitor question about microphones and am struggling to fully understand the mark scheme for it. Any help is appreciated
Attachment 751404
Mark scheme:

I understand that increasing capacitance will result in increasing the charge the capacitor will store but I don't understand how this will result in a p.d across the resistor, or how the direction of the charge flow will be reversed when the plates are moved apart.
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2 years ago
#2
The microphone circuit is attempting to keep a constant PD across the capacitor plates
capacitance is Q/V so as the capacitance is varied over time, charge must be moved to maintain the constant PD
moving charge is a current, current is detected by changes in the PD across the resistor (V=IR)
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#3
(Original post by Joinedup)
The microphone circuit is attempting to keep a constant PD across the capacitor plates
capacitance is Q/V so as the capacitance is varied over time, charge must be moved to maintain the constant PD
moving charge is a current, current is detected by changes in the PD across the resistor (V=IR)
I see, but how does the direction of the flow of charge reverse? Is it safe to say that the open switch in the diagram forces it to go in the opposite direction?

Also, the other possible response seems contradictory - if the charge stored by the capacitor increases then how can the p.d across it decrease?
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2 years ago
#4
(Original post by Jayc3)
I see, but how does the direction of the flow of charge reverse? Is it safe to say that the open switch in the diagram forces it to go in the opposite direction?
No. The switch is there to open the circuit and prevent the battery from discharging when the capacitor is not in use. i.e. it prolongs battery life.

Also, the other possible response seems contradictory - if the charge stored by the capacitor increases then how can the p.d across it decrease?
The microphone operation rests on the principle of force equilibrium: between the electric field stored between the capacitor plates and that of the e.m.f. generated by the battery.

Q = CV

V = Q/C

Capacitance is a function of the plate area and the distance (dielectric gap) between the plates.

Changes in air pressure, causes the distance between the capacitor plates to alter. i.e. the plate distance 'd', changes as a function of the sound pressure waves hitting the non-fixed plate.

Thus the capacitance alters dynamically as a facsimile of the sound.

Therefore the charge stored on the plates must also change as a facsimile of the sound pressure.

The battery charges the capacitor to a state of equilibrium. i.e. the charge stored on the capacitor (p.d. voltage pressure) exactly opposes the voltage pressure exerted by the battery and no further current can flow.

If the capacitance now changes (as a result of air pressure changes), the electrostatic field between the capacitor plates also alters. This either allows further charge to flow into the capacitor sourced from the battery (capacitance increased and current in one direction), or the electrostatic field between the capacitor plates becomes greater than that of the battery (the plates have moved further apart decreasing the capacitance) and excess charge on the capacitor flows in the opposite direction to reach equilibrium with the battery once again.

The a.c. waveform is a result of the movement of charge around the closed circuit as the electric force between the capacitor plates constantly equalises.

Since charge flow is via the resistor, a time varying p.d. across the resistor is developed in response to the sound pressure changes.
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#5
(Original post by uberteknik)
Q = CV

V = Q/C

Capacitance is a function of the plate area and the distance (dielectric gap) between the plates.

Changes in air pressure, causes the distance between the capacitor plates to alter. i.e. the plate distance 'd', changes as a function of the sound pressure waves hitting the non-fixed plate.

Thus the capacitance alters dynamically as a facsimile of the sound.

Therefore the charge stored on the plates must also change as a facsimile of the sound pressure.

The a.c. waveform is a result of the movement of charge around the closed circuit as the electric force between the capacitor plates constantly equalises.

Since charge flow is via the resistor, a time varying p.d. across the resistor is developed in response to the sound pressure changes.
I know that the equations say that charge will increase while p.d. decreases but I'm trying to understand it from a conceptual standpoint - how can both change at the same time, more charge should mean more potential difference right?
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#6
(Original post by uberteknik)
No. The switch is there to open the circuit and prevent the battery from discharging when the capacitor is not in use. i.e. it prolongs battery life.
Ah, I assumed from the diagram that the switch was just permanently open.
1
2 years ago
#7
(Original post by Jayc3)
I know that the equations say that charge will increase while p.d. decreases but I'm trying to understand it from a conceptual standpoint - how can both change at the same time, more charge should mean more potential difference right?

Think about how charge equilibrium across the capacitor plates is achieved.

Voltage is defined as joules per coulomb of charge. It's the amount of charge on the capacitor which creates the p.d. Charge the capacitor and take the battery away, the charge on the capacitor stays put including the p.d. across the plates. Now change the plate separation and the capacitance must change and with it, so does the p.d. across the plates - no battery needed.

It's sometimes useful to think of voltage like pressure in a liquid and current as the flow of that liquid. The voltage pressure already on the capacitor plates achieves equilibrium with the voltage pressure supplied by the battery. If the pressure exerted by either capacitor or battery changes, then current will flow to redistribute charge and achieve equilibrium once again.
1
2 years ago
#8
(Original post by Jayc3)
Ah, I assumed from the diagram that the switch was just permanently open.
There are two rules in physics:

Rule 1) Do not assume anything.

Rule 2) See Rule 1.

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#9
(Original post by uberteknik)
No. The switch is there to open the circuit and prevent the battery from discharging when the capacitor is not in use. i.e. it prolongs battery life.

The microphone operation rests on the principle of force equilibrium: between the electric field stored between the capacitor plates and that of the e.m.f. generated by the battery.

Q = CV

V = Q/C

Capacitance is a function of the plate area and the distance (dielectric gap) between the plates.

Changes in air pressure, causes the distance between the capacitor plates to alter. i.e. the plate distance 'd', changes as a function of the sound pressure waves hitting the non-fixed plate.

Thus the capacitance alters dynamically as a facsimile of the sound.

Therefore the charge stored on the plates must also change as a facsimile of the sound pressure.

The battery charges the capacitor to a state of equilibrium. i.e. the charge stored on the capacitor (p.d. voltage pressure) exactly opposes the voltage pressure exerted by the battery and no further current can flow.

If the capacitance now changes (as a result of air pressure changes), the electrostatic field between the capacitor plates also alters. This either allows further charge to flow into the capacitor sourced from the battery (capacitance increased and current in one direction), or the electrostatic field between the capacitor plates becomes greater than that of the battery (the plates have moved further apart decreasing the capacitance) and excess charge on the capacitor flows in the opposite direction to reach equilibrium with the battery once again.

The a.c. waveform is a result of the movement of charge around the closed circuit as the electric force between the capacitor plates constantly equalises.

Since charge flow is via the resistor, a time varying p.d. across the resistor is developed in response to the sound pressure changes.
Ah, I think I get what you're saying. So if the capacitance increases then the charge that can be stored for a given p.d. increases, causing charge to flow into the capacitor to take up this space. So the p.d. required across the capacitor falls as not as much 'pressure' is needed to maintain the charge but if the capacitance decreases then the capacitors ability to store charge at a given p.d. is worsened so it discharges through the resistor to get rid of the charge it cannot store and the p.d. across it will increase as it is in a sense harder for the charge to be kept on the capacitor. Is this understanding correct?
0
2 years ago
#10
(Original post by Jayc3)
Ah, I think I get what you're saying. So if the capacitance increases then the charge that can be stored for a given p.d. increases, causing charge to flow into the capacitor to take up this space.
Correct.

.......So the p.d. required across the capacitor falls as not as much 'pressure' is needed to maintain the same charge........
Correct.

but if the capacitance decreases then the capacitors ability to store charge at a given p.d. is worsened so it discharges through the resistor to get rid of the charge it cannot store
Correct

.....and the p.d. across it will increase as it is in a sense harder for the charge to be kept on the capacitor. Is this understanding correct?
The increased p.d. is a definition of joules per quantity of charge, because the existing charge force is now greater than that of the battery and hence charge flows in the opposite direction.

You got it!
1
#11
(Original post by uberteknik)
Correct.

Correct.

Correct

The increased p.d. is a definition of joules per quantity of charge, because the existing charge force is now greater than that of the battery and hence charge flows in the opposite direction.

You got it!
Niceee, didn't think I'd ever get my head around that. Thank you!
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