# Urgent A2 physics question!

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Thread starter 4 years ago
#1
Hi I have a really stupid physics question, it's been bugging me because it's practically GCSE or even lower and I feel really dumb asking this on TSR but I was too embarrassed to ask my teacher so here it is.

https://tavistockcollegescience.wiki...apacitance.pdf

It's on Page 20, Question 10 part a, i and ii.
I thought the answers would be the other way around but I don't see how it is this way?! Can someone please explain this in an understandable way? I always struggle with these types of questions

Thank you!

In before, you shouldn't be doing A2 physics if you don't know this
1
4 years ago
#2
(Original post by DedicatedWizard)
Hi I have a really stupid physics question, it's been bugging me because it's practically GCSE or even lower and I feel really dumb asking this on TSR but I was too embarrassed to ask my teacher so here it is.

https://tavistockcollegescience.wiki...apacitance.pdf

It's on Page 20, Question 10 part a, i and ii.
I thought the answers would be the other way around but I don't see how it is this way?! Can someone please explain this in an understandable way? I always struggle with these types of questions

Thank you!

In before, you shouldn't be doing A2 physics if you don't know this

in the parallel case you have 2 capacitors with their lower plates connected together... i.e. at the same potential as each other and their upper plates connected together... i.e also at the same potential as each other.
the ammeter A3 doesn't affect things here because ammeters are very low resistance and we're told the capacitors are fully charged.

the potential difference across the left capacitor must be equal to the potential difference across the right capacitor (this is true whether or not the capacitors have the same value of capacitance btw) and the left capacitor has 10V across it because it's connected across the poles of a 10V supply

in the series case you have 2 capacitors with their 'inner plates' connected together. Here the amount of charge on each capacitor is the same, the 'inner plates' are isolated from the rest of the circuit so charge carriers can't flow onto them, so any change in the amount of charge on one of the 'outer plates' must be matched by the opposite charge flowing onto the other 'outer plate'

if the series capacitors have different values of capacitance to each other they will each have different potential differences across their plates (recall that capacitance C=Q/V) but if they both have the same capacitance, the pd across both the capacitors will be equal.
The pd across both series capacitors together must sum up to the supply pd of 10V so the pd across either must be 5V

The parallel case is IMO the easiest to think about
1
4 years ago
#3
(Original post by DedicatedWizard)
Hi I have a really stupid physics question, it's been bugging me because it's practically GCSE or even lower and I feel really dumb asking this on TSR but I was too embarrassed to ask my teacher so here it is.

https://tavistockcollegescience.wiki...apacitance.pdf

It's on Page 20, Question 10 part a, i and ii.
I thought the answers would be the other way around but I don't see how it is this way?! Can someone please explain this in an understandable way? I always struggle with these types of questions

Thank you!

In before, you shouldn't be doing A2 physics if you don't know this
Joinedup has provided a comprehensive explanation but in addition here is my understanding for the series circuit.

In the series circuit, both of the capacitors have the same capacity for storing charge and, since the same amount of charge is passed through them, they both will share an equal amount of charge transferred from the battery. This is because you could consider the voltage, in general terms, to be the deriving force of the electrons and therefore when the battery pushes the electrons onto the negative plate of one of the capacitors, there will be an electric field between the isolated plates, which results in the electrons on the other (positive) plate being drawn and moved to the negative plate of the other capacitor. As a result, since both capacitors have the tendency to push back the electrons in the opposite direction, the total voltage across both of them must be equal to the total voltage across the battery. I should admit that I cannot explain the situation when the capacitors have a different capacitance, in which case the voltage across each capacitor is different. Maybe it's also late at night.
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