# A Level physics question on capacitors (2)

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#1
Hi everyone, I'm studying A2 Physics and I'm looking at the topic of capaictors, Here's the question I'm stuck on - any tips?

"Two caps of cap. C1 and C2 are connected in series, with a p.d. V = V1 + V2 across the combination. Show that the p.d.s V1 and V2 across the individual capacitors are

V1 = C2V/(C1 + C2) and V2 = C1V/(C1 + C2)."

I know the formula for capacitors in series - using that formula I found the total cap, C, in terms of C1 and C2, to be C = C1C2/(C1 + C2). But where do I go from here?
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6 years ago
#2
(Original post by Mr Cosine)
Hi everyone, I'm studying A2 Physics and I'm looking at the topic of capaictors, Here's the question I'm stuck on - any tips?

"Two caps of cap. C1 and C2 are connected in series, with a p.d. V = V1 + V2 across the combination. Show that the p.d.s V1 and V2 across the individual capacitors are

V1 = C2V/(C1 + C2) and V2 = C1V/(C1 + C2)."

I know the formula for capacitors in series - using that formula I found the total cap, C, in terms of C1 and C2, to be C = C1C2/(C1 + C2). But where do I go from here?

VT = V1 + V2 ..............(eq1)

QT = VTCT .................(eq2)

CT = C1C2/(C1+C2).......(eq3)

The key to this is to recognise that the two plates between the series capacitors are effectively isolated. The total charge is therefore limited by the charge capacity of the smallest plate size. i.e. because that is the only charge available to be moved between the central plates.

The voltage drop across each capacitor will be dependent on the value of the individual capacitances:

VC1 = QT/C1; VC2 = QT/C2;hence

VC1 = VTCT/C1 and VC1 = VTCT/C1

Substituting eq3

VC1 = VT/C1 x (C1C2/(C1+C2)

VC1 = VT x C2/(C1 + C2)

VC2 = VT x C1/(C1 + C2)
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#3
(Original post by uberteknik)

VT = V1 + V2 ..............(eq1)

QT = VTCT .................(eq2)

CT = C1C2/(C1+C2).......(eq3)

The key to this is to recognise that the two plates between the series capacitors are effectively isolated. The total charge is therefore limited by the charge capacity of the smallest plate size. i.e. because that is the only charge available to be moved between the central plates.

The voltage drop across each capacitor will be dependent on the value of the individual capacitances:

VC1 = QT/C1; VC2 = QT/C2;hence

VC1 = VTCT/C1 and VC1 = VTCT/C1

Substituting eq3

VC1 = VT/C1 x (C1C2/(C1+C2)

VC1 = VT x C2/(C1 + C2)

VC2 = VT x C1/(C1 + C2)
OK I can understand what you're saying. I think'll have to practise this a little bit more though to really get it. Thanks all the same though.
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