# Capacitors ppl

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#1
Ok TSR... need some help (again)
Spoiler:
Show

Initial state:
A capacitor C1 is connected to a supply. When a potential difference of 4.0 V is applied across the capacitor, it stores a charge of 0.80 nC.

ok ... it says : (ii) On the grid below, show how the charge stored on capacitor C2 varies with potential difference during this charge transfer process.

How do I know upto which voltage it will go to ?

thanks
0
11 years ago
#2
(Original post by antonfigo)
Ok TSR... need some help (again)
Spoiler:
Show

Initial state:
A capacitor C1 is connected to a supply. When a potential difference of 4.0 V is applied across the capacitor, it stores a charge of 0.80 nC.

ok ... it says : (ii) On the grid below, show how the charge stored on capacitor C2 varies with potential difference during this charge transfer process.

How do I know upto which voltage it will go to ?

thanks

The voltage of C1 and C2 must add up to 4V, right?
0
#3
(Original post by unamed)
The voltage of C1 and C2 must add up to 4V, right?
Thats what i thought... meaning 1 V ... but mark scheme shows that it must go upto 3V....i don't see why it should
0
11 years ago
#4
(Original post by antonfigo)
Thats what i thought... meaning 1 V ... but mark scheme shows that it must go upto 3V....i don't see why it should
Maybe because C1's discharges to 1V, charging C2 to 3V and then the C2 discharges? [or maybe I'm not making sense.]

I'll wait for an answer in that case.
0
11 years ago
#5
(Original post by unamed)
Maybe because C1's discharges to 1V, charging C2 to 3V and then the C2 discharges? [or maybe I'm not making sense.]

I'll wait for an answer in that case.
When the second capacitor was added charge was transferred from one to the other.

The p.d. across each capacitor became 3 V.

The p.d. across C1 dropped from 4V to 3V as some of the charge was transferred, and the p.d. across C2 went from 0V to 3V as it took up the charge that was transferred.

Hope that helps.
0
11 years ago
#6
(Original post by ghostwalker)
When the second capacitor was added charge was transferred from one to the other.

The p.d. across each capacitor became 3 V.

The p.d. across C1 dropped from 4V to 3V as some of the charge was transferred, and the p.d. across C2 went from 0V to 3V as it took up the charge that was transferred.

Hope that helps.
So, both capacitors have equal charge now?
0
11 years ago
#7
(Original post by unamed)
So, both capacitors have equal charge now?
No, they have equal voltage; their charge will depend on their capacitance.
0
11 years ago
#8
(Original post by ghostwalker)
No, they have equal voltage; their charge will depend on their capacitance.
Ghostwalker is right. Both capacitors will have three volts, but both capacitors will have a different charge because the capacitance is different.

I'll show you....

In an ideal world the capacitance of C1 is 200pF.

This is because C = Q / V, where C is capacitance, Q is charge, and V is volts.

ie: 0.8nC / 4V = 200 pF

When C2 is connected, there is a voltage drop to three volts on C1. This is because a charge of 0.2nC has transfered from C1 to C2, ie:

Charge transfered = 0.8nC - (200pF * 3V)

Therefore, C2 must be:

C2 = 0.2nC / 3V = 66pF

Why does only 0.2nC get transfered? Well it's simple, C2 is smaller and so cannot hold as much charge.

EDIT: Corrected typos
1
11 years ago
#9
Good explanation.
0
11 years ago
#10
Good explanation.
Just
Whoops, that will teach me for typing in a rush... lol. Thanks for spotting it
0
6 months ago
#11
(Original post by ghostwalker)
When the second capacitor was added charge was transferred from one to the other.

The p.d. across each capacitor became 3 V.

The p.d. across C1 dropped from 4V to 3V as some of the charge was transferred, and the p.d. across C2 went from 0V to 3V as it took up the charge that was transferred.

Hope that helps.
Hi I know it has been 10 years but i was doing this question recently and still can't wrap my head around why they both have a voltage of 3v we weren't give the capacitance of the two and had to first know that they both had 3v to work out the value of the capacitance .
0
6 months ago
#12
(Original post by m9999999999)
Hi I know it has been 10 years but i was doing this question recently and still can't wrap my head around why they both have a voltage of 3v we weren't give the capacitance of the two and had to first know that they both had 3v to work out the value of the capacitance .
Not sure what your issue is. You're told in the question that the p.d. across C1 becomes 3V, which must be the same as the p.d. across C2.

Initially when C2 is connected it will have 0V p.d. across it, and current will subsequently flow until they both have the same p.d. and that is 3V.
0
6 months ago
#13
(Original post by ghostwalker)
Not sure what your issue is. You're told in the question that the p.d. across C1 becomes 3V, which must be the same as the p.d. across C2.

Initially when C2 is connected it will have 0V p.d. across it, and current will subsequently flow until they both have the same p.d. and that is 3V.
so do we assume that the C1 no longer decreases as I don't know if the question states that clearly enough as it just says that it drops to 3V Thank you for responding BTW
0
6 months ago
#14
(Original post by m9999999999)
so do we assume that the C1 no longer decreases as I don't know if the question states that clearly enough as it just says that it drops to 3V Thank you for responding BTW
Yes. If you connect two capacitors like that, the change in voltage would be near instantaneous.
1
6 months ago
#15
(Original post by ghostwalker)
Yes. If you connect two capacitors like that, the change in voltage would be near instantaneous.
Oh thank you so much!
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