phy 5 question on capacitorsWatch

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Thread starter 13 years ago
#1
hi,pls can some1 explain how to carry out such an experiment to get 5 marks. im stuggling even to get 1. i dnt know how to do this.???

Describe how you would show experimentally that the charge stored on a 220 aF capacitor is proportional to the potential difference across the capacitor for a range of potential differences between 0 and 15 V.
Your answer should include a circuit diagram. [5]

im thinking V=CQ. draw a graph thru the origin of V against Q. shows its directly proportional. i only know that.

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13 years ago
#2
you want the circuit to have a two way switch, one for charging, and one for discharging. on one side, put the capacitor in series with a variable power supply, and put a voltmeter in parallel to it to get an accurate reading of V. then on the switch, have a coloumbmeter to record the charge passing when the capacitor is discharged.

you then want to get multiple readinds of V and corresponding values of Q. Using the equation Q=VC, as y=mx, draw Q against V and the capacitance will be the gradient.

for each value, take three readings and average.

check that the line is striaght and goes through the origin to confirm the directly proportional relationship.
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13 years ago
#3
is tehre such thing as a coulomb meter? i was always told to use ea stop clock and use ammeter - and use Q=IT

hmmm - better look int o that

pk
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13 years ago
#4
I thought discharging was almost instantaneous! Yeah we've used coulumbmetres in experiments before.
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13 years ago
#5
(Original post by mik1w)
I thought discharging was almost instantaneous! Yeah we've used coulumbmetres in experiments before.
so its true- but a pic i found showed it as a touch device - i.e. not connected in series like a component, but just touch it to a point in the circuit type gizmo - how does it actually work?
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13 years ago
#6
Isn't the graph for that exponential?
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13 years ago
#7
No, the exponential graph is againt time, and has a resistor in series with the capacitor - this is exponential because the current is proportional to the charge on the capacitor (dQ/dt = -kQ), when discharging, because as the capacitor loses charge, the voltage across it falls proportionally, therefore the voltage across the resistor increases and the current falls.
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13 years ago
#8
(Original post by Phil23)
so its true- but a pic i found showed it as a touch device - i.e. not connected in series like a component, but just touch it to a point in the circuit type gizmo - how does it actually work?
you can plug a wire into the ones we used, or simply use a flying lead instead of a switch and tough it against the power supply to charge and touch it on the coulombmetre to discharge.
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Thread starter 13 years ago
#9
(Original post by Lianah)
Isn't the graph for that exponential?
no it isnt. its a striaght line thru the origin. becuae V is directly proportionl to Q hence V=CQ
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13 years ago
#10
which board is this?
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Thread starter 13 years ago
#11
(Original post by mik1w)
you want the circuit to have a two way switch, one for charging, and one for discharging. on one side, put the capacitor in series with a variable power supply, and put a voltmeter in parallel to it to get an accurate reading of V. then on the switch, have a coloumbmeter to record the charge passing when the capacitor is discharged.

you then want to get multiple readinds of V and corresponding values of Q. Using the equation Q=VC, as y=mx, draw Q against V and the capacitance will be the gradient.

for each value, take three readings and average.

check that the line is striaght and goes through the origin to confirm the directly proportional relationship.

thanx a lot, my phy teacher gets carried away most of the time and we never really learn anything. we practically hve to teach ourselves or seek help elsewhere.

for this question can we use a joulmeter by any chance?
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13 years ago
#12
edexcel
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13 years ago
#13
(Original post by andyj72)
thanx a lot, my phy teacher gets carried away most of the time and we never really learn anything. we practically hve to teach ourselves or seek help elsewhere.

for this question can we use a joulmeter by any chance
hmm good question

the energy on a capacitor is 0.5*V*C, as you know the initial voltage, you should be able to find the value of C. so yes. never really thought about that though!

you may get some inaccuracies from the resistance in the wires (wires heating up as they form the only resistance) but this is probably negligable.
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Thread starter 13 years ago
#14
(Original post by mik1w)
No, the exponential graph is againt time, and has a resistor in series with the capacitor - this is exponential because the current is proportional to the charge on the capacitor (dQ/dt = -kQ), when discharging, because as the capacitor loses charge, the voltage across it falls proportionally, therefore the voltage across the resistor increases and the current falls.

i thought the voltage across the resistor fell. it says something like that in the NAS book during discharge and charging.

or is this a differnent senario?
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13 years ago
#15
voltage on capacitor = Q/C, as Q falls when the capacitor discharges, V must fall across capacitor, so the voltage across the resistor rises as total voltage is the same

I'm not 100% certain though, you made me doubt myself so Ill go and check. Id hate to be giving people the wrong info.
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Thread starter 13 years ago
#16
(Original post by mik1w)
hmm good question

the energy on a capacitor is 0.5*V*C, as you know the initial voltage, you should be able to find the value of C. so yes. never really thought about that though!

you may get some inaccuracies from the resistance in the wires (wires heating up as they form the only resistance) but this is probably negligable.
surprised myself, not really. hehe. yes thats a point, as the voltage increases it will heat up causing an increase in resistance. ill stick to ur 1st method cos it removes all the uncertainty factors.
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13 years ago
#17
mik1w method is probably the favoured method.

You can also do it this way

Use an ammeter, stop watch, capacitor, variable power supply.

take measurements of I and t at 1v increments
plot a graph of i vs t => area under graph = charge stored

do this serveral times and obtain an average

now plot Q vs V, expect a straight line through the origin [i think the gradient should be the capacitance of the capacitor].
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13 years ago
#18
do you want a resistor in series with the capacitor when discharging?
like I said above it takes maybe a millisecond for a capacitor to discharge I think, like a spark, you won't be able to time that.
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Thread starter 13 years ago
#19
(Original post by mik1w)
voltage on capacitor = Q/C, as Q falls when the capacitor discharges, V must fall across capacitor, so the voltage across the resistor rises as total voltage is the same

I'm not 100% certain though, you made me doubt myself so Ill go and check. Id hate to be giving people the wrong info.
ill hve a look. but ur argument is convincing. however, im thinking ur proposition is an example of kirchoffs 2nd law. in this case during capacitor discharge, it isnt conected to the emf. it will be connected to the resistor and act as a battery. so the initial V across the capacitor will equal the V across the resistor but as the capacitor discharges, the Q is dissipated in the resistor exponentially. (i think) so the current "I" decreases too.

just review my argument. its just that im not too sure. ive always thought about it in this way and seen it too.
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Thread starter 13 years ago
#20
[QUOTE=nas7232]mik1w method is probably the favoured method.

im in agreement. ur method is straight to the point. thanx.
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