Describe how to show charge is proportional to pd on a capacitor

AI Thread Summary
To demonstrate that the potential difference (pd) across a capacitor is proportional to the charge stored, one can use a voltmeter in parallel with the capacitor and an ammeter in series to measure current. A common experimental approach involves charging the capacitor through a large resistor at a constant rate while monitoring the current and pd. Techniques such as using a reed switch to alternately charge and discharge the capacitor can help measure the average current, allowing for the calculation of charge. The relationship can be confirmed by plotting charge against voltage, which should yield a straight line if the experiment is conducted correctly. Overall, careful measurement and analysis of current and voltage variations are crucial for accurate results.
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Homework Statement


Describe how you would show experimentally that the pd across a capacitor of fixed capacitance is proportional to the charge stored on the plates?

Homework Equations





The Attempt at a Solution


So the most obvious thing would be to place a voltmeter in parallel to the capacitor

To measure the current I could use an ammeter in series with the capacitor.

I know that Q = It but surely the charge stored on a charged capacitor is fixed?
Even If I had a variable resistor in the circuit the current in the circuit would change BUT SO WOULD THE potential difference.

The MS states the following points but it confuses me slightly:
- charge the capacitor through a large resistor at a constant rate
- by using a variable resistor to gradually reduce the resistancd in the circuit
- monitor the current with a micro-ammeter keeping the current constant
- record pd with a voltemter across capacitor
- pd should increase at a constant rate

HOW CAN THE CURRENT STAY CONSTANT - V = IR BUT the power sources does not change so the only thing that changes with R is I :S

please can someone explain to me where I am going wrong
 
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in fact would this be a sensible experiment:
- charge a capacitor to a particular p.d and check with a voltmeter
- using a coloumeter identify charge stored
- repeat for many different p.d (use a range of power sources)
- plot a graph
??
(repeat with other capacitors)
 
is the coloumeter is parallel or series with capacitor?
 
I have not met a coulomb meter!
 
technician said:
I have not met a coulomb meter!

well it is also called a voltameter
 
the technique I know uses a vibrating (reed) switch run from a fixed frequency supply to charge and discharge a capacitor. The current measured can be used to find the charge on the capacitor.
This is a standard textbook technique to determine capacitance/charge
 
technician said:
the technique I know uses a vibrating (reed) switch run from a fixed frequency supply to charge and discharge a capacitor. The current measured can be used to find the charge on the capacitor.
This is a standard textbook technique to determine capacitance/charge

how does it measure the charge on the capacitor?
 
This is the circuit arrangement. The voltage of the battery is known. The reed switch is operated by , maybe 100Hz AC. The reed switch alternately charges the capacitor from the battery and discharges the capacitor through a micro ammeter. The microammeter gives the average current flowing from the capacitor..i.e the charge per second from which the charge can be determined.
For accuracy the time constant of the discharge circuit must be much less than the time period of the AC driving the reed switch.
How does a vlotameter work?
 

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technician said:
This is the circuit arrangement. The voltage of the battery is known. The reed switch is operated by , maybe 100Hz AC. The reed switch alternately charges the capacitor from the battery and discharges the capacitor through a micro ammeter. The microammeter gives the average current flowing from the capacitor..i.e the charge per second from which the charge can be determined.
For accuracy the time constant of the discharge circuit must be much less than the time period of the AC driving the reed switch.
How does a vlotameter work?

as I understand, a voltameter uses a capacitor of ENORMOUS capacitace (known) and is placed in parallel with the desired capacitor.
The charge on the desired capacitor is shared equally according to the ratio of the capacitors so basically all charge goes to the known capacitor
using Q = CV we can find Q as we can find pd across capacitor

Although this does seem foolish as we are using the equation (Q=CV) for the equation we are trying to verify (Q=CV) :S

Why is the reed switch NOT in the circuit?
 
  • #10
so would this be a viable expt: (i have never used reed switch - not on syllabus either)
- monitor the discharge current with time simultaniously monitoring the discharge voltage (using a data logger for both)
- find the area under the current curve to get charge
- plot charge against voltage to give a straight line?
 
  • #11
I can see the logic behind your voltameter technique !
The vibrating switch has its limitations because of time constant implications but then every measurement in physics has some limitations.
The only other technique I know of for determining charge uses a "ballistic galvanometer"... I leave you Js to find out more about this !
I do know that the vibrating switch is a good technique, having used it regularly.
The purpose of the switch is only to connect the battery to the capacitor then the capacitor to the microammeter. It is not part of the "circuit"
 
  • #12
details of the technique:
Voltage...this is no problem, it is the battery voltage. Let's say 9V
As an example, if you have a reed switch vibrating at 100Hz and measure a steady current of 100microAmps then 1 micro coulomb is transferred each second.
If you ignore time constant considerations this means that 9V charges a capacitor with 1 microCoulomb of charge
So the capacitance must be C=Q/V = (1x10^-6)/9 = 1.1x10^-7F
You can manipulate these figures to find any of the unknown quantities.
These days reed switches are readily available and this is an easy experiment to do.
If you need any more help with this technique let me know.

PS a signal generator is needed to drive the reed switch at a known frequency
 
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  • #13
technician said:
I can see the logic behind your voltameter technique !
The vibrating switch has its limitations because of time constant implications but then every measurement in physics has some limitations.
The only other technique I know of for determining charge uses a "ballistic galvanometer"... I leave you Js to find out more about this !
I do know that the vibrating switch is a good technique, having used it regularly.
The purpose of the switch is only to connect the battery to the capacitor then the capacitor to the microammeter. It is not part of the "circuit"

it seems a ballistic galvonometer measures the charge DISCHARGED through it

so if I know how voltage varies with time, I can find charge dishcarged with time and then plot a graph

OR I could use the suggestion of monitoring change in current and integrating curve the nplotting resutls against voltage
 
  • #14
You are correct regarding the ballistic galvanometer. If you have access to one it is a possible technique.
You need to obtain measurements of voltage applied (easy with a multimeter) and charge (several techniques now available to you!)
Then plot a graph of volts against charge...should be a straight line
 
  • #15
technician said:
You are correct regarding the ballistic galvanometer. If you have access to one it is a possible technique.
You need to obtain measurements of voltage applied (easy with a multimeter) and charge (several techniques now available to you!)
Then plot a graph of volts against charge...should be a straight line

yey..sorry to nag but was this method also correct:
-monitor the discharge current with time simultaniously monitoring the discharge voltage (using a data logger for both)
- find the area under the current curve to get charge
- plot charge against voltage to give a straight line?
 
  • #16
That sounds OK but I think I would use the total area under the current against time graph to give the charge and use the steady value of the applied voltage.
You have to be very careful analysing the information from graphs showing the voltage AND the current varying... I would be inclined to keep one constant...the charging voltage each time
 
  • #17
technician said:
That sounds OK but I think I would use the total area under the current against time graph to give the charge and use the steady value of the applied voltage.
You have to be very careful analysing the information from graphs showing the voltage AND the current varying... I would be inclined to keep one constant...the charging voltage each time

I guess that makes sense
thanks
 
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