Show that Q is proportional to V for capacitor

AI Thread Summary
To experimentally demonstrate that the charge (Q) stored on a 220 µF capacitor is proportional to the potential difference (V), one can charge the capacitor while varying the resistance of a variable resistor to maintain a constant current. The charge can be calculated using the formula Q = It, where I is the current and t is the time taken for the capacitor to charge. The experiment should confirm that Q/V remains constant across the tested voltage range of 0 to 15 V. Concerns were raised about measuring charge accurately, as traditional coulombmeters may not suffice due to the capacitor's high capacitance. Maintaining a constant current could be challenging unless the charging process is sufficiently slow.
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Homework Statement


Describe how you would show experimentally that the charge stored on a 220 µF 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.


Homework Equations





The Attempt at a Solution



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- charge the capacitor for a given Voltage, V
- ensure the capacitor is charging at a constant rate by varying the resistance of the variable resistor --> ensures current remains constant
- Q=It (current * time take for capacitor to charge)
- capacitor is charged when V on meter = V of supply
- Q/V = constant for range of values of V

Is this method correct - would it work?

p.s: I am aware my diagram is not very well drawn - i will obviously improve this in the exam
 
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V is easy to measure...do you know of any way to measure Q?
 
truesearch said:
V is easy to measure...do you know of any way to measure Q?

my syllabus only talks about coloumbmeters but the MS said the a coloumbeter would be insufficient for this experiment because the capacitance is too high.

I was simply wondering, would my method work?
 
Your method sounds good to me - although it might be a bit tricky keeping the current constant unless the capacitor was charging very slowly.
 
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