Calculating electric charge from graph (capacitor)

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To calculate the electric charge on a capacitor from a voltage versus time graph, integration from zero to the fully charged time is necessary. The initial voltage is 10 V, with a capacitance of 2 µF and a resistance of 1 MΩ. There is confusion regarding whether the process is charging or discharging, as the initial voltage suggests discharging while the context implies charging. The calculated solution from the graph program is approximately 236 Vs, but its relevance remains unclear. Clarification on the charging status and correct terminology for capacitance and resistance is also discussed.
krisu334
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Homework Statement
When charging a capacitor we obtained a graph of voltage in terms of time. From the graph, find the amount of electronic charge on the capacitor.
Relevant Equations
Initial voltage: 10 V
Capacity: 2*10^(-6) Fahr.
Resistance: 1*10^6 Ohm
Apparently, we need to integrate the functions from 0 to the time when it is fully charged. However, I integrated in terms of t so the soultion (according to a graph programme) should be around 236 Vs but I don’t see how this could help me.
 
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Hello @krisu334 ,
:welcome: ##\ ##!​
What is the expected relationship between ##V## and ##t##?
 
krisu334 said:
Homework Statement: When charging a capacitor we obtained a graph of voltage in terms of time. From the graph, find the amount of electronic charge on the capacitor.
Relevant Equations: Initial voltage: 10 V
Capacity: 2*10^(-6) Fahr.
Resistance: 1*10^6 Ohm

Apparently, we need to integrate the functions from 0 to the time when it is fully charged. However, I integrated in terms of t so the soultion (according to a graph programme) should be around 236 Vs but I don’t see how this could help me.
Hi @krisu334. In addition to @BvU ’s question:

Presumably V is the voltage across the capacitor. Are you charging or discharging? You say “Initial voltage: 10 V” which implies you are discharging. But you also say “from 0 to the time when it is fully charged” which implies charging.

Minor points, for information:
The unit of capacitance (not “capacity”) is the ‘farad’ (lower case), symbol ‘F’.
The unit of resistance is the ‘ohm’ (lower case), symbol ‘Ω’.
 
Well, what is the status?
 
Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

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