How Is Capacitance Calculated in a Charged Parallel-Plate Capacitor?

AI Thread Summary
Capacitance in a charged parallel-plate capacitor is calculated using the formula C = Q/V, where Q is the charge and V is the potential difference. In this scenario, the charges on the plates are +900uC and +100uC, resulting in a potential difference of 4V. The initial calculation using +900uC gives a capacitance of 225uC, but considering the total charge transferred (400uC), the correct capacitance is 100uC. The discussion also touches on the need to understand Gauss' law for further insights into capacitance. Understanding these concepts is essential for accurately calculating capacitance in similar problems.
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Homework Statement


The two plates of a parallel-plate capacitor initially carry equal amount of positive charge. If some charges are transferred from one plate to another, the charges on the plates are respectively +900uC and +100uC. The potential difference across the plates becomes 4V. What is the capacitance of the capacitor?

Homework Equations



C=Q/V

The Attempt at a Solution


I just pick randomly pick +900uC instead of +100uC:
C=900/4=225uC

4. Then I check the answer:

Totally 400uC moves away from one to the metal plate and move to another plate, so:
C=400/4=100uC

According to what I have learn, Q is the charge of one of the metal plate. In here, why should I subsitute 400uC, the amount of charge that is transferred?
 
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Consider the relative charge...

Have you met Gauss' law?
 
Last edited:
I haven't heard about it before. I will try to learn it by myself. Thank you:rolleyes:
 
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