Parallel plate capacitor's charge

by livewire5
Tags: capacitators, charge, electric field, surface charge
livewire5 is offline
Feb11-12, 10:52 AM
P: 7

In this question the charge is calculated by isolating Q in E=surface charge density/absalon not

My question is, the electric field of an infinite charged plane is E=surface charge density/(2*absalon naught)

because the electric field in center of parallel plate capacitor are parallel we add

surface charge density/(2*absalon naught)+surface charge density/(2*absalon naught)

to give us E between the electrodes to be surface charge density/absalon not

but when that question uses that formula to find the charge, shouldnt the charge found be divided by 2 to get the charge on each electrode since that formula we initially created by adding the E of both electrodes n this question asks to find charge on each electrode...

so what i am really saying is that why isnt the Q in that formula (in that link) charge of both electrodes (their magnitude ofcourse) why is that Q the charge on one electrode?
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BruceW is offline
Feb12-12, 06:35 AM
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P: 3,337
As you said, the electric field due to one plate is surface charge density/(2*absalon naught) So this is equal to Q/(2A*epsilon naught) (where A is area, and Q is charge on one plate). And the total electric field is twice this (since there are two oppositely charged plates), so it is Q/(A*epsilon naught)

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