Related questions with answers
Suppose that you charge a 1 F capacitor in a circuit containing two 1.5 V batteries, so the final potential difference across the plates is 3 V. How much charge is on each plate? How many excess electrons are on the negative plate?
Solutions
VerifiedGiven
We are given two batteries connected to the capacitor where the emf of one battery is 1.5 V. So, the two batteries in series will produce emf = 2(1.5 V) = 3 V.
Solution
The capacitor is charged with two batteries 3 , this means the potential difference between the two parallel plates is . Where the potential difference is related to the capacitance and the charge on the plates by
Now we can plug our values for and into equation (1) to get the magnitude of the charge on each plate
The charges on the plates is due to the number of electrons or positive charges. So, the number of excess electrons equals the charge per the charge of one electron
Now we can plug our values for and to get the number of excess electrons
From the definition of capacitance.
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