The plates of a parallel plate capacitor are square with a side of d = 0.78 cm a

Question

The plates of a parallel plate capacitor are square with a side of d = 0.78 cm and the voltage across the plates is 3.46 V. The distance between the plates is 2.00 mm. One capacitor plate has an excess of electrons and the other has a matching deficit of electrons. What is the number of excess electrons?

The plates of a parallel plate capacitor are square with a side of d = 0.78 cm and the voltage across the plates is 3.46 V. The distance between the plates is 2.00 mm. One capacitor plate has an excess of electrons and the other has a matching deficit of electrons. What is the number of excess electrons?

Explanation / Answer

Apply Q = CV

First we need C which is found from C = kEA/d

C = (1)(8.85 X 10^-12)(.0078)(.0078)/(2 X 10^-3)

C = 2.69 X 10^-13 F

Q = (2.69 X 10^-13)(3.46)

Q = 9.31 X 10^-13

Divide that by the charge per electron

9.31 X 10^-13/1.6 X 10^-19 = 5.82 X 10^6 electrons

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