Answer: The plates of a parallel-plate capacitor are 3.28

Chapter 24, Problem 2E

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QUESTION:

The plates of a parallel-plate capacitor are 3.28 mm apart, and each has an area of \(12.2\mathrm{\ cm}^2\). Each plate carries a charge of magnitude \(4.35\times 10^{-8}\mathrm{\ C}\). The plates are in vacuum.

(a) What is the capacitance?

(b) What is the potential difference between the plates?

(c) What is the magnitude of the electric field between the plates?

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QUESTION:

The plates of a parallel-plate capacitor are 3.28 mm apart, and each has an area of \(12.2\mathrm{\ cm}^2\). Each plate carries a charge of magnitude \(4.35\times 10^{-8}\mathrm{\ C}\). The plates are in vacuum.

(a) What is the capacitance?

(b) What is the potential difference between the plates?

(c) What is the magnitude of the electric field between the plates?

ANSWER:

Step 1 of 4

The capacitance of a parallel plate capacitor is given by , where  is vacuum permittivity,  is the area of each plate and  is the separation between the plates.

Given,

Area

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