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A parallel-plate air capacitor has a capacitance of 920

Chapter 24, Problem 24E

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QUESTION:

A parallel-plate air capacitor has a capacitance of 920 pF. The charge 011 each plate is 2.55 µC. (a) What is the potential difference between the plates? (b) If the charge is kept constant, what will be the potential difference between the plates if the separation is doubled? (c) How much work is required to double the separation?

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QUESTION:

A parallel-plate air capacitor has a capacitance of 920 pF. The charge 011 each plate is 2.55 µC. (a) What is the potential difference between the plates? (b) If the charge is kept constant, what will be the potential difference between the plates if the separation is doubled? (c) How much work is required to double the separation?

ANSWER:

Solution 24E Step 1: Electric field is the negative gradient of the potential difference. E=(V/d) The charge in a parallel plate capacitor Q=CV

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