Capacitor Potential, Separation, and Work Calculations

Chapter 24, Problem 24.24

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QUESTION:

A parallel-plate air capacitor has a capacitance of 920 pF. The charge on each plate is \(2.55 \mu \mathrm{C}\). (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 on each plate is \(2.55 \mu \mathrm{C}\). (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:

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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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Capacitor Potential, Separation, and Work Calculations
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Explore the concept of capacitance and learn how to calculate potential difference, the impact of separation, and the work required for a parallel-plate air capacitor.


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