9. Show that each plate of a parallel-plate capacitor exerts a force on the 1 Q² 2€,A F = " other equal to by calculating the rate at which the work changes by increasing the separation, W. Why does using F = QE, with E being the electric field between the plates, give the wrong answer? dW dr

9. Show that each plate of a parallel-plate capacitor exerts a force on the 1 Q² 2€,A F = " other equal to by calculating the rate at which the work changes by increasing the separation, W. Why does using F = QE, with E being the electric field between the plates, give the wrong answer? dW dr

Physics for Scientists and Engineers, Technology Update (No access codes included)

9th Edition

ISBN:9781305116399

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter26: Capacitance And Dielectrics

Section: Chapter Questions

Problem 26.14OQ: You charge a parallel-plate capacitor, remove it from the battery, and prevent the wires connected...

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Water constitutes about 70% of earth. Some important distinguishing properties of water are high molar concentration, small dissociation constant and high dielectric constant.

Electrostatic Potential and Capacitance

An electrostatic force is a force caused by stationary electric charges /fields. The electrostatic force is caused by the transfer of electrons in conducting materials. Coulomb’s law determines the amount of force between two stationary, charged particles. The electric force is the force which acts between two stationary charges. It is also called Coulomb force.

Question

9. Show that each plate of a parallel-plate capacitor exerts a force on the
1 Q²
2€,A'
F =
other equal to
by calculating the rate at which the work changes by increasing the
separation, Why does using F = QE, with E being the electric
field between the plates, give the wrong answer?
dW
da

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