Asked by Amanda
The electric potential increases from 10 V to 70 V from the bottom plate to the top plate of a parallel-plate capacitor. We are going to move a charge of +5 x 10-4 C from the bottom plate to the top plate. What is the magnitude of the change in potential energy of this charge? Do not enter any (-) sign in your answer.
Answers
Answered by
Damon
voltage is potential energy per unit charge.
because
The force on a charged particle in an electric field is q E
between the plates of your capacitor E= V/d
so the work done on the charge = Force * distance = q E d = = q V
so
5*10^-4*60 = 300 *10^-4 = 3*10^-2 or .03J
because
The force on a charged particle in an electric field is q E
between the plates of your capacitor E= V/d
so the work done on the charge = Force * distance = q E d = = q V
so
5*10^-4*60 = 300 *10^-4 = 3*10^-2 or .03J
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