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.
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
To find the magnitude of the change in potential energy of the charge, we can use the equation:
ΔPE = q * ΔV
Where:
ΔPE is the change in potential energy
q is the charge
ΔV is the change in electric potential
In this case, the charge is +5 x 10^-4 C and the change in electric potential is 70 V - 10 V = 60 V.
Now we can calculate the magnitude of the change in potential energy:
ΔPE = (+5 x 10^-4 C) * (60 V)
= 3 x 10^-2 J
Therefore, the magnitude of the change in potential energy of the charge is 3 x 10^-2 J.