In the previous problem (Question 4), if the distance between the two plates of the capacitor is 3 cm, what is the magnitude of the uniform electric field between the two plates?
d = .03 meters
E = V/d
so... E=.03J/3m=0.01J/m??
no
I do not know what question 4 is
but divide the voltage by .03 meters to get E
in a capacitor E = volts/ distance
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.
so E=v/d
E=.03/3
E=.01?
V = 60 volts
d = .03 meters
E = 60/.03 = 2000
To find the magnitude of the uniform electric field between the two plates of a capacitor, you need to know the voltage across the plates and the distance between the plates. In this case, you are given the distance between the plates as 3 cm, but no information about the voltage is provided.
The electric field between the plates of a parallel plate capacitor is given by the equation:
E = V / d
Where E is the magnitude of the electric field, V is the voltage across the plates, and d is the distance between the plates.
Without the voltage information, it is not possible to calculate the magnitude of the electric field. You would need to know the voltage or have some additional information to determine the electric field between the plates.