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?

Answer 1

d = .03 meters
E = V/d

Answer 2

so... E=.03J/3m=0.01J/m??

Answer 3

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

Answer 4

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.

Answer 5

so E=v/d

E=.03/3
E=.01?

Answer 6

V = 60 volts
d = .03 meters

E = 60/.03 = 2000