The plates of a parallel plate capacitor each have an area of 0.40 m2 and are separated by a distance of 0.02 m. They are charged until the potential difference between the plates is 3000 V. The charged capacitor is then isolated.

Question

How much work is required to move a –4.0 ìC charge from the negative plate to the positive plate of this system?

Work= force*distance= Eq*distance=Volt/distance*q*distance= Vq

So the work is Volts times charge.

thanks i though it was somnethin differnt cause they gave the distance and the area

Answer 1

W=deltaV*Q
W=(0-3000)*(-4*10^-6)
W=1.2*10^-2 J