Asked by kevin
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.
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
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
Answers
Answered by
Ajay Sharma
W=deltaV*Q
W=(0-3000)*(-4*10^-6)
W=1.2*10^-2 J
W=(0-3000)*(-4*10^-6)
W=1.2*10^-2 J
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