Question

The capacitor of capacitance C=(2.0±0.1)μF is charged to potential diffrence V=(20±0.2)v . what is tha charge on the capacitor?

Answer 1

Q=CV

Q=(2.0+-0.1)×10^-6(20+-0.2)

Q=(40+-0.3)×10^-6

Answer 2

Answer:

solution. (40±0.3)*10–⁶