Question

HURRY!!!!!!!

A capacitor has a capacitance of 0.40 µF at a voltage of 9.0 V. What is the charge on each plate of the capacitor?

Answer 1

3.6

Step-by-step explanation:

Answer 2

The basic relationship between capacitance (C), charge (Q) and voltage (V) for a parallel-plate capacitor is

`C= (Q)/(V)`
in our problem,
`C=0.40 \mu F=0.40 \cdot 10^(-6) F` and V=9.0 V, so the charge on each plate is

`Q=CV=(0.40 \cdot 10^(-6)F)(9.0 V)=3.6 \cdot 10^(-6) C`