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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?

User Alena
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2 Answers

1 vote

Answer:

3.6

Step-by-step explanation:

User Allicarn
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3 votes
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
User Jed Anderson
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