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jekas [21]
2 years ago
6

HURRY!!!!!!!

Physics
2 answers:
saul85 [17]2 years ago
7 0
The basic relationship between capacitance (C), charge (Q) and voltage (V) for a parallel-plate capacitor is
C= \frac{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
zzz [600]2 years ago
5 0

Answer:

3.6

Explanation:

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<em></em>

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