3 years ago

HURRY!!!!!!!

2 answers:

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]3 years ago

5 0

Answer:

3.6

Explanation:

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