Question

A 0.40-μF capacitor is connected to a 5.0-V battery. How much charge is on each plate of the capacitor? Express your answer using two significant figures.

Answer 1

Answer:

$Q=2.0\times10^{-6}C$

Explanation:

The equation of charge on the capacitor given its capacitance and a voltage applied to it is Q=CV.

Our capacitance is expressed in microfarads, where micro stands for $\times10^{-6}$, so in S.I. we sill have:

$Q=CV=(0.4\times10^{-6}F)(5V)=0.000002C=2.0\times10^{-6}C$

Where the results has been expresed with two significant figures.