Question

What is the charge on a 0.25 μF capacitor when the capacitor is connected to a 9.0 V battery?Group of answer choices1.2x10^-12 C2.3x10^-6 C3.6x10^-7 C2.8x10^-8 C

Answer 1

Step-by-step explanation

the charge is given by the expression

`\begin{gathered} Q=CV \\ \end{gathered}`

Step 1

Let

`\begin{gathered} Q=CV \\ \end{gathered}`
`\begin{gathered} Capaci\tan ce;0.25\cdot10^(-6)F \\ \text{Voltage}=\text{ 9.0 V} \end{gathered}`

hence

`\begin{gathered} Q=0.25\cdot10^(-6)F\cdot9.0\text{ V} \\ Q=2.25\cdot10^(-6)\text{ C} \\ \text{rounded} \\ Q=2.3\cdot10^(-6)\text{ C} \end{gathered}`

therefore, the answer is

`\begin{gathered} B)2.3\cdot10^(-6)\text{ C} \\ \end{gathered}`

I hope this helps you