Question

A 11 volt battery is connected to a capacitor of 2.3 uF. What is the charge in of the capacitor?

Answer 1

`Q=2.53*10^(-5)C`

Step-by-step explanation: The amount of charge that moves into the capacitor plates depends upon the capacitance and the applied voltage, it is given by the following formula:

`Q=CV\Rightarrow(1)`

To find the charge on the capacitor we need to identify the knows and unknows as follows:

`\begin{gathered} C=2.3\mu F\Rightarrow C=2.3*10^(-6)F \\ V=11V \end{gathered}`

Plugging these values in formula (1) we get the answer as follows:

`\begin{gathered} Q=(2.3*10^(-6)F)\cdot(11V)=0.0000253=2.53*10^(-5)C \\ Q=2.53*10^(-5)C \end{gathered}`

Therefore the charge on the capacitor is:

`Q=2.53*10^(-5)C`