Question

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

Answer 1

Answer:

$Q=2.53\times10^{-5}C$

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:

$\begin{gathered} C=2.3\mu F\Rightarrow C=2.3\times10^{-6}F \\ V=11V \end{gathered}$

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

$\begin{gathered} Q=(2.3\times10^{-6}F)\cdot(11V)=0.0000253=2.53\times10^{-5}C \\ Q=2.53\times10^{-5}C \end{gathered}$

Therefore the charge on the capacitor is:

$Q=2.53\times10^{-5}C$