Answer:
13.2\mu C
Step-by-step explanation:
The capacitance of the capacitor C = 2.2 microfarad
![=2.2* 10^(-6)F](https://img.qammunity.org/2020/formulas/physics/college/vk5rtkodufn408qfig18fh7qyc4haitp5h.png)
Voltage at the capacitor =6 V
The charge on the capacitor is equal to the product of capacitance and the voltage , so charge = capacitance ×voltage
So charge
![Q=CV=2.2* 10^(-6)* 6=13.2* 10^(-6)C=13.2\mu C](https://img.qammunity.org/2020/formulas/physics/college/usgjefv8cirmwribu8lhe2fmtu472fehno.png)