Answer:
Current in the capacitor will be equal to 0.01 A
Step-by-step explanation:
Capacitance of the capacitor
![C=20nF=20* 10^(-9)F](https://img.qammunity.org/2021/formulas/physics/college/d7uuqb5tzma1llwxfh96o3h72uchu621uv.png)
Capacitor voltage is change to 5 volt
So change in capacitor voltage
![dV=5volt](https://img.qammunity.org/2021/formulas/physics/college/ue3trkv0eupwriv6mdtysuv63xk78eur6o.png)
Time taken to change in capacitor voltage
![dt=10\mu sec=10^(-5)sec](https://img.qammunity.org/2021/formulas/physics/college/qqo5i7xe9ihe5bbwmx8wc5y3vb7b5dlu2i.png)
Current in the capacitor is equal to
![i=C(dV)/(dt)=20* 10^(-9)* (5)/(10^(-5))=0.01A](https://img.qammunity.org/2021/formulas/physics/college/a8sizy2ml434ikeeby8b5kenf26z6st4sy.png)
So current in the capacitor will be equal to 0.01 A