A parallel plate capacitor of capacitance 2 `muF` is fully charged to a potential difference of 500 volts and then disconnected from the battery. Using insulating handles, separation between the paltes is made thrice the initial value. What will be the new potential difference between the plates of the capacitor and by what amount stored energy changes in this process?

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the initial charge stored in the capacitor
The charge \( Q \) stored in a capacitor is given by the formula:
\[ Q = C \times V \]
Where:
...