A parallel-plate capacitor with plate area 2.6 cm^2 and air-gap separation 0.25 mm is connected to a 30 V battery, and fully charged. The battery is then disconnected. I Solved most of the parts of this q's but not this one:

The plates are now pulled to a separation of 0.55 mm. What is the charge on the capacitor now?

note:

I found 125.5pC and the answer unit is pC

276.12 pC

Explanation:

We are given that

Area,A=

Where

Potential difference, V=30  V

We have to find the charge on the capacitor when the plates are pulled to a separation of 0.55 mm.

We know that

Charge ,

Where

Using the formula

When the plates are now pulled to a separation of 0.55 mm.Then, the charge on the plates remain same because the  battery has been disconnected.

Therefore, charge on the capacitor=276.12 pC

less than one you can't create energy