Question

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

Answer 1

276.12 pC

Step-by-step explanation:

We are given that

Area,A=
`2.6 cm^2=2.6* 10^(-4)m^2`

Where
`1 cm^2=10^(-4) m^2`

`d=0.25 mm=0.25* 10^(-3) m`

`1mm=10^(-3) m`

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 ,
`Q=(\epsilon_0 A V)/(d)`

Where
`\epsilon_0=8.85* 10^(-12)`

Using the formula

`Q=(8.85* 10^(-12)* 2.6* 10^(-4)* 30)/(0.25* 10^(-3))`

`Q=2.7612* 10^(-10) C`

`Q=276.12p C`

`1 pC=10^(-12) C`

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