Answer:
276.12 pC
Explanation:
We are given that
Area,A=[tex]2.6 cm^2=2.6\times 10^{-4}m^2[/tex]
Where [tex]1 cm^2=10^{-4} m^2[/tex]
[tex]d=0.25 mm=0.25\times 10^{-3} m[/tex]
[tex]1mm=10^{-3} m[/tex]
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 ,[tex]Q=\frac{\epsilon_0 A V}{d}[/tex]
Where [tex]\epsilon_0=8.85\times 10^{-12}[/tex]
Using the formula
[tex]Q=\frac{8.85\times 10^{-12}\times 2.6\times 10^{-4}\times 30}{0.25\times 10^{-3}}[/tex]
[tex]Q=2.7612\times 10^{-10} C[/tex]
[tex]Q=276.12p C[/tex]
[tex]1 pC=10^{-12} C[/tex]
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
