Question

Derive an expression for the capacitance of a parallel plate air Capacitor.
How the capacitance will change if a dielectric slab of thickness 't' is inserted
between the plates of capacitor.​

Answer 1

$\huge\boxed{\fcolorbox{red}{yellow}{Shailesh}}$

Let A be the plate area and d be the distance between the plates,

1) A dielectric slab of thickness t is inserted,

Due to polarization inside the dielectric electric field will reduce,

So outside of dielectric, field = E₀

inside of dielectric, field = E = E₀/k

where k is the dielectric constant,

So the net potential will be,

V = Et + E₀(d-t)

=> V = E₀t/k + E₀(d-t)

=> V = E₀(d - t + t/k)

we know that, E₀ = q/ε₀A

=> V = q(d - t + t/k)/ε₀A

=> C = q/V

=> C = ε₀A/(d - t + t/k)

which is the required expression for capacitance,

2) A metallic plate of thickness t is inserted,

when a conducting metallic plate is inserted then electric field inside the plate will be zero

=> E = E₀/k = 0

=> k = ∞

hence putting the value of k in the above equation we get,

C = ε₀A/(d - t)

which is the required expression for capacitance with metallic plate

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