Question

Derive an expression for the capacitance of a parallel plate capacitor having a dielectric slab between its plate sand discuss the special cases.

Answer 1

Determine the area of parallel plate capacitor in the air if the capacitance is 25 nF and separation between the plates is 0.04m. Therefore, area of parallel plate capacitor is 117.09 m2.

Answer 2

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