A dielectric slab of thickness ‘t’ is kept in between the plates, each of area ‘A’, of a parallel plate capacitor separated by a distance ‘d’, Derive an expression for the capacitance of this capacitor for t << d.

Solution

Let, A be the area of the two parallel plate capacitor and, d be the seperation between them.

Given, a dielectric of thickness t<d is inserted in between the plates.

The total electric field inside the dielectric slab will be,

E = \frac{E_{0}}{K} = E_{0} - E',

where,
E_o is the initial electric field without the dielectric slab and,
E’ is the opposite field developed inside the slab due to polarisation of slab.

Total potential difference between the plates,

V = E_{0} (d - t) + Et = \frac{σ}{ε_{0}} (d - t) + \frac{σ}{{kε}_{0}} t = \frac{σ}{ε_{0}} [(d - t) + \frac{t}{k}]

V = \frac{q}{{Aε}_{0}} [(d - t) + \frac{t}{k}]

where, q is the charge on each plate,

σ

is the surface charge density,

ε_{\circ}

is the permittivity of medium
k is the dielectric constant.

Now, capacitance,

C = \frac{q}{V}

\Rightarrow

C = \frac{q}{\frac{q}{{Aε}_{0}} [(d - t) + \frac{t}{k}]}

C = \frac{{Aε}_{0}}{[(d - t) + \frac{t}{k}] .}

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Electrostatic Potential And Capacitance

A dielectric slab of thickness ‘t’ is kept in between the plates, each of area ‘A’, of a parallel plate capacitor separated by a distance ‘d’, Derive an expression for the capacitance of this capacitor for t << d.

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