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प्रश्न
A dielectric slab of thickness 't’ is kept between the plates of a parallel plate capacitor with plate separation 'd' (t < d). Derive the expression for the capacitance of the capacitor.
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उत्तर
P and Q are two parallel plates. The distance between them is d.
A dielectric material of thickness t(t < d) and dielectric constant K is inserted in between the plates.
In the remaining space (d - t) there is air.
The electric field in the dielectric-filled portion = `E_1 = sigma/(epsi_0K)`
The electric field in the air-filled portion = `E_2 = sigma/(epsi_0)`
The potential difference between the plates = V
V = E2 × (d - t) + E1 × t
Or, V = `sigma/(2epsi_0) xx (d - t) + sigma/(epsi_0K) xx t`
Or, V = `q/(Aepsi_0) xx [(d - t) + t/K]`
Now, capacitance = C = `q/V`
Or, C = `(epsi_0A)/((d - t) + t/K)`
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