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Question
A slab of material of dielectric constant K has the same area as that of the plates of a parallel plate capacitor but has the thickness 2d/3, where d is the separation between the plates. Find out the expression for its capacitance when the slab is inserted between the plates of the capacitor.
Solution
Initially when there is vacuum between the two plates, the capacitance of the two parallel plates is, ,`C_0 = (epsi_0A)/d` where, A is the area of parallel plates.
Suppose that the capacitor is connected to a battery, an electric field E0 is produced.
Now if we insert the dielectric slab of thickness `t = (2d)/3`the electric field reduces to E.
Now the gap between plates is divided in two parts, for distance t there is electric field E and for the remaining distance (d–t) the electric field is E0.
If V be the potential difference between the plates of the capacitor, then V=Et + E0(d–t)
`V = (2Ed)/3 +E_0 (d-(2d)/3) = (2Ed)/3 +(E_0d)/3 = d/3 (2E + E_0) (because t = d/2)`
`or , V = d/3 ((2E_0)/K + E_0) = (dE_0)/(3K) (K+2) (As,E_0/E = K)`
`Now , E_0 = σ/epsi_0 = q/(epsi_0A) => V = d/(3K) q/(epsi_0A) (K+2)`
`therefore C = q/V = (3K_(epsi_0 A))/(d(K+2))`.
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