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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.
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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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