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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 d/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 plate 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=d/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 = (Ed)/3 + E_0 (d-d/3)=(Ed)/3 +E_0((2d)/3) = d/3 (E +2E_0) (therefore t=d/2)`
`or, V = d/3 (E_0/K +2E_0) (dE_0)/(3K)(2K +1) (As,E_0/E =K)`
Now,`E_0 = σ/epsi_0 = q/(epsi_0A) => V = d/(3K )q/(epsi_0A) (2K +1)`
We know, `C = q/V = (3Kepsi_0A)/(d(2K+1))`
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