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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/2, 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/2 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)/2 + (E_0d)/2 =d/2 (E +E_0) (therefore t =d/2)`
`=> V=d/2 ((E_0)/K +E_0) (dE_0)/(2K) (K+1) (As,(E_0)/E = K)`
Now, `E_0 =σ/epsi_0 = q/(epsiA )=>V = d/(2K) q/(epsi_0A) (K +1)`
we know, `C = q/V = (2Kepsi_0A)/(d(K+1))`
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