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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 .

Solution
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Initially when there is vacuum between the two plates, the capacitance of the capictor is
C0=ε0Ad
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 reduce to E
Eis producedow, if we insert the dielectric slab of thickness t=
t=d2the electric field reduces to E
Now, the gap between plates is divided in two parts, for distance there is electric field E and for the remaining distance
(d-t) the electric field is E0
0
ε0
If V be the potential difference between the plates of the capacitor, then

V=Et+E0(dt)V=Ed2+E0d2=d2(E+E0)(t=d2)

V=d2(E0K+E0)=dE02K(K+1)(asE0E=K)

nowE0=σε0=qε0AV=d2Kqε0A(K+1)weknowC=qV=2Kε0A(K+1)d

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