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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 seperation b/w the plates. Find out expression for its capacitance when slab is inserted b/w the plates of capacitor |
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Answer» hareStudy laterVIDEO EXPLANATIONANSWERInitially when there is vacuum between the two PLATES, the capacitance of the capictor is C 0 = dε 0 A Where, A is the area of PARALLEL plates Suppose that the capacitor is connected to a battery, an electric field E 0 is producednow if we insert the dielectric slab of thickness t=d/2 the electric field reduce to EEis producedow, if we insert the dielectric slab of thickness t=t=d2the electric field REDUCES to ENow, 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 E 0 0ε0 If V be the potential DIFFERENCE between the plates of the capacitor, thenV=E t +E 0 (d−t)V= 2Ed + 2E 0 d = 2d (E+E 0 )(t= 2d ) V= 2d ( KE 0 +E 0 )= 2KdE 0 (K+1)(as EE 0 =K)nowE 0 = ε 0 σ = ε 0 Aq ⇒V= 2Kd ε 0 Aq (K+1)weknowC= Vq = (K+1)d2Kε 0 A |
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