A slab of material with a dielectric constant | vedclass

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(C) In the absence of a dielectric,the electric field between the two plates is $E_0 = \frac{V_0}{d}$.When the dielectric slab is inserted,the electri

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$\frac{V_0}{2}$

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(C) In the absence of a dielectric,the electric field between the two plates is $E_0 = \frac{V_0}{d}$.When the dielectric slab is inserted,the electric field inside the dielectric becomes $E = \frac{E_0}{K}$,where $K=3$ is the dielectric constant.The thickness of the dielectric slab is $t = \frac{3}{4}d$,and the thickness of the air gap is $d - t = d - \frac{3}{4}d = \frac{1}{4}d$.The new potential difference $V$ is the sum of the potential drops across the air gap and the dielectric slab:$V = E_0 \left(\frac{1}{4}dight) + E \left(\frac{3}{4}dight)$$V = E_0 \left(\frac{1}{4}dight) + \frac{E_0}{K} \left(\frac{3}{4}dight)$$V = E_0 d \left[ \frac{1}{4} + \frac{3}{4K} ight]$Since $E_0 d = V_0$ and $K = 3$:$V = V_0 \left[ \frac{1}{4} + \frac{3}{4 	imes 3} ight]$$V = V_0 \left[ \frac{1}{4} + \frac{1}{4} ight]$$V = V_0 \left( \frac{2}{4} ight) = \frac{V_0}{2}$

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