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The two plates of a parallel plate capacitor are $4\;mm$ apart. A slab of dielectric constant 3 and thickness 3 mm is introduced between the plates with its faces parallel to them. The distance between the plates is so adjusted that the capacitance of the capacitor becomes (2/3) rd of its original value. What is the new distance between the plates?

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$C_1 \bigg( \large\frac{K_1+K_2}{2} \bigg)$
Hence A is the correct answer.
answered Jun 16, 2014 by meena.p

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