Question

A parallel plate condenser has initially air medium between the plates. If a slab of dielectric constant 5 having thickness half the difference of separation between the plates is introduced, the percentage increase in its capacity is :

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Solution

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Correct option is B)

$t=2d ,K=25$

$C=[(d−t)+kt ]ε_{o}A $

$C=dε_{o}A $

Initially capacitance is C

$C_{1}=[d−2d +2(5)d ]ε_{o}A $

$C_{1}=2d [1+51 ]ε_{o}A $

$C_{1}=d[1+51 ]ε_{o}A $

$C_{1}=d[1+51 ]ε_{o}A $

Therefore % increase =$[610 −1]×100=66.7%$

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