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Capacitance of capacitor becomes $\large\frac{4}{3}$ times its original value if a dielectric slab of thickness $\large\frac{d}{2}$ is inserted b/w the plates (d= seperation between the plates) . The dielectric constant of the slab is:

$(A)\;2 \\ (B)\;4 \\ (C)\;6 \\ (D)\;8 $

1 Answer

Original capacity $C_0= \large\frac{\in _09 A}{d}$
On introducing dielectric slab of thickness $\large\frac{d}{2}$ . the capacity becomes
$C= \large\frac{\in _0 A}{\Large\frac{d}{2k} + \bigg(d -\Large\frac {d}{2}\bigg ) }$
$\quad= \large\frac{ \in _0 A}{\Large\frac{d}{2} \bigg( \Large\frac{1}{K} +1\bigg) }$
As $c=\large\frac{4}{3}$
$\quad= \large\frac{ \in _0 A}{\Large\frac{d}{2} \bigg(1+ \Large\frac{1}{K} \bigg) }$
$\qquad= \large\frac {\in _0 A}{d} \times \large\frac{4}{3}$
=> $1+ \large\frac{1}{k}=\large\frac{3}{2}$
=> $\large\frac{1}{k} =\frac{1}{2}$
$K=2$
Hence A is the correct answer.
answered Jan 8, 2014 by meena.p
 

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