MCQ
A capacitor has capacitance $5 \mu F$ when it's parallel plates are separated by air medium of thickness $d$. A slab of material of dielectric constant $1.5$ having area equal to that of plates but thickness $\frac{ d }{2}$ is inserted between the plates. Capacitance of the capacitor in the presence of slab will be $..........\mu F$
  • A
    $5$
  • $6$
  • C
    $4$
  • D
    $3$

Answer

Correct option: B.
$6$
b
$C _{\text {new }}=\frac{\in_0 A }{\frac{\left(\frac{ d }{2}\right)}{1.5}+\frac{\left(\frac{ d }{2}\right)}{1}}$

$=\frac{\in_0 A}{\left(\frac{d}{3}+\frac{d}{2}\right)}=\frac{6 \in_0 A}{5 d}$

$=\frac{6}{5} \times 5 \mu F =6 \mu F$

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