Question

Capacity of a parallel plate condenser is $10\mu F$ when the distance between its plates is $8cm$. If the distance between the plates is reduced to $4cm$, its capacity will be :

• A. $10\mu F$
• B. $15\mu F$
• C. $20\mu F$
• D. $40\mu F$

Answer 1

Answer: (C)

$20\mu F$

The capacitance $C$ is given by,
$C=\frac{{\epsilon }_r{\epsilon }_{o}A}{d}$, where ${\epsilon }_r$ and ${\epsilon }_o$ are the permitivity of the material and free space respectively.

For a $10\mu F$ capacitance we can write,
$C_1=\frac{{\epsilon }_r{\epsilon }_{o}A}{8\times {10}^{-2}}F$
$C_2=\frac{{\epsilon }_r{\epsilon }_{o}A}{4\times {10}^{-2}}F$
$⟹\frac{C_1}{C_2}=\frac{4\times {10}^{-2}}{8\times {10}^{-2}}$
$⟹\frac{10\mu }{C_2}=\frac{4\times {10}^{-2}}{8\times {10}^{-2}}$
$⟹C_2=20\mu F$.