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Q. The capacitance of an air filled parallel plate capacitor is $10\, \mu F$. The separation between the plates is doubled and the space between the plates is then filled with wax giving the capacitance a new value of $40 \times 10^{-12}$ farads. The dielectric constant of wax is

Electrostatic Potential and Capacitance

Solution:

$C_{1}=\frac{\varepsilon_{0} A}{d}$
and $C_{2}=\frac{K \varepsilon_{0} A}{2 d}$
$\Rightarrow \frac{C_{2}}{C_{1}}=\frac{K}{2}$
$\Rightarrow \frac{40 \times 10^{-12}}{10 \times 10^{-12}}=\frac{K}{2}$
$\Rightarrow K=8$