Question

$A6\times {10}^{-4}$ F parallel plate air capacitor is connected to a 500 V battery. When air is replaced by another dielectric material, $7.5\times {10}^{-4}C$ charge flows into the capacitor. The value of the dielectric constant of the material is:

• A. 1.5
• B. 2
• C. 1.0025
• D. 3.5

Answer 1

The correct option is C 1.0025

Let the dielectric constant be $K$.
Capacitance of parallel plate air capacitor $C=6\times {10}^{-4}F$
Voltage of the battery $V=500$ volts
Thus charge stored on air capacitor $Q_a=CV=(7.5\times {10}^{-4})\left(500\right)=3000\times {10}^{-4}$ C
Extra charge flown $Q=7.5\times {10}^{-4}$ C
Thus total charge stored on dielectric capacitor $Q^{\prime }=(3000+7.5)\times {10}^{-4}=3007.5\times {10}^{-4}$ C
Thus capacitance of parallel plate dielectric capacitor $C=\frac{Q^{\prime }}{V}=\frac{3007.5\times {10}^{-4}}{500}=6.015\times {10}^{-4}F$
Capacitance of parallel plate dielectric capacitor $C^{\prime }=KC$
$\therefore$ $.015\times {10}^{-4}=K(6\times {10}^{-4})$
$⟹K=1.0025$