A parallel plate capacitor of area A = 30 cm² and separation d = 5 mm is charged by a battery of 60 V. If the air between the plates is replaced by a dielectric of κ = 4 with the battery still connected, then what is the ratio of the initial charge on the plates divided by the final charge on the plates?

Question

Quizplus · Accepted Answer

The capacitance of a parallel plate capacitor is given by

$C=\frac{\kappa \epsilon_0 A}{d}$

where κ is the dielectric constant, ε0 is the permittivity of free space, A is the area of the plates, and d is the separation between the plates.

Initially, the capacitance of the parallel plate capacitor with air is

$C_1=\frac{\epsilon_0 A}{d}=\frac{(8.85	imes10^{-12}\,\mathrm{F/m})\,(0.003^2\,\mathrm{m}^2)}{0.005\,\mathrm{m}}=1.06	imes10^{-10}\,\mathrm{F}$

When the air is replaced with the dielectric, the capacitance becomes

$C_2=\frac{\kappa \epsilon_0 A}{d}=\frac{(4)(8.85	imes10^{-12}\,\mathrm{F/m})\,(0.003^2\,\mathrm{m}^2)}{0.005\,\mathrm{m}}=4.25	imes10^{-10}\,\mathrm{F}$

The charge on the parallel plate capacitor is given by

$Q=CV$

where V is the applied voltage. Since the battery is still connected, the voltage remains at 60 V. Thus, the initial charge on the capacitor is

$Q_1=C_1V=(1.06	imes10^{-10}\,\mathrm{F})(60\,\mathrm{V})=6.36	imes10^{-9}\,\mathrm{C}$

and the final charge on the capacitor is

$Q_2=C_2V=(4.25	imes10^{-10}\,\mathrm{F})(60\,\mathrm{V})=2.55	imes10^{-8}\,\mathrm{C}$

The ratio of the initial charge to the final charge is therefore

$\frac{Q_1}{Q_2}=\frac{6.36	imes10^{-9}\,\mathrm{C}}{2.55	imes10^{-8}\,\mathrm{C}}=0.25$

Therefore, the answer is D.

A Parallel Plate Capacitor of Area a = 30 Cm2

A Parallel Plate Capacitor of Area a = 30 Cm²