Question

A parallel plate capacitor of capacitance 5 $\mu F$ and plate separation 6 cm is connected to a 1 V battery and charged. A dielectric of dielectric constant 4 and thickness 4 cm is introduced between the plates of the capacitor. The additional charge that flows into the capacitor from the battery is

• A. 2 $\mu C$
• B. 3 $\mu C$
• C. 5 $\mu C$
• D. 10 $\mu C$

Answer 1

Answer: (C)

5 $\mu C$

Charge on capacitor plates without the dielectric is
$=CV=(5\times {10}^{-6}F)\times 1V=5\times {10}^{-6}C=5\mu C$
The capacitance after the dielectric is introduced is
$C^{\prime }=\frac{{\epsilon }_{0}A}{d-(t-\frac{t}{K})}=\frac{{\epsilon }_{0}A/d}{1-\left(\frac{t-\frac{t}{K}}{d}\right)}=\frac{C}{1-\left(\frac{t-\frac{t}{K}}{d}\right)}=\frac{5\mu F}{1-\left(\frac{4cm-\frac{4cm}{4}}{6cm}\right)}=\frac{5\mu F}{1-\left(\frac{4-1}{6}\right)}=10\mu F$
$\therefore$ Charge on capacitor plate now will be
$Q^{\prime }=C^{\prime }V=10\mu F\times 1V=10\mu C$
Additional charge transferred $=Q^{\prime }-Q=10\mu C-5\mu C=5\mu C$