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