12.5 mathrm{pF} સંધારકતા ધરાવતા એક સમાંતર પ્લ | vedclass

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Question

Before inserting dielectric capacitance is given $\mathrm{C}_0=12.5 \mathrm{pF}$ and charge on the capacitor $\mathrm{Q}=\mathrm{C}_0 \mathrm{~V}$ Aft

Vedclass · Accepted Answer

$750$

Vedclass · Answer

Before inserting dielectric capacitance is given $\mathrm{C}_0=12.5 \mathrm{pF}$ and charge on the capacitor $\mathrm{Q}=\mathrm{C}_0 \mathrm{~V}$ After inserting dielectric capacitance will become $\epsilon_{\mathrm{s}} \mathrm{C}_0$.

Change in potential energy of the capacitor

$=\mathrm{E}_{\mathrm{i}}-\mathrm{E}_{\mathrm{r}}$

$=\frac{\mathrm{Q}^2}{2 \mathrm{C}_{\mathrm{i}}}-\frac{\mathrm{Q}^2}{2 \mathrm{C}_{\mathrm{f}}}=\frac{\mathrm{Q}^2}{2 \mathrm{C}_0}\left[1-\frac{1}{\epsilon_{\mathrm{r}}}ight]$

$=\frac{\left(\mathrm{C}_0 \mathrm{~V}ight)^2}{2 \mathrm{C}_0}\left[1-\frac{1}{\epsilon_{\mathrm{r}}}ight]=\frac{1}{2} \mathrm{C}_0 \mathrm{~V}^2\left[1-\frac{1}{\epsilon_{\mathrm{r}}}ight]$

Using $\mathrm{C}_0=12.5 \mathrm{pF}, \mathrm{V}=12 \mathrm{~V}, \epsilon_{\mathrm{r}}=6$

$=\frac{1}{2}(12.5) 	imes 12^2\left[1-\frac{1}{6}ight]=\frac{1}{2}(12.5) 	imes 12^2 	imes$

$\frac{5}{6}$

$=750 \mathrm{pJ}=750 	imes 10^{-12} \mathrm{~J}$

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