A capacitor has two circular plates,each of r | vedclass

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(B) The area of the plates is $A = \pi r^2 = \pi 	imes (8 	imes 10^{-2})^2 \approx 0.0201\,m^2$ and the distance is $d = 1\,mm = 1 	imes 10^{-3}\,m

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$1.2 	imes 10^{-5}\,J$

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(B) The area of the plates is $A = \pi r^2 = \pi 	imes (8 	imes 10^{-2})^2 \approx 0.0201\,m^2$ and the distance is $d = 1\,mm = 1 	imes 10^{-3}\,m$.The capacitance of the capacitor with the dielectric is given by $C = \frac{K \varepsilon_0 A}{d}$.Substituting the values: $C = \frac{6 	imes 8.854 	imes 10^{-12} 	imes 0.0201}{1 	imes 10^{-3}} \approx 1.068 	imes 10^{-9}\,F$.The potential difference is $V = 150\,V$.The energy stored in the capacitor is $U = \frac{1}{2}CV^2$.$U = \frac{1}{2} 	imes (1.068 	imes 10^{-9}) 	imes (150)^2$.$U = 0.5 	imes 1.068 	imes 10^{-9} 	imes 22500$.$U \approx 1.2 	imes 10^{-5}\,J$.

$A$ capacitor has two circular plates,each of radius $8\,cm$,separated by a distance of $1\,mm$. $A$ dielectric slab (dielectric constant $K = 6$) is inserted between the plates. Calculate the energy stored in the capacitor when it is connected to a $150\,V$ potential difference.

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