A parallel plate capacitor with width 4,cm,le | vedclass

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Question

(A) The capacitor can be considered as two capacitors in parallel: one with air and one with the dielectric slab.Area of the air part $A_1 = (7\,cm

Vedclass · Accepted Answer

$240$

Vedclass · Answer

(A) The capacitor can be considered as two capacitors in parallel: one with air and one with the dielectric slab.Area of the air part $A_1 = (7\,cm 	imes 4\,cm) = 28\,cm^2 = 28 	imes 10^{-4}\,m^2$.Area of the dielectric part $A_2 = (1\,cm 	imes 4\,cm) = 4\,cm^2 = 4 	imes 10^{-4}\,m^2$.Separation $d = 4\,mm = 4 	imes 10^{-3}\,m$.Capacitance of the air part $C_1 = \frac{\epsilon_0 A_1}{d} = \frac{\epsilon_0 (28 	imes 10^{-4})}{4 	imes 10^{-3}} = 0.7 \epsilon_0\,F$.Capacitance of the dielectric part $C_2 = \frac{K \epsilon_0 A_2}{d} = \frac{5 \epsilon_0 (4 	imes 10^{-4})}{4 	imes 10^{-3}} = 0.5 \epsilon_0\,F$.Total effective capacitance $C_{	ext{eff}} = C_1 + C_2 = 0.7 \epsilon_0 + 0.5 \epsilon_0 = 1.2 \epsilon_0\,F$.Electrostatic energy $U = \frac{1}{2} C_{	ext{eff}} V^2 = \frac{1}{2} (1.2 \epsilon_0) (20)^2 = 0.6 \epsilon_0 	imes 400 = 240 \epsilon_0\,J$.

Similar Questions

$A$ dielectric slab is inserted between the plates of an isolated charged capacitor. Which of the following quantities will remain the same?

$A$ parallel plate capacitor has a plate area of $40\,cm^2$ and a plate separation of $2\,mm$. The space between the plates is filled with a dielectric medium of thickness $1\,mm$ and dielectric constant $5$. The capacitance of the system is:

If $q_{f}$ is the free charge on the capacitor plates and $q_{b}$ is the bound charge on the dielectric slab of dielectric constant $k$ placed between the capacitor plates,then the bound charge $q_{b}$ can be expressed as:

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