Capacitors

JEE Main level, numerical answer, medium

82 of 180

Question 82 of 180

A parallel plate capacitor with plate area AA and plate separation dd is filled with a dielectric material of dielectric constant K=4.K=4. The thickness of the dielectric material is x,x, where x<d.x\lt d.
Let C1{C}_{1} and C2{C}_{2} be the capacitance of the system for x=13dx=\dfrac{1}{3}d and x=2d3,x=\dfrac{2d}{3}, respectively. If C1=2μF{C}_{1}=2\mu F, the value of C2{C}_{2} is _____ μF.\mu F.

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Answer: 3

The formula to calculate the capacitance of a parallel plate capacitor having thickness dd and with a dielectric of thickness xx between the plates is given by
C=ϵ0A(d−x)+xK...(1)C=\dfrac{{\epsilon }_{0}A}{\left(d-x\right)+\dfrac{x}{K}}...(1)
Substitute the values of the parameters for first case into equation (1) to obtain the capacitance.
C1=ϵ0A2d3+d34=4ϵ0A3d...(2)\begin{matrix}{C}_{1} & = & \dfrac{{\epsilon }_{0}A}{\dfrac{2d}{3}+\dfrac{\dfrac{d}{3}}{4}} \\ & = & \dfrac{4{\epsilon }_{0}A}{3d}...(2)\end{matrix}
Similarly, substitute the parameters into equation (1) for the second case to obtain the capacitance
C2=ϵ0Ad3+2d34=4ϵ0A2d=2ϵ0Ad...(3)\begin{matrix}{C}_{2} & = & \dfrac{{\epsilon }_{0}A}{\dfrac{d}{3}+\dfrac{\dfrac{2d}{3}}{4}} \\ & = & \dfrac{4{\epsilon }_{0}A}{2d} \\ & = & \dfrac{2{\epsilon }_{0}A}{d}...(3)\end{matrix}
Divide equation (3) by equation (2) to obtain the capacitance in the second case.
C2C1=2ϵ0Ad4ϵ0A3d=32C2=32C1...(4)\begin{matrix}\dfrac{{C}_{2}}{{C}_{1}} & = & \dfrac{\dfrac{2{\epsilon }_{0}A}{d}}{\dfrac{4{\epsilon }_{0}A}{3d}} \\ & = & \dfrac{3}{2} \\ {C}_{2} & = & \dfrac{3}{2}{C}_{1}...(4)\end{matrix}
Substitute the value of the capacitance in the first case into equation (4) to obtain the value of the capacitance in the second case.
C2=32×2μF=3μF\begin{matrix}{C}_{2} & = & \dfrac{3}{2}\times 2\mu F \\ & = & 3\mu F\end{matrix}

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