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JEE Main
Exam
JEE Main
Subject
Physics
Chapter
Capacitance

JEE Main · Physics · Capacitance

Consider a parallel plate capacitor with plates in the shape of...

Consider a parallel plate capacitor with plates in the shape of square and in XYX Y-plane. The gap between the plates is filled with dielectric material. The dielectric constant kk varies with XX-axis as k(x)=[1+(xL)α],wherek(x)=\left[1+ \left(\frac{x}{L}\right)^\alpha\right], where\alphaisaconstant.Letis a constant. LetC_d and CaC_a are capacitance in the presence of dielectric and air, respectively. If the ratio \frac{C_d}{C_a}=\frac{7}{6}, then the value of α\alpha must beConsider a parallel plate capacitor with plates in the shape of square and in X Y -plane. The gap between the plates is

Options

  1. A3
  2. B5
  3. C7
  4. D9

Correct answer

B. 5

Step-by-step solution

Consider a parallel plate capacitor with plates in the shape of square and in X Y -plane. The gap between the plates is
Area of square plate, A=L2A=L^2 Consider an elemental capacitor at distance xx from above of thickness dxd x. Capacitance of elemental capacitor, d C=kε0dAd=kε0(L⋅dx)d\frac{k \varepsilon_0 d A}{d}=\frac{k \varepsilon_0(L \cdot d x)}{d}All such elemental capacitors will be in parallel and equivalent capacitance will be sum of aII, so Cd=ε0Ld∫0L1+(xL)a\cdotdx and Cd=ε0Ld[x+(xL)1+α(1+α)⋅1L]0L and Cd=ε0Ld[L+L1+α]\RightarrowCd=ε0L2d(2+α1+α)\begin{aligned}& C_d=\frac{\varepsilon_0 L}{d} \int_0^L 1+\left(\frac{x}{L}\right)^a \cdotd x \\ \text{ and } C_d=\frac{\varepsilon_0 L}{d}\left[x+\frac{\left(\frac{x}{L}\right)^{1+\alpha}}{(1+\alpha) \cdot \frac{1}{L}}\right]_0^L \\ \text{ and } C_d=\frac{\varepsilon_0 L}{d}\left[L+\frac{L}{1+\alpha}\right] \RightarrowC_d=\frac{\varepsilon_0 L^2}{d}\left(\frac{2+\alpha}{1+\alpha}\right) \end{aligned}It is given, CdCa=76\frac{C_d}{C_a}=\frac{7}{6} where, Ca=C_a= capacitance in absence of dielectric.⇒ε0L2d(2+α1+α)ε0L2d=76 and ⇒2+α1+α=76⇒12+6α=7+7α and ⇒α=5\begin{array}{ccc} \Rightarrow \quad\frac{\frac{\varepsilon_0 L^2}{d}\left(\frac{2+\alpha}{1+\alpha}\right)}{\frac{\varepsilon_0 L^2}{d}}=\frac{7}{6} \text{ and } \Rightarrow \frac{2+\alpha}{1+\alpha}=\frac{7}{6}\\ \Rightarrow \quad 12+6 \alpha=7+7 \alpha \text{ and } \Rightarrow \quad \alpha=5 \end{array}
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