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NEET Physics QUESTION #1063
Question 1
Parallel plate capacitor, plates separated by \(d\). Two dielectric slabs: \(K_1\) with thickness \(\frac{3d}{8}\) and \(K_2\) with thickness \(\frac{d}{2}\). Capacitance becomes twice that of air capacitor. If \(K_1=1.25K_2\), value of \(K_1\) is:
  • 1.60
  • 1.33
  • 2.66✔️
  • 2.33
Correct Answer Explanation
Air gap remaining: \(d - \frac{3d}{8} - \frac{d}{2} = \frac{d}{8}\). Effective capacitance (slabs in series): \(\frac{1}{C_{eff}}=\frac{3d/8}{K_1\varepsilon_0 A}+\frac{d/2}{K_2\varepsilon_0 A}+\frac{d/8}{\varepsilon_0 A}\). Setting \(C_{eff}=2C_0=\frac{2\varepsilon_0 A}{d}\): \(\frac{d}{2\varepsilon_0 A}=\frac{3d/8}{K_1\varepsilon_0 A}+\frac{d/2}{K_2\varepsilon_0 A}+\frac{d/8}{\varepsilon_0 A}\). \(\frac{1}{2}=\frac{3}{8K_1}+\frac{1}{2K_2}+\frac{1}{8}\). With \(K_1=1.25K_2\): solving gives \(K_1\approx2.66\).