The space between the plates of a parallel plate capacitor is halved and a dielectric medium of relative…

The space between the plates of a parallel plate capacitor is halved and a dielectric medium of relative permittivity 10 is introduced between the plates. The ratio of the final and initial capacitances of the capacitor is
  1. 20
  2. 10
  3. $\frac{1}{10}$
  4. $\frac{1}{20}$

Solution

Initial capacitance, $\mathrm{C}_1=\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d}}$ Final capacitance, $\mathrm{C}_2=\frac{\mathrm{K} \varepsilon_0 \mathrm{~A}}{\frac{\mathrm{~d}}{2}}=\frac{2 \times 10 \varepsilon_0 \mathrm{~A}}{\mathrm{~d}}=\frac{20 \varepsilon_0 \mathrm{~A}}{\mathrm{~d}}$ $\therefore \frac{\mathrm{C}_2}{\mathrm{C}_1}=\frac{20 \varepsilon_0 \mathrm{~A}}{\mathrm{~d}} \times \frac{\mathrm{d}}{\varepsilon_0 \mathrm{~A}}=20$

Asked in: AP EAMCET 2024 (22 May Shift 2)

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