
A capacitor is made of a flat plate of area \(A\) and a second plate of stair-like structure as shown in the…

- \(\frac{\varepsilon_0 A}{3 d}\)
- \(\frac{6 \varepsilon_0 A}{11 d}\)
- \(\frac{3 \varepsilon_0 A}{d}\)
- \(\frac{11 \varepsilon_0 A}{18 d}\)
Solution

Given, area of each stair \(=\frac{A}{3}\) and height of each stair \(=d\) \(\therefore\) Capacitance of parallel plate capacitor is given as \(C=\frac{\varepsilon_0 A}{d}\) First capacitance of capacitor, \(C_1=\frac{\varepsilon_0 A}{3 d}\) second capacitance of capacitor, \(C_2=\frac{\varepsilon_0 A}{3(2 d)}=\frac{\varepsilon_0 A}{6 d}\) and third capacitance of capacitor, \(C_3=\frac{\varepsilon_0 A}{3(3 d)}=\frac{\varepsilon_0 A}{9 d}\) Equivalent capacitance of capacitor, \(\begin{gathered} C_{e q}=C_1+C_2+C_3 \\ C_{e q}=\frac{\varepsilon_0 A}{3 d}+\frac{\varepsilon_0 A}{6 d}+\frac{\varepsilon_0 A}{9 d} \\ C_{e q}=\frac{\left(6 \varepsilon_0+3 \varepsilon_0+2 \varepsilon_0\right) A}{18 d}=\frac{11 \varepsilon_0 A}{18 d} \end{gathered}\) Hence, the capacitance of the arrangement is \(\frac{11 \varepsilon_0 A}{18 d}\)
Asked in: AP EAMCET 2019 (20 Apr Shift 1)