A capacitor is made of a flat plate of area A | vedclass

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(D) The arrangement can be considered as three parallel plate capacitors connected in parallel,each having an area of $\frac{A}{3}$.Let the distance b

Vedclass · Accepted Answer

$\frac{11 \varepsilon_0 A}{18 d}$

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(D) The arrangement can be considered as three parallel plate capacitors connected in parallel,each having an area of $\frac{A}{3}$.Let the distance between the flat plate and the first,second,and third stairs be $d$,$2d$,and $3d$ respectively.The capacitance of a parallel plate capacitor is given by $C = \frac{\varepsilon_0 A}{d}$.For the first part,$C_1 = \frac{\varepsilon_0 (A/3)}{d} = \frac{\varepsilon_0 A}{3d}$.For the second part,$C_2 = \frac{\varepsilon_0 (A/3)}{2d} = \frac{\varepsilon_0 A}{6d}$.For the third part,$C_3 = \frac{\varepsilon_0 (A/3)}{3d} = \frac{\varepsilon_0 A}{9d}$.Since these capacitors are connected in parallel,the equivalent capacitance $C_{eq}$ is the sum of individual capacitances:$C_{eq} = C_1 + C_2 + C_3$$C_{eq} = \frac{\varepsilon_0 A}{3d} + \frac{\varepsilon_0 A}{6d} + \frac{\varepsilon_0 A}{9d}$Taking the common denominator as $18d$:$C_{eq} = \frac{6\varepsilon_0 A + 3\varepsilon_0 A + 2\varepsilon_0 A}{18d} = \frac{11\varepsilon_0 A}{18d}$

$A$ capacitor is made of a flat plate of area $A$ and a second plate of stair-like structure as shown in the figure. The area of each stair is $\frac{A}{3}$ and the height is $d$. The capacitance of the arrangement is

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