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

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(A) The arrangement can be viewed as three capacitors connected in parallel,each having an area of $\frac{A}{3}$ but different plate separations.For t

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

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(A) The arrangement can be viewed as three capacitors connected in parallel,each having an area of $\frac{A}{3}$ but different plate separations.For the first capacitor,the separation is $d$. Its capacitance is $C_1 = \frac{\varepsilon_0 (A/3)}{d} = \frac{\varepsilon_0 A}{3d}$.For the second capacitor,the separation is $2d$. Its capacitance is $C_2 = \frac{\varepsilon_0 (A/3)}{2d} = \frac{\varepsilon_0 A}{6d}$.For the third capacitor,the separation is $3d$. Its capacitance is $C_3 = \frac{\varepsilon_0 (A/3)}{3d} = \frac{\varepsilon_0 A}{9d}$.Since they are in parallel,the equivalent capacitance is $C_{eq} = C_1 + C_2 + C_3$.$C_{eq} = \frac{\varepsilon_0 A}{3d} + \frac{\varepsilon_0 A}{6d} + \frac{\varepsilon_0 A}{9d} = \frac{\varepsilon_0 A}{d} \left( \frac{1}{3} + \frac{1}{6} + \frac{1}{9} \right)$.Taking the least common multiple of $3, 6, 9$,which is $18$:$C_{eq} = \frac{\varepsilon_0 A}{d} \left( \frac{6 + 3 + 2}{18} \right) = \frac{11 \varepsilon_0 A}{18 d}$.

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

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