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(B) The capacitance of a parallel plate capacitor is given by $C = \frac{\varepsilon_0 A'}{d'}$,where $A'$ is the area and $d'$ is the separation.Give

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

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(B) The capacitance of a parallel plate capacitor is given by $C = \frac{\varepsilon_0 A'}{d'}$,where $A'$ is the area and $d'$ is the separation.Given that each capacitor has an area $A' = \frac{A}{3}$,the individual capacitances are:$C_1 = \frac{\varepsilon_0 (A/3)}{d} = \frac{\varepsilon_0 A}{3d}$$C_2 = \frac{\varepsilon_0 (A/3)}{2d} = \frac{\varepsilon_0 A}{6d}$$C_3 = \frac{\varepsilon_0 (A/3)}{3d} = \frac{\varepsilon_0 A}{9d}$Since the capacitors are connected in parallel,the equivalent capacitance $C_{eq}$ is the sum of the individual capacitances:$C_{eq} = C_1 + C_2 + C_3$$C_{eq} = \frac{\varepsilon_0 A}{3d} + \frac{\varepsilon_0 A}{6d} + \frac{\varepsilon_0 A}{9d}$$C_{eq} = \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 $18$:$C_{eq} = \frac{\varepsilon_0 A}{d} \left( \frac{6 + 3 + 2}{18} \right) = \frac{11 \varepsilon_0 A}{18d}$

Three parallel plate air capacitors are connected in parallel. Each capacitor has plate area $\frac{A}{3}$ and the separation between the plates is $d, 2d$ and $3d$ respectively. The equivalent capacity of the combination is ($\varepsilon_0 =$ absolute permittivity of free space).

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