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(A) In Figure $A$, the capacitor is divided into two capacitors in parallel, each with area $A/2$ and plate separation $d$. One has air $(K=1)$ and th

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$2d/3$

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(A) In Figure $A$, the capacitor is divided into two capacitors in parallel, each with area $A/2$ and plate separation $d$. One has air $(K=1)$ and the other has dielectric $(K=2)$.$C_A = C_1 + C_2 = \frac{\varepsilon_0 (A/2)}{d} + \frac{K \varepsilon_0 (A/2)}{d} = \frac{\varepsilon_0 A}{2d} (1 + K) = \frac{\varepsilon_0 A}{2d} (1 + 2) = \frac{3 \varepsilon_0 A}{2d}$.In Figure $B$, the capacitor is equivalent to two capacitors in series, one with air of thickness $(d-t)$ and one with dielectric of thickness $t$.$C_B = \frac{\varepsilon_0 A}{d - t + t/K} = \frac{\varepsilon_0 A}{d - t + t/2} = \frac{\varepsilon_0 A}{d - t/2}$.Equating $C_A = C_B$:$\frac{3 \varepsilon_0 A}{2d} = \frac{\varepsilon_0 A}{d - t/2} \implies \frac{3}{2d} = \frac{1}{d - t/2}$.$3(d - t/2) = 2d \implies 3d - 3t/2 = 2d \implies d = 3t/2 \implies t = 2d/3$.

$A$ capacitor is half-filled with a dielectric $(K=2)$ as shown in Figure $A$. If the same capacitor is to be filled with the same dielectric as shown in Figure $B$, what would be the thickness $t$ of the dielectric so that the capacitor still has the same capacity?

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