A parallel plate capacitor has the space between its plates filled by two slabs of thickness \((\mathrm{d} / 2)\) each and dielectric constant \(\mathrm{K}_{1}\) and \(\mathrm{K}_{2}\) If \(\mathrm{d}\) is the plate separation of the capacitor, then capacity of the capacitor is .......... (A) $\left[\left(2 \mathrm{~d} \in_{0}\right) / \mathrm{A}\right]\left[\left(\mathrm{K}_{1}+\mathrm{K}_{2}\right) /\left(\mathrm{K}_{1} \mathrm{~K}_{2}\right)\right]$ (B) $\left[\left(2 \mathrm{~A} \in_{0}\right) / \mathrm{d}\right]\left[\left(\mathrm{K}_{1} \mathrm{~K}_{2}\right) /\left(\mathrm{K}_{1}+\mathrm{K}_{2}\right)\right]$ (C) $\left[\left(2 \mathrm{Ad} \epsilon_{0}\right) / \mathrm{d}\right]\left[\left(\mathrm{K}_{1}+\mathrm{K}_{2}\right) /\left(\mathrm{K}_{1} \mathrm{~K}_{2}\right)\right]$ d] \(\left(K_{1}+K_{2}\right)\) (D) \(\left[\left(2 \mathrm{~A} \in_{0}\right) /\right.\)

Step by step solution

01

Understanding the problem

The key to this problem is understanding that the two dielectric slabs share the capacitor space evenly. Given that each slab has a thickness of \(d / 2\), they totally cover up the space between the capacitor plates. Since the slabs have different dielectric constants, we need to compute the capacitance contribution of each slab separately, and then compute the overall capacitance.

02

Finding the capacitance of each slab

The capacitance for each capacitor slab can be computed using the formula for capacitance of a parallel plate capacitor, which is \(C = \frac{\kappa \epsilon_0 A}{d}\). Since the dielectric fills half the distance between the plates for each slab, the thickness \(d/2\) should be used in the formula. Therefore, the capacitance for the dielectrics with dielectric constants \(K_1\) and \(K_2\) would be \(C_1 = \frac{K_1 \epsilon_0 A}{d/2}\) and \(C_2 = \frac{K_2 \epsilon_0 A}{d/2}\).

03

Computing the total capacitance

The total capacitance is the sum of the capacitances from each dielectric slab. However, we have to remember that these capacitances are in series, not in parallel. In series circuit, the total capacitance is given by the reciprocal of the sum of the reciprocals of the individual capacitances. Hence, the total capacitance \(C_{total}\) is given by: \(C_{total} = \frac{1} {(1/C_1)+(1/C_2)}\) Plugging in the expressions obtained in Step 2 and simplifying, one can find that \(C_{total} = \frac{(K_1 K_2) / (K_1 + K_2)}{\epsilon_0 A / d}\). This corresponds to answer choice (B).

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