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Step-by-Step Solution
Step 1: Identify the Portions of the Capacitor
The capacitor is effectively split into two parts, each occupying half the total plate area A:
One part (area A/2) has no dielectric filling (air or vacuum).
The other part (area A/2) is filled with two dielectric slabs of dielectric constants $k_1$ and $k_2$ placed in series along the thickness, each slab having thickness $d/2$.
Because these two parts lie side by side (each covering half the plate area), they act as capacitors in parallel.
Step 2: Capacitance of the Air-Filled Portion
For the portion with area A/2 and no dielectric (i.e., dielectric constant = 1), the capacitance is:
$C_{\text{air}} \;=\; \dfrac{\varepsilon_0 \, (A/2)}{d} \;=\; \dfrac{\varepsilon_0 A}{2d}\,.$
Step 3: Capacitance of the Portion with Two Dielectrics
The other half of the plate area (A/2) has two dielectrics $k_1$ and $k_2$ in series, each of thickness $d/2$. The net thickness along that path is still $d$, but we must account for each slab’s contribution in series. For an area A/2, the equivalent capacitance of these two dielectrics in series is given by:
$C_{k_1 k_2} \;=\; \dfrac{\varepsilon_0 \, (A/2)}{\left(\frac{d}{2\,k_1} \;+\; \frac{d}{2\,k_2}\right)}\,.$
Simplify the denominator:
$\dfrac{d}{2\,k_1} \;+\; \dfrac{d}{2\,k_2} \;=\; \dfrac{d}{2}\,\left(\dfrac{1}{k_1} \;+\; \dfrac{1}{k_2}\right)
\;=\; \dfrac{d\,(k_1 + k_2)}{2\,k_1\,k_2}\,.$
Hence the expression becomes:
$C_{k_1 k_2} \;=\; \dfrac{\varepsilon_0 \, (A/2)}{\frac{d\,(k_1 + k_2)}{2\,k_1\,k_2}}
\;=\; \dfrac{\varepsilon_0 \, (A/2)\,2\,k_1\,k_2}{d\,(k_1 + k_2)}
\;=\; \dfrac{\varepsilon_0 A\, k_1\,k_2}{d\, (k_1 + k_2)}\,.$
Step 4: Total (Equivalent) Capacitance
Since these two parts (the air-filled and the two-dielectric–filled regions) lie in parallel, their capacitances add directly:
$C_{\text{eq}} \;=\; C_{\text{air}} \;+\; C_{k_1 k_2}
\;=\; \dfrac{\varepsilon_0 A}{2d} \;+\; \dfrac{\varepsilon_0 A\, k_1\,k_2}{d\, (k_1 + k_2)}\,.$
Factor out $\dfrac{\varepsilon_0 A}{d}$:
$C_{\text{eq}} \;=\; \dfrac{\varepsilon_0 A}{d}
\left(\dfrac{1}{2} \;+\; \dfrac{k_1 k_2}{k_1 + k_2}\right)\,.
Step 5: Final Result
The capacitance of the capacitor is:
$\boxed{\,C_{\text{eq}} \;=\; \dfrac{\varepsilon_0 A}{d}\left(\dfrac{1}{2} \;+\; \dfrac{k_1 k_2}{k_1 + k_2}\right)\,}.
