JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Capacitance

From the figure the dielectrics split the gap into a left half (with $K_1$ stacked over $K_3$) and a right half (with $K_2$ over $K_4$), the two halves sitting side by side across the plate. Within each half, stacking two slabs through the thickness behaves like two capacitors in series, so the left half contributes an effective constant tied to $K_1+K_3$ and the right half to $K_2+K_4$. The two halves, side by side, add like capacitors in parallel. Carrying the series-then-parallel algebra through to a single equivalent constant gives $K=\frac{(K_1+K_3)(K_2+K_4)}{K_1+K_2+K_3+K_4}$.