Physics · Electrostatic Potential And Capacitance · NEET
Look at how each slab touches the plates. If a slab covers the FULL plate area but only part of the gap d (slabs stacked vertically between plates), the field passes through them one after another, so they are in SERIES: use 1/C = 1/C1 + 1/C2. If a slab covers part of the AREA but the full gap d (slabs placed side by side), the charge splits between them, so they are in PARALLEL: use C = C1 + C2. Same physical dielectric can behave either way depending on placement.
If a slab of dielectric constant K and thickness t (t < d) is placed across the full area, the rest of the gap is air. This is a series of the slab-part and the air-part. The result is C = e0*A / (d - t + t/K). Check: if t = 0 (no slab), C = e0*A/d (plain capacitor); if t = d (fully filled), C = K*e0*A/d. This one formula solves most partial-slab NEET problems.
Because 'half-filled' is ambiguous. If the slab fills half the GAP (thickness d/2, full area), the slabs are in SERIES and C = 2K*e0*A/[d(K+1)]. If the slab fills half the AREA (full gap d), they are in PARALLEL and C = e0*A(K+1)/(2d). Always draw the picture and ask: is the plate AREA split, or is the gap d split? That decides series vs parallel.
For the standard NEET formula the slab spans the region it occupies, and any leftover space is treated as air (K = 1). A slab of thickness t in a gap d gives C = e0*A/(d - t + t/K) no matter where in the gap it sits, because the air gaps above and below add up to (d - t) and are in series with the slab. Position inside the gap does not change C; only the total air thickness and the slab thickness matter.
Break the capacitor into simple blocks. Slabs sharing the same gap but different areas are in parallel (add their C). Blocks stacked over the full area are in series (add 1/C). In the NEET 2016 figure, three slabs k1, k2, k3 fill the top half (each over area A/3, in parallel) and k4 fills the bottom half (full area). So the top block is in series with k4. Combine step by step, never all at once.
A parallel-plate capacitor of area A, plate separation d and capacitance C is filled with four dielectric materials k1, k2, k3 and k4 as shown: k1, k2, k3 each fill one-third of the area in the top half of the gap, and k4 fills the full area in the bottom half. If a single dielectric of constant k gives the same capacitance C, then k is given by:
The plates of a parallel plate capacitor are separated by d. Two slabs of dielectric constant K1 and K2 with thickness 3d/8 and d/2 (both full area) are inserted. The capacitance becomes two times larger than with nothing between the plates. If K1/K2 = 1.25, the value of K1 is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
For a single slab of dielectric constant K and thickness t inside a gap d (full area), C = e0*A / (d - t + t/K). The (d - t) is the air part and t/K is the slab part, combined in series. This covers most partial-slab NEET questions.
Side-by-side slabs (each covering part of the plate AREA, full gap) are in parallel, so you ADD their capacitances: C = C1 + C2. Stacked slabs (full area, part of the gap) are in series, so you add reciprocals: 1/C = 1/C1 + 1/C2.
Yes. Any dielectric has K > 1, so it always raises C compared with air in that region. The amount depends on K, the thickness, and whether the slab is in series or parallel with the rest.
Because the air spaces above and below the slab simply add up to a total air thickness (d - t) and are in series with the slab. Moving the slab up or down changes the two air gaps but not their sum, so C stays the same.
Multi-dielectric and partial-slab problems appear almost every year (2016 and 2025 among recent papers). They are high-value because one formula plus a clean series-or-parallel decision solves them fast, so mastering this gives reliable marks.
