Q.A parallel plate capacitor with air between the plates has a capacitance of $8\ \text{pF}$ ($1\ \text{pF} = 10^{-12}\ \text{F}$). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?

Showing the 12 most recent of 27 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.A parallel plate capacitor of plate area 3A and plate separation 3d is filled with three dielectric slabs with dielectric constants $K_1=2$, $K_2=4$ and $K_3=6$ as shown in the figure. The capacitance of the capacitor is (Figure: a parallel plate capacitor with total plate separation 3d and total plate area 3A. Dielectric $K_1$ fills the left region of thickness $d$ across the entire plate area 3A. The remaining thickness $2d$ (to the right) is divided along the area into two parts: the upper part, of area 2A, is filled with dielectric $K_2$; the lower part, of area A, is filled with dielectric $K_3$.) (A) $\dfrac{42A\epsilon_o}{13d}$ (B) $\dfrac{54A\epsilon_o}{7d}$ (C) $\dfrac{54A\epsilon_o}{13d}$ (D) $\dfrac{32A\epsilon_o}{7d}$

   ›Reveal solutionSolution

   One dielectric fills a full-area layer of thickness $d$ (in series with the rest); the remaining $2d$ is split by area into two dielectrics side by side (in parallel with each other). Combine parallel first, then series with the first layer, to get $C=\dfrac{42A\epsilon_0}{13d}$.

   Concept and Intuition

   When a capacitor is sliced by a plane parallel to the plates, the two slices behave as capacitors in series (same charge flows through, different fields in each layer). When a capacitor is divided by a plane perpendicular to the plates (side by side, same thickness), the two parts behave as capacitors in parallel (same voltage across each, charges add). Here the top-to-bottom split (K1 vs. the K2/K3 layer) is a series split; the left-right split within the lower $2d$ layer (K2 vs K3) is a parallel split.

   Step-by-Step Solution

   1. K1 region: full area $3A$, thickness $d$: $C_1 = \dfrac{K_1\epsilon_0(3A)}{d} = \dfrac{2\epsilon_0(3A)}{d} = \dfrac{6\epsilon_0 A}{d}$.
   2. K2 region: area $2A$, thickness $2d$: $C_2 = \dfrac{K_2\epsilon_0(2A)}{2d} = \dfrac{4\epsilon_0(2A)}{2d} = \dfrac{4\epsilon_0 A}{d}$.
   3. K3 region: area $A$, thickness $2d$: $C_3 = \dfrac{K_3\epsilon_0(A)}{2d} = \dfrac{6\epsilon_0 A}{2d} = \dfrac{3\epsilon_0 A}{d}$. …

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2. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.A parallel plate capacitor with dielectric is charged completely and the battery is then disconnected. Now if the dielectric is slowly pulled out of the capacitor then the variation of the potential

   (v) of the capacitor with respect to the length

   (x) of the dielectric pulled out is represented by (A) [FIGURE: V–x graph — a straight line of positive slope; V starts at a positive value at x = 0 and increases linearly as x increases] (B) [FIGURE: V–x graph — a horizontal straight line; V stays constant at a positive value as x increases] (C) [FIGURE: V–x graph — a curve that starts at a positive V, dips slightly to a shallow minimum, and then rises, curving upward increasingly steeply as x increases] (D) [FIGURE: V–x graph — a curve that starts at a positive V and rises smoothly and continuously with x, the value of V increasing throughout as x increases]

   ›Reveal solutionSolution

   At constant charge, C decreases linearly as the dielectric is withdrawn, so V = Q/(C0 - kx) is a concave-up, accelerating curve (option C).

   Concept and Intuition

   Once the battery is disconnected the charge Q is trapped on the plates and stays constant. A partially inserted dielectric behaves like two capacitors in parallel: the still-covered part (permittivity K) and the exposed air part. As the dielectric is pulled out by length x, the high-permittivity area shrinks and the air area grows, so the total capacitance falls. With Q fixed, V = Q/C means the voltage rises as C falls.

   Step-by-Step Solution

   1. Model the plate (length L, width w, separation d) as two parallel capacitors. After pulling out length x: C(x) = (epsilon0 w/d)[K(L - x) + x] = (epsilon0 w/d)[KL - (K - 1)x].
   2. So C decreases linearly with x: C(x) = C0 - k x, where C0 = (epsilon0 w/d)KL and k = (epsilon0 w/d)(K - 1) > 0.
   3. Charge is constant, so V = Q/C = Q / (C0 - k x).
   4. At x = 0, V = Q/C0, a positive intercept (not zero) — matching all four graphs starting above the origin.
   5. Slope: dV/dx = Qk / (C0 - k x)^2, which is positive and grows as x increases (denominator shrinks). Hence V rises with increasing steepness — concave up, accelerating.
   6. This matches the concave-up, increasingly-steep curve (option C), not the straight line (A), the flat line (B), or the saturating concave-down curve (D).

   Common Mistakes …

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3. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.In a parallel plate capacitor, each plate cross-sectional area is A and distance between plates is d. If a dielectric slab of area A, thickness $\dfrac{3d}{4}$ and dielectric constant 3 is placed between the parallel plates of the capacitor, then the new capacitance will be _____ times the initial capacitance (A) $\dfrac{1}{2}$ (B) $\dfrac{1}{3}$ (C) 3 (D) 2

   ›Reveal solutionSolution

   A partially-filled parallel-plate capacitor (dielectric slab covering full plate area but only part of the gap) is modeled as two capacitors in series — the dielectric-filled section and the remaining air gap. Here both series capacitances turn out equal, giving a clean overall factor of 2. Answer: 2 times $C_0$.

   Concept and Intuition

   When a dielectric slab of the same cross-sectional area as the plates but less than the full gap thickness is inserted, the capacitor behaves as two capacitors stacked in series along the field direction: one region has the dielectric (thickness $t$, constant $K$), and the remaining region ($d-t$) is air. This is because the electric displacement/charge is the same all the way across (series-like, same "current" of field lines through both), while the potential drops add up across the two regions — exactly the signature of a series combination. Each region individually is just a simple parallel-plate capacitor with its own thickness playing the role of "plate separation."

   Step-by-Step Solution

   1. Original (no dielectric) capacitance: $C_0 = \dfrac{\varepsilon_0 A}{d}$.
   2. Dielectric-filled region: thickness $t=\dfrac{3d}{4}$, dielectric constant $K=3$:

   $C_1 = \frac{K\varepsilon_0 A}{t} = \frac{3\varepsilon_0 A}{3d/4} = \frac{3\varepsilon_0 A \times 4}{3d} = \frac{4\varepsilon_0 A}{d} = 4C_0$

   3. Remaining air gap: thickness $d-t = \dfrac{d}{4}$: $C_2 = \frac{\varepsilon_0 A}{d/4} = \frac{4\varepsilon_0 A}{d} = 4C_0$ …

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4. AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
   Q.When dielectric is inserted between the plates of a capacitor with battery connected, energy increases because of (A) increase in charge (B) increase in voltage (C) increase in field (D) increase in separation between the plates

   ›Reveal solutionSolution

   Tests understanding of the battery-connected (constant-voltage) case for a capacitor with a dielectric — voltage stays fixed, capacitance and charge rise, and energy grows because of the extra charge the battery pushes in.

   Concept and Intuition

   There are two classic capacitor-with-dielectric scenarios, and they behave oppositely: if the capacitor is isolated (charge fixed), inserting a dielectric drops the voltage and the stored energy decreases. But if the battery stays connected (voltage fixed), the battery does extra work pushing more charge onto the plates as capacitance rises, and the stored energy increases — driven by that extra charge, not by any change in voltage (which the battery holds constant by definition).

   Step-by-Step Solution

   1. With the battery connected, $V$ across the capacitor is fixed by the battery (that's what "battery connected" means) — it cannot increase.
   2. Inserting a dielectric of constant $K>1$ raises the capacitance: $C \to C' = KC$.
   3. Since $Q = CV$ and $V$ is unchanged, the charge rises: $Q' = C'V = KCV = KQ > Q$.
   4. Stored energy $U = \tfrac12 QV$ (equivalently $\tfrac12CV^2$); with $V$ fixed and $Q$ risen to $KQ$, $U' = \tfrac12(KQ)V = K\,U > U$ — energy increases. …

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5. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.As shown in the figure two capacitors $C_1$ & $C_2$ each having same gap between the plates x filled with different media of dielectric constants 3K and 6K respectively. If these two capacitors are connected to a battery, the ratio of potential differences across dielectric layers of $C_1$ and $C_2$ is [FIGURE] (two adjoining parallel-plate capacitors $C_1$ and $C_2$ sharing a common middle plate; each capacitor's plate gap is x (combined width $2x$); $C_1$'s gap is filled with a dielectric of constant 3K, and $C_2$'s gap is filled with a dielectric of constant 6K) (A) $2$ (B) $6$ (C) $4$ (D) $8$

   ›Reveal solutionSolution

   Two capacitors sharing a middle plate are in series, so they carry equal charge; the potential drop across each is inversely proportional to its capacitance, giving a ratio of $2$.

   Concept and Intuition

   When two parallel-plate capacitors share a common (floating) middle plate and the outer plates connect to the battery, the arrangement is electrically a series combination: the same charge $Q$ flows through both. Since $V=Q/C$ for each capacitor at fixed $Q$, whichever capacitor has the smaller capacitance ends up with the larger potential difference across it.

   Step-by-Step Solution

   1. Both capacitors have identical geometry (same gap $x$, same plate area $A$), differing only in dielectric constant: $C_1=\dfrac{\varepsilon_0(3K)A}{x}$, $C_2=\dfrac{\varepsilon_0(6K)A}{x}$.
   2. Since the two capacitors share the middle plate, they are in series and carry the same charge $Q$. …

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6. AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
   Q.A parallel plate capacitor charged and disconnected from the battery then a dielectric is inserted between the plates of the capacitor. The energy of capacitor (A) Increases (B) decreases (C) remains same (D) zero

   ›Reveal solutionSolution

   Charge is conserved once the battery is removed, so the relevant energy formula is $U=Q^2/(2C)$; since $C$ increases with the dielectric, $U$ falls.

   Concept and Intuition

   A parallel plate capacitor's energy can be written two ways: $U=\dfrac{1}{2}CV^2$ (useful when voltage is fixed, i.e. still connected to a battery) or $U=\dfrac{Q^2}{2C}$ (useful when charge is fixed, i.e. isolated). After the battery is disconnected, the capacitor is an isolated system — no external circuit can change $Q$. So $Q$ is the conserved quantity here, and we must use $U=Q^2/2C$.

   Step-by-Step Solution

   1. Before inserting the dielectric: capacitance $C_0$, charge $Q$ (fixed after disconnection), energy $U_0=\dfrac{Q^2}{2C_0}$.
   2. Insert a dielectric of constant $K>1$: new capacitance $C=KC_0$.
   3. Charge $Q$ cannot change (isolated system), so new energy $U=\dfrac{Q^2}{2KC_0}=\dfrac{U_0}{K}$.
   4. Since $K>1$, $U<U_0$ — the energy decreases. …

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7. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.A capacitor of capacitance $2\,\mu\mathrm{F}$ is charged to 50 V and then disconnected from the source. Later the gap between the plates of the capacitor is filled with a dielectric material. If the energy stored in the capacitor is decreased by 25% of its initial value, then the dielectric constant of the dielectric material is (A) $\dfrac{2}{3}$ (B) $\dfrac{4}{3}$ (C) $\dfrac{3}{4}$ (D) $\dfrac{3}{2}$

   ›Reveal solutionSolution

   This tests energy stored in an isolated (disconnected) capacitor as a dielectric is inserted. Answer: $K=\dfrac{4}{3}$.

   Concept and Intuition

   Once a charged capacitor is disconnected from its source, its charge $Q$ is fixed (nowhere for charge to go). Inserting a dielectric of constant $K$ multiplies the capacitance by $K$: $C'=KC$. Since $U=\dfrac{Q^2}{2C}$ with $Q$ fixed, increasing $C$ by $K$ decreases the stored energy by the same factor $K$ — energy is not conserved here because work is done pulling the dielectric in (or released as the field does work on it).

   Step-by-Step Solution

   1. Initial energy: $U_i=\dfrac{Q^2}{2C}$.
   2. After the dielectric is inserted: $C'=KC$, so $U_f=\dfrac{Q^2}{2KC}=\dfrac{U_i}{K}$.
   3. Given: energy decreases by 25%, so $U_f=0.75\,U_i$.
   4. Therefore $\dfrac{U_i}{K}=0.75\,U_i \Rightarrow K=\dfrac{1}{0.75}=\dfrac{4}{3}$. …

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8. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.The radii of the inner and outer spheres of a spherical capacitor are 8 cm and 9 cm respectively. The outer sphere is earthed and the inner sphere is charged. If the space between the concentric spheres is filled with a liquid of dielectric constant 5, the capacitance of the capacitor is (A) 400 pF (B) 40 pF (C) $400\ \mu\mathrm{F}$ (D) $40\ \mu\mathrm{F}$

   ›Reveal solutionSolution

   Applying the spherical-capacitor formula with a dielectric filling the gap between the 8 cm and 9 cm radii spheres gives a capacitance of exactly 400 pF.

   Concept and Intuition

   A spherical capacitor consists of two concentric conducting spherical shells; the field exists only in the gap between them (zero inside the inner sphere and outside the outer, by Gauss's law, since the outer sphere is earthed). Filling that gap with a dielectric of constant $K$ increases the capacitance by that same factor $K$ compared to vacuum, because the dielectric weakens the field for a given charge, allowing more charge to be stored per unit potential difference.

   Step-by-Step Solution

   1. Capacitance of a spherical capacitor (vacuum): $C_0 = 4\pi\varepsilon_0\dfrac{ab}{b-a}$, where $a$ = inner radius, $b$ = outer radius.
   2. With dielectric of constant $K$ filling the gap: $C = 4\pi\varepsilon_0 K\dfrac{ab}{b-a} = \dfrac{K}{9\times10^9}\times\dfrac{ab}{b-a}$ (using $\dfrac{1}{4\pi\varepsilon_0}=9\times10^9$ SI).
   3. Convert: $a=8\,\text{cm}=0.08$ m, $b=9\,\text{cm}=0.09$ m, so $ab = 0.0072\ \mathrm{m^2}$ and $b-a=0.01$ m. …

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9. AP EAPCET 2025Set ap-2025-05-19-FN1 markMCQ
   Q.To decrease the capacitance of a parallel plate capacitor (A) a dielectric material is to be introduced between the plates (B) the area of the plates is to be increased (C) the area of the plates is to be increased and distance between them is to be decreased (D) the distance between the plates is to be increased

   ›Reveal solutionSolution

   Since $C \propto A/d$ (times $\varepsilon_r$ for any dielectric present), capacitance decreases only when the separation $d$ is increased (or the area is decreased, or a dielectric removed) — introducing a dielectric or increasing area both increase $C$.

   Concept and Intuition

   The parallel plate capacitor formula $C = \dfrac{\varepsilon_0\varepsilon_r A}{d}$ shows $C$ is directly proportional to plate area $A$ and dielectric constant $\varepsilon_r$, and inversely proportional to plate separation $d$. To decrease $C$, we must either shrink $A$, shrink $\varepsilon_r$ (remove/weaken a dielectric), or grow $d$.

   Step-by-Step Solution

   1. $C = \dfrac{\varepsilon_0 \varepsilon_r A}{d}$.
   2. Introducing a dielectric increases $\varepsilon_r$, so option (A) increases $C$ — wrong direction.
   3. Increasing area $A$ (option B) increases $C$ — wrong direction.
   4. Increasing area while decreasing distance (option C) increases $C$ on both counts — wrong direction. …

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10. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
    Q.A parallel plate capacitor has plates of area $0.4\pi \mathrm{~m^2}$ and spacing of 0.5 mm. If a slab of thickness 0.5 mm and dielectric constant 4.5 is introduced in between the plates of the capacitor, then the capacitance of the capacitor is (A) 100 nF (B) 60 pF (C) 100 pF (D) 60 nF

    ›Reveal solutionSolution

    Because the dielectric slab fills the entire gap, the capacitor's capacitance is simply $K\varepsilon_0A/d$; this evaluates to 100 nF.

    Concept and Intuition

    When a dielectric slab of thickness $t$ less than the full gap $d$ is inserted, the effective capacitance formula has an air-gap term $(d-t)$ and a dielectric term $t/K$ added in series. But when $t=d$ (the slab exactly fills the gap, as here — both are 0.5 mm), that air-gap term vanishes and the capacitor behaves exactly like a parallel-plate capacitor entirely filled with the dielectric: $C = \dfrac{K\varepsilon_0 A}{d}$.

    Step-by-Step Solution

    1. Note slab thickness $=0.5\,\mathrm{mm}=$ full plate spacing $d$, so the dielectric fills the whole gap.
    2. Use $C = \dfrac{K\varepsilon_0 A}{d}$ with $K=4.5$, $A=0.4\pi\ \mathrm{m^2}$, $d=5\times10^{-4}\ \mathrm m$.
    3. First find $C_0=\dfrac{\varepsilon_0 A}{d} = \dfrac{8.85\times10^{-12}\times0.4\pi}{5\times10^{-4}} \approx 2.224\times10^{-8}\ \mathrm F$ (the capacitance without dielectric). …

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11. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
    Q.If a dielectric slab of dielectric constant 3 is introduced between the plates of a capacitor having electric field $1.5\pi\ \mathrm{N\,C^{-1}}$, then the electric displacement is (A) $125\times10^{-12}\ \mathrm{C\,m^{-2}}$ (B) $125\times10^{-9}\ \mathrm{C\,m^{-2}}$ (C) $250\times10^{-12}\ \mathrm{C\,m^{-2}}$ (D) $250\times10^{-9}\ \mathrm{C\,m^{-2}}$

    ›Reveal solutionSolution

    This tests the relation between electric displacement, permittivity, dielectric constant, and the field inside a dielectric, $D = \varepsilon_0 K E$, giving $125\times10^{-12}\,\mathrm{C\,m^{-2}}$.

    Concept and Intuition

    The electric displacement field $D$ relates to the free charge alone and is connected to the actual field $E$ inside a linear dielectric medium by $D = \varepsilon_0 K E = \varepsilon E$, where $\varepsilon = \varepsilon_0 K$ is the medium's permittivity. Unlike $E$, which is reduced by the dielectric, $D$ stays tied to the free surface charge density and can be obtained directly once $K$ and $E$ (in the dielectric) are known.

    Step-by-Step Solution

    1. Given: $K = 3$, $E = 1.5\pi\,\mathrm{N\,C^{-1}}$.
    2. $D = \varepsilon_0 K E = (8.85\times10^{-12})(3)(1.5\pi)$.
    3. $(8.85\times10^{-12})(3) = 26.55\times10^{-12}$. …

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12. AP EAPCET 2024Set ap-2024-05-16-FN1 markMCQ
    Q.A parallel plate capacitor has two square plates of side 5 cm separated by 15 mm. A pyrex glass slab of dielectric constant $K = 4.5$ and thickness 10 mm is placed between the plates. The capacitance of the capacitor is (A) $3.11\times10^{-12}$ F (B) $311\ \mu\mathrm{F}$ (C) 311 PF (D) 3.11 nF

    ›Reveal solutionSolution

    This tests the formula for capacitance when a dielectric slab only partially fills the gap between the plates — the slab's thickness effectively shrinks by a factor of $K$ in the "equivalent vacuum gap" calculation.

    Concept and Intuition

    When a dielectric slab of thickness $t$ (less than the full plate separation $d$) is inserted, it's equivalent to replacing that slab with a vacuum gap of thickness $t/K$ (since the dielectric increases capacitance, it behaves as if it were a "thinner" vacuum gap). The remaining vacuum gap $(d-t)$ stays as is. Adding these gives an effective total separation to use in the ordinary parallel-plate formula $C=\varepsilon_0A/d_{eff}$.

    Step-by-Step Solution

    1. Plate side $=5$ cm $=0.05$ m, so area $A=(0.05)^2=2.5\times10^{-3}\ \text{m}^2$.
    2. Total separation $d=15$ mm $=0.015$ m; dielectric thickness $t=10$ mm $=0.01$ m; remaining air gap $=d-t=0.005$ m.
    3. Effective separation: $d_{eff}=(d-t)+\dfrac{t}{K}=0.005+\dfrac{0.01}{4.5}=0.005+0.00222=0.00722\ \text{m}$. …

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