A parallel plate capacitor is filled equally (half) with two dielectrics of dielectric constant 1 and 2, as shown in…

Year	2026	2025	2024	Total
Questions	24	39	16	79

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

Capacitance with dielectric:

C = (εᵣ ε₀ A)/(d)

Series Capacitors:

Ceq = (Cₐ Cb)/(Cₐ + Cb)

Parallel Capacitors:

Ceq = Cₐ + Cb

Core Logic

Let $C₀ = (ε₀ A)/(d)$ $C_0 = \frac{\varepsilon_0 A}{d}$ be the capacitance without any dielectric.

First Configuration (Series connection):

The dielectrics split the gap vertically, so the effective thickness of each slab is $d/2$ $d/2$, and the area remains $A$ $A$.

Cₐ = (ε₁ ε₀ A)/(d/2) = 2ε₁ C₀

Cb = (ε₂ ε₀ A)/(d/2) = 2ε₂ C₀

Since they are in series:

C₁ = (Cₐ Cb)/(Cₐ + Cb) = ((2ε₁ C₀)(2ε₂ C₀))/(2ε₁ C₀ + 2ε₂ C₀) = (4ε₁ε₂ C₀²)/(2C₀(ε₁ + ε₂)) = (2ε₁ε₂)/(ε₁ + ε₂) C₀

Second Configuration (Parallel connection):

The dielectrics split the area horizontally, so the effective area of each slab is $A/2$ $A/2$, and the distance remains $d$ $d$.

Cc = (ε₁ ε₀ (A/2))/(d) = (ε₁ C₀)/(2)

Cd = (ε₂ ε₀ (A/2))/(d) = (ε₂ C₀)/(2)

Since they are in parallel:

C₂ = Cc + Cd = (ε₁ + ε₂) (C₀)/(2)

Series equivalent circuit of dielectric capacitor for Q8 — Illustrates two parallel plate configurations: stacked horizontally (series) and stacked vertically (parallel).

Step 1: Calculating the Ratio

Now, compute $(C₁)/(C₂)$ $\frac{C_1}{C_2}$:

(C₁)/(C₂) = (((2ε₁ε₂)/(ε₁ + ε₂)) C₀)/(((ε₁ + ε₂)/(2)) C₀) = (4ε₁ε₂)/((ε₁ + ε₂)²)

Pattern Recognition

For dielectric-filled capacitors: splitting the gap ( $d/2$ $d/2$) leads to a series combination, while splitting the plate area ( $A/2$ $A/2$) leads to a parallel combination. Shortcut: $Cₛₑᵣᵢₑₛ = harmonic mean$ , $Cparallel = arithmetic mean$ . The ratio $(C₁)/(C₂)$ $\frac{C_1}{C_2}$ is always the ratio of the harmonic mean of the dielectric constants to their arithmetic mean!

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Q jee_main_2026_21_jan_morning Capacitance with Dielectric

A parallel plate capacitor has capacitance C, when there is vacuum within the parallel plates. A sheet having thickness

((1)/(3))rd

of the separation between the plates and relative permittivity K is introduced between the plates. The new capacitance of the system is :

A. $3KC2K + 1$ $\frac{3\mathrm{KC}}{2\mathrm{K} + 1}$
B. $CK2 + K$ $\frac{\mathrm{CK}}{2 + \mathrm{K}}$
C. $3CK²(2K + 1)²$
D. $4KC3K - 1$ $\frac{4\mathrm{KC}}{3\mathrm{K} - 1}$

Solution

Related Formula

C = (ε₀ A)/(d - t + (t)/(K))

Alternatively, treat as two capacitors in series: $Ceq = (C₁ C₂)/(C₁ + C₂)$ $C_{eq} = \frac{C_1 C_2}{C_1 + C_2}$

Core Logic

Initial capacitance (vacuum): $C = A ε₀d$ $C = \frac{A \epsilon_{0}}{d}$.

Capacitance with Dielectric diagram for Q34 - JEE Main 2026 Morning

The system can be modelled as two capacitors in series:

A vacuum capacitor of thickness $d - (d)/(3) = (2d)/(3)$ $d - \frac{d}{3} = \frac{2d}{3}$
A dielectric capacitor of thickness $(d)/(3)$ $\frac{d}{3}$ and dielectric constant $K$ $K$.

Step 1: Capacitors in Series

The capacitances are:

C₁ = (ε₀ A)/(((2d)/(3))) = (3)/(2) ((ε₀ A)/(d)) = (3)/(2) C

C₂ = (K ε₀ A)/(((d)/(3))) = 3K ((ε₀ A)/(d)) = 3KC

Now, equivalent capacitance in series:

Ceq = (C₁ C₂)/(C₁ + C₂) = (((3)/(2) C) \times (3KC))/((3)/(2) C + 3KC)

Ceq = ((9)/(2)KC²)/((3)/(2)C(1 + 2K))

Ceq = (3KC)/(2K + 1)

Pattern Recognition

Partial dielectric filling of thickness $t$ $t$: Use formula $Cnew = (ε₀ A)/(d - t + t/K)$ . Plugging $t = d/3$ $t = d/3$ directly gives $(ε₀ A)/(d - d/3 + d/3K) = (ε₀ A)/((2d)/(3) + (d)/(3K)) = (3K ε₀ A)/(2Kd + d) = (3KC)/(2K+1)$ .

Chapter Mix

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Q45 jee_main_2026_21_jan_morning Electric Potential Energy

A point charge of

10⁻⁸

C is placed at origin. The work done in moving a point charge 2

μ

C from point A(4, 4, 2) m to B(2, 2, 1) m is ____ J.

( 14πε₀=9\times10⁹ in SI units)

A. $45 \times 10⁻⁶$ $45 \times 10^{-6}$
B. $0$ $0$
C. $30 \times 10⁻⁶$ $30 \times 10^{-6}$
D. $15 \times 10⁻⁶$ $15 \times 10^{-6}$

Solution

Related Formula

Wₑₓₜ = Δ U = Uf - Uᵢ

U = (1)/(4πε₀) (q₁ q₂)/(r)

Core Logic

Work done by external agent: $Wₑₓₜ = Δ U$ $W_{\text{ext}} = \Delta U$, where $Δ U$ $\Delta U$ is the change in potential energy.

Wₑₓₜ = 14π ε₀ q₁ q₂rf - 14π ε₀ q₁ q₂rᵢ

Calculate the distances of points A and B from the origin: $rᵢ = |A| = \sqrt(4² + 4² + 2²) = \sqrt(16+16+4) = \sqrt(36) = 6 m$ $rf = |B| = \sqrt(2² + 2² + 1²) = \sqrt(4+4+1) = \sqrt(9) = 3 m$

Step 1: Calculate Work Done

Wₑₓₜ = (9 \times 10⁹) \times (10⁻⁸ \times 2 \times 10⁻⁶) [ (1)/(3) - (1)/(6) ]

Wₑₓₜ = 18 \times 10⁻⁵ \times ((2-1)/(6))

Wₑₓₜ = 18 \times 10⁻⁵ \times (1)/(6) = 3 \times 10⁻⁵ J = 30 \times 10⁻⁶ J

Pattern Recognition

Electric field is conservative. Work done simply equals change in $kqq/r$ $kqq/r$ from initial to final radial coordinate. No path dependence.

Chapter Mix

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Q26 jee_main_2026_21_jan_evening Electric Potential Energy

Consider two identical metallic spheres of radius R each having charge Q and mass m. Their centers have an initial separation of 4R. Both the spheres are given an initial speed of u towards each other. The minimum value of u, so that they can just touch each other is: (Take

k= 14πε₀

and assume

kQ²>Gm²

where G is the Gravitational constant)

A. $kQ²4mR(1- Gm²kQ²)$
B. $kQ²4mR(1+ Gm²kQ²)$
C. $kQ²2mR(1- Gm²kQ²)$
D. $kQ²2mR(1- Gm²2kQ²)$

Solution

Related Formula

Kᵢ + Uᵢ = Kf + Uf

U = (kq₁q₂)/(r) - (Gm₁m₂)/(r)

Core Logic

Using energy conservation from the initial state (separation $4R$ $4R$) to the final state (just touching, so center-to-center separation is $2R$ $2R$). Both spheres have mass $m$ $m$ and speed $u$ $u$.

Initial Energy:

Eᵢ = 2((1)/(2)mu²) - (Gm²)/(4R) + (kQ²)/(4R)

Final Energy (just touching implies final velocity is zero):

Ef = - (Gm²)/(2R) + (kQ²)/(2R)

Step 1: Equating Energies

mu² - (Gm²)/(4R) + (kQ²)/(4R) = - (Gm²)/(2R) + (kQ²)/(2R)

mu² = ((kQ²)/(2R) - (kQ²)/(4R)) - ((Gm²)/(2R) - (Gm²)/(4R))

mu² = (kQ²)/(4R) - (Gm²)/(4R)

u² = (1)/(4mR)(kQ² - Gm²)

Step 2: Final Conclusion

u = \sqrt((kQ²)/(4mR)(1 - (Gm²)/(kQ²)))

Pattern Recognition

When dealing with two forces (electrostatic repulsion and gravitational attraction), their potentials simply superimpose linearly. The change in total potential energy equals the loss in kinetic energy.

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Q27 jee_main_2026_21_jan_evening Capacitance

The charge stored by the capacitor C in the given circuit in the steady state is ________

Capacitance diagram for Q27 - JEE Main 2026 Evening — Circuit diagram showing a 5 microfarad capacitor connected in parallel with branches containing resistors and diodes.

A. $12.5$ $12.5$
B. $10$ $10$
C. $7.5$ $7.5$
D. $5$ $5$

Solution

Related Formula

$Q = C Vc$ $Q = C V_c$ where $Vc$ $V_c$ is the steady-state voltage across the capacitor.

Core Logic

In steady state, the capacitor acts as an open circuit (blocks DC current). We must analyze the active branches.

Solution diagram for Q27 - JEE Main 2026 Evening — Circuit diagram showing a 5 microfarad capacitor connected in parallel with branches containing resistors and diodes.

The branch with the reversed-biased diode will carry no current. Current flows through the outer loop via the forward-biased diode. Total active resistance $Req = 1 Ω + 4 Ω = 5 Ω$ .

Current in the steady state:

i = VReq = (2.5)/(5) = 0.5 A

Step 1: Voltage Calculation

The voltage across the capacitor $Vc$ $V_c$ is equal to the voltage drop across the $4 Ω$ $4\,\Omega$ resistor, because the branch is connected in parallel.

Vc = i \times 4 Ω

Vc = 0.5 \times 4 = 2 V

Step 2: Final Conclusion

The charge stored is: $Q = C Vc$ $Q = C V_c$

Q = 5 \times 2 V = 10

Pattern Recognition

Capacitor in steady-state DC = open wire. Analyze only the paths where current can physically flow, check diode polarity, find nodal voltage across the capacitor terminals, apply Q=CV.

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Q28 jee_main_2026_22_january_morning Electric Field and Superposition

Six point charges are kept

60circ

apart from each other on the circumference of a circle of radius R as shown in figure. The net electric field at the centre of the circle is \_\_\_\_. (

εₒ

is permittivity of free space)

Electrostatics diagram for Q28 - JEE Main 2026 January Morning — Six point charges arranged at 60 degree intervals along a circular circumference.

A. $- 5Q8πε₀R²( i+\sqrt(3) j)$
B. $- Q4πε₀R²(\sqrt(3) i- j)$
C. $-( 5Q8πε₀R²)( i-3 j)$
D. $Q4πε₀R²(\sqrt(3) i- j)$

Solution

Related Formula

E = (kQ)/(r²)

Core Logic

Solution vector diagram for Q28 - JEE Main 2026 Morning — Six point charges arranged at 60 degree intervals along a circular circumference.

By symmetry and vector addition of electric fields due to the point charges placed on the circle:

Eₙₑₜ = 2E₀ 30°(- i) + 2E₀ 30°( j)

= (2kQ)/(r²)[ \sqrt(3)2(- i) + (1)/(2) j]

= -(Q)/(4πε₀ R²)(\sqrt(3) i - j)

Pattern Recognition

Sees: Symmetrical charge distribution on a circle. Shortcut: Resolve components symmetrically and combine vector contributions. Check: Matches option (2). ✓

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Solution & Explanation

Related Formula

Core Logic

Step 1: Calculating the Ratio

Pattern Recognition

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Solution

Related Formula

Core Logic

Step 1: Capacitors in Series

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Work Done

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Equating Energies

Step 2: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Voltage Calculation

Step 2: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

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