An electron is moving under the influence of the electric field of a uniformly charged infinite plane sheet S having…

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

Class 12 Physics: Electrostatics

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

Class 12 Physics: Electrostatics

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.

Chapter Mix

Class 12 Physics: Electrostatics Class 11 Physics: Gravitation

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.

Chapter Mix

Class 12 Physics: Electrostatics Class 12 Physics: Current Electricity Class 12 Physics: Semiconductor Electronics

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). ✓

Chapter Mix

Class 12 Physics: Electrostatics

Solution & Explanation

Related Formula

Core Logic

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Step 1: Apply Second Equation of Motion

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