Two metal spheres of radius R and 3R have same surface charge density sigma. If they are brought in contact and then…

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Q40 jee_main_2026_23_january_morning Electric Flux and Gauss's Law

Two point charges 2q and q are placed at vertex A and centre of face CDEF of the cube as shown in figure. The electric flux passing through the cube is :

Electric Flux and Gauss's Law diagram for Q40 - JEE Main 2026 Morning — A cube marking vertex A with 2q and face CDEF with q.

A. $(3q)/(ε₀)$ $\frac{3q}{\varepsilon_0}$
B. $(q)/(ε₀)$ $\frac{q}{\varepsilon_0}$
C. $(3q)/(2ε₀)$ $\frac{3q}{2\varepsilon_0}$
D. $(3q)/(4ε₀)$ $\frac{3q}{4\varepsilon_0}$

Solution

Related Formula

φ = Qenclosedε₀

Core Logic

By Gauss's law, we consider the fraction of each charge that is geometrically "inside" the cube. A charge at a vertex is shared equally among 8 adjoining cubes. So its contribution to this specific cube is $1/8$ $1/8$. A charge at the center of a face is shared equally between 2 adjoining cubes. So its contribution to this specific cube is $1/2$ $1/2$.

Step 1: Calculate Enclosed Charge

Charge at vertex A = $2q$ $2q$. Inside portion = $(1)/(8)(2q) = (q)/(4)$ $\frac{1}{8}(2q) = \frac{q}{4}$. Charge at face center = $q$ $q$. Inside portion = $(1)/(2)(q) = (q)/(2)$ $\frac{1}{2}(q) = \frac{q}{2}$.

Qᵢₙ = (q)/(4) + (q)/(2) = (3q)/(4)

Step 2: Final Flux Calculation

φ = Qᵢₙε₀ = (3q)/(4)ε₀ = 3q4ε₀

Pattern Recognition

Sees: "flux through cube" + "charges on boundaries" → strictly evaluate solid angle fractions. Vertex = 1/8, Edge = 1/4, Face = 1/2.

Chapter Mix

Class 12 Physics: Electrostatics

Q50 jee_main_2026_23_january_morning Capacitors and Capacitance

The space between the plates of a parallel-plate capacitor of capacitance C (without any dielectric) is now filled with three dielectric slabs of dielectric constants

K₁=2

,

K₂=3

, and

K₃=5

(as shown in the figure). If new capacitance is

(n)/(3)C

then the value of n is ____.

Capacitors and Capacitance diagram for Q50 - JEE Main 2026 Morning — A capacitor filled with K1 covering half the thickness and full area, and K2, K3 dividing the remaining half thickness across their widths.

Numerical Answer. Answer: 8 to 8

Solution

Related Formula

C = ε₀Ad

Cdielectric = K ε₀A'd'

Series: $Ceq = ( 1C₁ + 1C₂)⁻¹$ Parallel: $Ceq = C₁ + C₂$ $C_{\text{eq}} = C_{1} + C_{2}$

Core Logic

The original capacitance is $C = ε₀Ad$ $C = \frac{\varepsilon_{0}A}{d}$. We mentally split the system into 4 distinct block capacitors based on area and thickness splits. Upper half contains $K₁$ $K_{1}$, acting on full Area but half depth. It can be split into two parallel halves to match the lower blocks. Left branch: Top block $C₁$ $C_{1}$ (area $A/2$ $A/2$, depth $d/2$ $d/2$, $K=2$ $K=2$) in series with Bottom block $C₂$ $C_{2}$ (area $A/2$ $A/2$, depth $d/2$ $d/2$, $K=3$ $K=3$). Right branch: Top block $C₄$ $C_{4}$ (area $A/2$ $A/2$, depth $d/2$ $d/2$, $K=2$ $K=2$) in series with Bottom block $C₃$ $C_{3}$ (area $A/2$ $A/2$, depth $d/2$ $d/2$, $K=5$ $K=5$).

Step 1: Calculate Block Capacitances

C₁ = 2 ε₀ (A/2)d/2 = 2 ε₀ Ad = 2C

C₂ = 3 ε₀ (A/2)d/2 = 3 ε₀ Ad = 3C

C₄ = 2 ε₀ (A/2)d/2 = 2 ε₀ Ad = 2C

C₃ = 5 ε₀ (A/2)d/2 = 5 ε₀ Ad = 5C

Step 2: Resolve Left and Right Series Combinations

Wait, the provided solution assigns block $C₁ = 5C$ $C_1 = 5C$ and $C₂=2C$ $C_2=2C$. It splits it differently: upper block $Cupper = 2 ε₀ Ad/2 = 4C$ . Lower left $CLL = 3 ε₀ A/2d/2 = 3C$ . Lower right $CLR = 5 ε₀ A/2d/2 = 5C$ . Then Lower parallel $CL = 3C + 5C = 8C$ $C_{L} = 3C + 5C = 8C$. Then $Ceq = (4C \times 8C)/(4C + 8C) = (32)/(12)C = (8)/(3)C$ . Let's check the PDF solution logic: the solution calculates $C₁ = 5C, C₂ = 2C$ $C_1 = 5C, C_2 = 2C$ in series. It seems it mismatched K values to the variable names but the math output is $Ceq = (92)/(35)C \approx 2.62C = (7.9)/(3)C n \approx 8$ . Wait, the solution logic says: $Ceq = (92)/(35)C \approx 2.62 C$ . The correct pure method: Upper block $K₁=2$ $K_1=2$. $C₁ = 2 ε₀ Ad/2 = 4C$ . Bottom blocks are parallel: $C₂ = 3 ε₀ A/2d/2 = 3C$ , $C₃ = 5 ε₀ A/2d/2 = 5C$ . Equivalent bottom $= 3C + 5C = 8C$ $= 3C + 5C = 8C$. Total $Ceq = (4C \cdot 8C)/(4C + 8C) = (32)/(12)C = (8)/(3)C$ . So exact $n = 8$ $n = 8$. The PDF solution makes an arithmetic convolution splitting it vertically fully into two branches but correctly arrives at approximately 8. We will use the direct correct layout that yields exactly 8.

Step 3: Final Capacitance

Ceq = (8)/(3) C

We are given $Ceq = (n)/(3) C$ $C_{\text{eq}} = \frac{n}{3} C$. Comparing both, we get $n = 8$ $n = 8$.

Pattern Recognition

Sees: "composite dielectrics with splits" → Treat series divisions (depth cuts) as series capacitors and parallel divisions (area cuts) as parallel capacitors. Horizontal boundary = Series. Vertical boundary = Parallel.

Chapter Mix

Class 12 Physics: Electrostatics

Q31 jee_main_2026_23_january_evening Electric Dipole

Two shorts dipoles

(A, B)

, A having charges

\pm 2μ C

and length 1 cm and B having charges

\pm 4μ C

and length 1 cm are placed with their centres 80 cm apart as shown in the figure. The electric field at a point P, equi-distant from the centres of both dipoles is ____ N/C.

Electric Dipole diagram for Q31 - JEE Main 2026 Evening — Configuration of two short electric dipoles and an equidistant point P.

A. $(9)/(16)\sqrt(2) \times 10⁵$ $\frac{9}{16}\sqrt{2} \times 10^{5}$
B. $4.5 \sqrt(2) \times 10⁴$ $4.5 \sqrt{2} \times 10^{4}$
C. $9 \sqrt(2) \times 10⁴$ $9 \sqrt{2} \times 10^{4}$
D. $(9)/(16)\sqrt(2) \times 10⁴$ $\frac{9}{16}\sqrt{2} \times 10^{4}$

Solution

Related Formula

Electric field due to a short dipole on its axial line:

Eaxial = (2Kp)/(r³)

Electric field due to a short dipole on its equatorial line:

Eequatorial = -(Kp)/(r³)

Core Logic

Point P is at a distance of $r = 40 cm = 0.4 m$ from the centre of both dipoles. For dipole A, point P lies on the axial line. For dipole B, point P lies on the equatorial line.

Dipole moments:

P₁ = q \times d = 2 \times 10⁻⁶ \times 10⁻² = 2 \times 10⁻⁸ C\cdot m

P₂ = q \times d = 4 \times 10⁻⁶ \times 10⁻² = 4 \times 10⁻⁸ C\cdot m

Step 1: Calculate Field Components

E₁ = -(2KP₁)/(r³) i

Wait, the orientation in the figure sets the directions. From the diagram, A is oriented left-to-right (minus to plus) so $E₁$ $\vec{E}_1$ on its axis is along $+ i$ $+\hat{i}$ or $- i$ $-\hat{i}$ depending on precise placement. Based on standard convention:

E₁ = (2KP₁)/(r³) i

$E₂ = -(KP₂)/(r³) j$ (Equatorial field points opposite to dipole vector)

Net Field vector based on solution structure:

Eₙₑₜ = 2 \times 9 \times 10⁹ \times 2 \times 10⁻⁸(0.4)³ i - 9 \times 10⁹ \times 4 \times 10⁻⁸(0.4)³ j

Step 2: Find Magnitude of Net Field

Eₙₑₜ = 9 \times 10⁹ \times 10⁻⁸(0.4)³ [ (2 \times 2) i - 4 j ]

Wait, the solution simplifies it as:

Eₙₑₜ = 9 \times 10⁹ \times 4 \times 10⁻⁸(0.4)³ [ i - j ]

Magnitude:

| Eₙₑₜ| = (360)/((0.064)) \sqrt(1² + (-1)²) = 360 \sqrt(2)0.064

| Eₙₑₜ| = (360 \times 10³)/(64) \sqrt(2) = (90)/(16) \times 10³ \sqrt(2) = (9)/(16) \sqrt(2) \times 10⁴ N/C

Pattern Recognition

Identify the relationship of the observation point to each dipole (axial vs equatorial). The axial field has a factor of 2, while the equatorial does not. Vector addition is mandatory since the fields are perpendicular.

Chapter Mix

Class 12 Physics: Electrostatics

Q32 jee_main_2026_23_january_evening Electrostatic Potential Energy

Two charges

7μ C

and

-2μ C

are placed at

(-9,0,0)

cm and

(9,0,0)

cm respectively in an external field

E= Ar² r

, where

A=9\times10⁵ N/C.m²

. Considering the potential at infinity is 0, the electrostatic energy of the configuration is ____ J.

A. $1.4$ $1.4$
B. $-90.7$ $-90.7$
C. $49.3$ $49.3$
D. $24.3$ $24.3$

Solution

Related Formula

dV = - E \cdot d r

Utotal = Uself + Uinteraction = q₁ V₁ + q₂ V₂ + kq₁ q₂r₁₂

Core Logic

First, derive the potential function $V(r)$ $V(r)$ from the given external electric field.

V(r) = - \int\infty^r (A)/(r²) dr = - [ -(A)/(r) ]\infty^r = (A)/(r)

Both charges are at distance $r = 9 cm = 0.09 m$ from the origin (where the field points radiate from).

Step 1: Calculate Energy of Individual Charges in External Field

Energy of $q₁$ $q_1$ in the external field:

U₁ = q₁ V₁ = (7 \times 10⁻⁶) ( A9 \times 10⁻² )

Energy of $q₂$ $q_2$ in the external field:

U₂ = q₂ V₂ = (-2 \times 10⁻⁶) ( A9 \times 10⁻² )

Sum of self energies:

Uself = 5 \times 10⁻⁶ \times 9 \times 10⁵9 \times 10⁻² = (4.5)/(0.09) = 50 J

Step 2: Calculate Interaction Energy Between Charges

Distance between charges $r₁₂ = 9 - (-9) = 18 cm = 0.18 m$ .

Uinteraction = k q₁ q₂r₁₂

Uinteraction = 9 \times 10⁹ \times (7 \times 10⁻⁶) \times (-2 \times 10⁻⁶)18 \times 10⁻²

Uinteraction = - 9 \times 14 \times 10⁻³0.18 = -0.7 J

Step 3: Total Energy

Utotal = Uself + Uinteraction

Utotal = 50 - 0.7 = 49.3 J

Pattern Recognition

When an external field is present, total energy is the sum of the potential energies of each charge independently in that external field, plus the mutual interaction energy between the charges themselves.

Chapter Mix

Class 12 Physics: Electrostatics

Q34 jee_main_2026_23_january_evening Capacitance with Dielectrics

A parallel plate capacitor with plate separation 5 mm is charged by a battery. On introducing a mica sheet of 2 mm and maintaining the connections of the plates with the terminals of the battery, it is found that it draws 25% more charge from the battery. The dielectric constant of mica is ____.

A. $2.5$ $2.5$
B. $2.0$ $2.0$
C. $1.5$ $1.5$
D. $1.0$ $1.0$

Solution

Related Formula

C = (ε₀ A)/(d)

Ceq = (ε₀ A)/((d - t) + (t)/(K))

$Q = CV$ $Q = CV$

Core Logic

Capacitance with Dielectrics diagram for Q34 - JEE Main 2026 Evening

Initially, capacitance is

C = (ε₀ A)/(5)

. The battery maintains constant voltage

V

. Initial charge

Q₁ = CV

.

When a $2 mm$ $2 \, \mathrm{mm}$ mica sheet is introduced, it acts as two capacitors in series: air gap ( $3 mm$ $3 \, \mathrm{mm}$) and mica gap ( $2 mm$ $2 \, \mathrm{mm}$). New charge $Q₂ = 1.25 Q₁$ $Q_2 = 1.25 Q_1$ (since it draws 25% more). Thus, $Ceq = 1.25 C$ $C_{eq} = 1.25 C$.

Step 1: Calculate Equivalent Capacitance

Ceq = (C₁ C₂)/(C₁ + C₂) = ((ε₀ A)/(3) \times (K ε₀ A)/(2))/((ε₀ A)/(3) + (K ε₀ A)/(2))

Ceq = (ε₀ A)/((3)/(1) + (2)/(K)) = (K ε₀ A)/(3K + 2)

Step 2: Equate to New Charge Requirement

1.25 \times (ε₀ A)/(5) = (K ε₀ A)/(3K + 2)

0.25 = (K)/(3K + 2)

0.25 (3K + 2) = K

$0.75K + 0.5 = K$ $0.75K + 0.5 = K$ $0.25K = 0.5$ $0.25K = 0.5$ $K = 2$ $K = 2$

Pattern Recognition

When a dielectric slab of thickness $t$ $t$ is inserted, equivalent capacitance is directly modeled as $C' = (ε₀ A)/((d - t) + t/K)$ . Constant voltage implies $Q \propto C$ $Q \propto C$, directly yielding the ratio equation.

Chapter Mix

Class 12 Physics: Electrostatics

Solution & Explanation

Related Formula

Core Logic

Step 1: Ratio Calculation

Pattern Recognition

Chapter Mix

Reference Study Guides

More Electrostatics Previous-Year Questions — Page 3

Solution

Related Formula

Core Logic

Step 1: Calculate Enclosed Charge

Step 2: Final Flux Calculation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Block Capacitances

Step 2: Resolve Left and Right Series Combinations

Step 3: Final Capacitance

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Field Components

Step 2: Find Magnitude of Net Field

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Energy of Individual Charges in External Field

Step 2: Calculate Interaction Energy Between Charges

Step 3: Total Energy

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Equivalent Capacitance

Step 2: Equate to New Charge Requirement

Pattern Recognition

Chapter Mix

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