Match List-I with List-II. List-I List-II (A) Electric field inside (distance r > 0 from center) of a uniformly charged spherical shell with surface charge density σ, and radius R. (I) sigma / epsilon0 (B) Electric field at distance r > 0 from a uniformly charged infinite plane sheet with surface charge density σ. (II) sigma / 2epsilon0 (C) Electric field outside (distance r > 0 from center) of a uniformly charged spherical shell with surface charge density σ, and radius R (III) 0 (D) Electric field between 2 oppositely charged infinite plane parallel sheets with uniform surface charge density σ. (IV) sigma R2 / epsilon0 r2 Choose the correct answer from the options given below:

### Core Logic Mapping electrostatics equations via Gauss's law applications : * (A) Inside a shell, enclosed charge is zero E = 0 (III) . * (B) Near an infinite sheet, E = (sigma/2epsilon0) (II) . * (C) Outside a shell, E = (kQ/r2) = (sigma R2/epsilon0 r2) (IV) . * (D) Between opposite sheets, fields add up: (sigma/2epsilon0) + (sigma/2epsilon0) = (sigma/epsilon0) (I) . Hence, the proper combination sequence is (A)-(III), (B)-(II), (C)-(IV), (D)-(I). ### Chapter Mix Class 12 Physics: Electrostatics

Match List-I with List-II. List-I List-II (A) Electric field inside (distance r > 0 from center) of a uniformly charged spherical shell with surface charge density σ, and radius R. (I) sigma / epsi

### Core Logic Mapping electrostatics equations via Gauss's law applications : * (A) Inside a shell, enclosed charge is zero E = 0 (III) . * (B) Near an infinite sheet, E = (sigma/2epsilon0) (II) . * (C) Outside a shell, E = (kQ/r2) = (sigma R2/epsilon0 r2) (IV) . * (D) Between opposite sheets, fields add up: (sigma/2epsilon0) + (sigma/2epsilon0) = (sigma/epsilon0) (I) . Hence, the proper combination sequence is (A)-(III), (B)-(II), (C)-(IV), (D)-(I). ### Chapter Mix Class 12 Physics: Electrosta

Match List-I with List-II. List-I List-II (A) Electric field inside (distance r > 0 from center) of a uniformly charged spherical shell with surface charge density σ, and radius R. (I) sigma / epsilon0 (B) Electric field at distance r > 0 from a uniformly charged infinite plane sheet with surface charge density σ. (II) sigma / 2epsilon0 (C) Electric field outside (distance r > 0 from center) of a uniformly charged spherical shell with surface charge density σ, and radius R (III) 0 (D) Electric field between 2 oppositely charged infinite plane parallel sheets with uniform surface charge density σ. (IV) sigma R2 / epsilon0 r2 Choose the correct answer from the options given below:

### Core Logic Mapping electrostatics equations via Gauss's law applications : * (A) Inside a shell, enclosed charge is zero E = 0 (III) . * (B) Near an infinite sheet, E = (sigma/2epsilon0) (II) . * (C) Outside a shell, E = (kQ/r2) = (sigma R2/epsilon0 r2) (IV) . * (D) Between opposite sheets, fields add up: (sigma/2epsilon0) + (sigma/2epsilon0) = (sigma/epsilon0) (I) . Hence, the proper combination sequence is (A)-(III), (B)-(II), (C)-(IV), (D)-(I). ### Chapter Mix Class 12 Physics: Electrostatics

Match List-I with List-II. List-I List-II (A) Electric field inside (distance r > 0 from center) of a uniformly charged spherical shell with surface charge density σ, and radius R. (I) sigma / epsi

### Core Logic Mapping electrostatics equations via Gauss's law applications : * (A) Inside a shell, enclosed charge is zero E = 0 (III) . * (B) Near an infinite sheet, E = (sigma/2epsilon0) (II) . * (C) Outside a shell, E = (kQ/r2) = (sigma R2/epsilon0 r2) (IV) . * (D) Between opposite sheets, fields add up: (sigma/2epsilon0) + (sigma/2epsilon0) = (sigma/epsilon0) (I) . Hence, the proper combination sequence is (A)-(III), (B)-(II), (C)-(IV), (D)-(I). ### Chapter Mix Class 12 Physics: Electrosta

Match List-I with List-II. List-I List-II (A) Electric field inside (distance r

Year	2026	2025	2024	Total
Questions	24	39	16	79

List-I	List-II
(A) Electric field inside (distance r > 0 from center) of a uniformly charged spherical shell with surface charge density σ, and radius R.	(I) $σ / ε₀$ $\sigma / \epsilon_0$
(B) Electric field at distance r > 0 from a uniformly charged infinite plane sheet with surface charge density σ.	(II) $σ / 2ε₀$ $\sigma / 2\epsilon_0$
(C) Electric field outside (distance r > 0 from center) of a uniformly charged spherical shell with surface charge density σ, and radius R	(III) 0
(D) Electric field between 2 oppositely charged infinite plane parallel sheets with uniform surface charge density σ.	(IV) $σ R² / ε₀ r²$ $\sigma R^2 / \epsilon_0 r^2$

Q44 jee_main_2026_28_january_morning Potential Energy of System of Charges

Two point charges of 1 nC and 2 nC are placed at the two corners of equilateral triangle of side 3 cm. The work done in bringing a charge of 3 nC from infinity to the third corner of the triangle is ____

.

14πε₀=9\times10⁹ N\cdot m²/C²

A. $2.7$ $2.7$
B. $5.4$ $5.4$
C. $3.3$ $3.3$
D. $27$ $27$

Solution

Related Formula

W = q₃ \cdot Vsystem

V = (kq₁)/(r₁) + (kq₂)/(r₂)

Core Logic

Work done is equal to the potential energy acquired by the 3 nC charge in the electric field of the other two charges.

Step 1: Formula Setup

W = ( (kq₁)/( ) + (kq₂)/( ) ) q₃

Step 2: Value Substitution

q₁ = 1 \times 10⁻⁹ ~C

q₂ = 2 \times 10⁻⁹ ~C

q₃ = 3 \times 10⁻⁹ ~C

= 3 \times 10⁻² ~m

W = 9 \times 10⁹3 \times 10⁻² ( 1 \times 10⁻⁹ + 2 \times 10⁻⁹ ) \times 3 \times 10⁻⁹

Step 3: Final Calculation

W = 3 \times 10¹¹ \times 3 \times 10⁻⁹ \times 3 \times 10⁻⁹

W = 27 \times 10⁻⁷ ~J = 2.7 \times 10⁻⁶ ~J = 2.7

Pattern Recognition

Factor out common terms ( $k/$ $k/\ell$) before substituting micro/nano orders to prevent simple arithmetic exponent errors.

Chapter Mix

Class 12 Physics: Electrostatic Potential and Capacitance

Q35 jee_main_2026_28_january_evening Capacitance and Electric Field

Identify the correct statements : A. Effective capacitance of a series combination of capacitors is always smaller than the smallest capacitance of the capacitor in the combination. B. When a dielectric medium is placed between the charged plates of a capacitor, displacement of charges cannot occur due to insulation property of dielectric. C. Increasing of area of capacitor plate or decreasing of thickness of dielectric is an alternate method to increase the capacitance. D. For a point charge, concentric spherical shells centered at the location of the charge are equipotential surfaces. Choose the correct answer from the options given below.

A. $A, B and C only$ $\text{A, B and C only}$
B. $C and D only$ $\text{C and D only}$
C. $A, C and D only$ $\text{A, C and D only}$
D. $B and D only$ $\text{B and D only}$

Solution

Core Logic

Let's analyze each statement:

Statement A: For a series combination, $1Ceq = (1)/(C₁) + (1)/(C₂) +$ This means $Ceq$ $C_{\text{eq}}$ is always less than the smallest individual capacitance. This statement is correct.

Statement B: When a dielectric is placed in an electric field, macroscopic displacement of free charges doesn't happen, but microscopic 'displacement' (polarization) does occur. Charges bound to atoms slightly shift, creating bound surface charges. Thus, saying 'displacement of charges cannot occur' is strictly false in the context of polarization. The statement is incorrect.

Statement C: Capacitance $C = (k ε₀ A)/(d)$ $C = \frac{k \varepsilon_0 A}{d}$. Increasing area ( $A$ $A$) or decreasing the thickness ( $d$ $d$, which acts as plate separation if fully filled) increases the capacitance. This statement is correct.

Statement D: The potential of a point charge is $V = (1)/(4πε₀) (q)/(r)$ . For a constant $r$ $r$ (concentric spherical shells), $V$ $V$ is constant. This statement is correct.

Step 1: Conclusion

Statements A, C, and D are correct.

Chapter Mix

Class 12 Physics: Electrostatics

Q38 jee_main_2026_28_january_evening Electric Potential

Which one of the following is not a measurable quantity?

A. $Voltage difference$ $\text{Voltage difference}$
B. $Resistance$ $\text{Resistance}$
C. $Voltage$ $\text{Voltage}$
D. $Displacement current$ $\text{Displacement current}$

Solution

Core Logic

The term 'voltage' here implies absolute electric potential at a point. In physics, the absolute potential at any single point is meaningless on its own—it requires a reference point (usually taken as infinity for convenience). Thus, only the potential difference between two points can be directly measured experimentally.

Step 1: Analyzing Options

Resistance, voltage difference (potential difference), and displacement current are all directly or indirectly measurable physical quantities. Absolute voltage (potential) is relative.

Pattern Recognition

Like potential energy, absolute potential is arbitrary and unmeasurable without setting a reference baseline.

Chapter Mix

Class 12 Physics: Electrostatics

Q41 jee_main_2026_28_january_evening Electric Field Lines and Gauss's Law

Identify the correct statements: A. Electrostatic field lines form closed loops. B. The electric field lines point radially outward when charge is greater than zero. C. The Gauss-Law is valid only for inverse-square force. D. The workdone in moving a charged particle in a static electric field around a closed path is zero. E. The motion of a particle under Coulomb's force must take place in a plane. Choose the correct answer from the options given below:

A. $A, B, D, E Only$ $\text{A, B, D, E Only}$
B. $A, B, C, D Only$ $\text{A, B, C, D Only}$
C. $B, C, D, E Only$ $\text{B, C, D, E Only}$
D. $A, C, E Only$ $\text{A, C, E Only}$

Solution

Core Logic

Statement A: Electrostatic field lines never form closed loops. This indicates the conservative nature of the electrostatic field. (Incorrect)

Statement B: For $q > 0$ $q > 0$, electric field lines point radially outward. (Correct)

Statement C: Gauss's Law fundamentally relies on the surface area of a sphere $4π r²$ $4\pi r^2$ canceling the $1/r²$ $1/r^2$ dependence of the force. If the force were not an inverse-square law, Gauss's theorem would not hold true. (Correct)

Statement D: Electrostatic force is conservative, meaning work done around any closed path is zero. (Correct)

Statement E: Coulomb's force is a central force. Motion under any central force always takes place in a plane (conservation of angular momentum direction). (Correct)

Step 1: Final Evaluation

Statements B, C, D, and E are correct.

Chapter Mix

Class 12 Physics: Electrostatics

Q1 jee_main_2025_02_april_evening Dielectrics and Polarization

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): Net dipole moment of a polar linear isotropic dielectric substance is not zero even in the absence of an external electric field. Reason (R): In absence of an external electric field, the different permanent dipoles of a polar dielectric substance are oriented in random directions. In the light of the above statements, choose the most appropriate answer from the options given below:

A. $(A) is correct but (R) is not correct$
B. $Both (A) and (R) are correct but (R) is not the correct explanation of (A)$
C. $Both (A) and (R) are correct and (R) is the correct explanation of (A)$
D. $(A) is not correct but (R) is correct$

Solution

Related Formula

Pₙₑₜ = Σ pᵢ

where: $Pₙₑₜ$ $\vec{P}_{\text{net}}$ = net dipole moment of the dielectric $pᵢ$ $\vec{p}_i$ = dipole moment of the individual $i$ $i$-th molecule

Core Logic

No external electric field is present ( $Eₑₓₜ = 0$ $E_{\text{ext}} = 0$). Due to thermal agitation, all molecular permanent dipoles are randomly oriented in space:

Pₙₑₜ = 0 when Eₑₓₜ = 0

Thus:

Assertion (A) is false because it claims the net dipole moment is non-zero even without an external field.
Reason (R) is true because it correctly describes that different permanent dipoles are randomly oriented.

Step 1: Final Conclusion

Therefore, (A) is not correct but (R) is correct.

Pattern Recognition

Sees: "polar dielectric" + "no external field" → net bulk dipole moment is always zero. Trap: Confusing the molecular level with the macroscopic level. Each molecule in a polar dielectric has a permanent dipole moment, but the macro substance has zero net moment due to random thermal orientations. Shortcut: No external field means vectors cancel globally, which implies zero net moment. Thus (A) is false immediately.

Chapter Mix

Class 12 Physics: Electrostatics

Solution & Explanation

Core Logic

Chapter Mix

More Electrostatics Previous-Year Questions — Page 5

Solution

Related Formula

Core Logic

Step 1: Formula Setup

Step 2: Value Substitution

Step 3: Final Calculation

Pattern Recognition

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Solution

Core Logic

Step 1: Conclusion

Chapter Mix

Solution

Core Logic

Step 1: Analyzing Options

Pattern Recognition

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Solution

Core Logic

Step 1: Final Evaluation

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Step 1: Final Conclusion

Pattern Recognition

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