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Electrostatics appeared 79 times across 3 years — 9.1% of Physics. This question is from Electric Field due to Continuous Charge Distribution.

Year 2026 2025 2024 Total
Questions 24 39 16 79

A metallic ring is uniformly charged as shown in figure. AC and BD are two mutually perpendicular diameters. Electric field due to arc AB to 'O' is 'E' is magnitude. What would be the magnitude of electric field at 'O' due to arc ABC?
Uniformly charged ring with perpendicular diameters
A circle showing perpendicular axes AC and BD dividing it into quadrants.

Solution & Explanation

Related Formula

The electric field due to a circular arc subtending an angle φ at the center is given by:

Earc = (2kλ)/(R) ((φ)/(2))
Core Logic

Arc AB subtends 90^° (one quadrant) at the center. The electric field due to it is given as E. Arc ABC consists of two independent quadrants: arc AB and arc BC. Each quadrant independently creates an electric field of magnitude E pointing along the bisector of that specific quadrant.

Step 1: Vector Addition

The electric field EAB is directed at 45^° away from both axes into the third quadrant. The electric field EBC is directed at 45^° towards the matching opposite quadrant. Since EAB and EBC are perpendicular to each other, their resultant magnitude is: Eₙₑₜ = √(E² + E²) = √(2)E

Vector field components due to charged arcs
A circle showing perpendicular axes AC and BD dividing it into quadrants.
Vector field components due to charged arcs
A circle showing perpendicular axes AC and BD dividing it into quadrants.

Pattern Recognition

Symmetric components of a ring create orthogonal vector fields. Each 90^° arc produces a field of magnitude E directed along its angular bisector. Two adjacent quadrants have bisectors separated by 90^°, hence use orthogonal vector addition.

Chapter Mix

Class 12 Physics: Electrostatics

Reference Study Guides

More Electrostatics Previous-Year Questions — Page 5

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=9×10⁹ N· m²/C²
  • A. 2.7
  • B. 5.4
  • C. 3.3
  • D. 27

Solution

Related Formula
W = q₃ · 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.

Equilateral triangle charge diagram
Equilateral triangle charge diagram

Step 1: Formula Setup
W = ( (kq₁)/( ) + (kq₂)/( ) ) q₃
Step 2: Value Substitution
q₁ = 1 × 10⁻⁹ ~C q₂ = 2 × 10⁻⁹ ~C q₃ = 3 × 10⁻⁹ ~C = 3 × 10⁻² ~m W = 9 × 10⁹3 × 10⁻² ( 1 × 10⁻⁹ + 2 × 10⁻⁹ ) × 3 × 10⁻⁹
Step 3: Final Calculation
W = 3 × 10¹¹ × 3 × 10⁻⁹ × 3 × 10⁻⁹ W = 27 × 10⁻⁷ ~J = 2.7 × 10⁻⁶ ~J = 2.7
Pattern Recognition

Factor out common terms (k/) 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
  • B. C and D only
  • C. A, C and D only
  • D. 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 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). Increasing area (A) or decreasing the thickness (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 (concentric spherical shells), 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
  • B. Resistance
  • C. Voltage
  • D. 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
  • B. A, B, C, D Only
  • C. B, C, D, E Only
  • D. 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, electric field lines point radially outward. (Correct)

Statement C: Gauss's Law fundamentally relies on the surface area of a sphere 4π r² canceling the 1/r² 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 = net dipole moment of the dielectric pᵢ = dipole moment of the individual i-th molecule

Core Logic

No external electric field is present (Eₑₓₜ = 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

More Electrostatics Questions — jee_main_2025_04_april_evening

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JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)