60 JEE Main-style MCQs focused on ratios, graphs, region selection and formula traps.
Exam-style question • Q1
Question: JEE Main calculation/concept MCQ: For Two Parallel Infinite Sheets, identify the correct graph clue.
- A. Zero-constant-zero step graph.
- B. Do not use σ/2ε0 for the final between-field of equal opposite sheets.
- C. Using only qenclosed for both plates without direction loses superposition information.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. Zero-constant-zero step graph.
Solution: If outside is not zero for equal opposite sheets, a sign has been missed.
Exam-style question • Q2
Question: JEE Main calculation/concept MCQ: For Conducting Solid Sphere, select the formula that matches the stated region.
- A. E = 0 inside; E = Q/(4πε0r2) outside.
- B. Do not use the non-conducting inside formula Qr/(4πε0R3).
- C. An off-centre sphere ruins the constant-E symmetry.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. E = 0 inside; E = Q/(4πε0r2) outside.
Solution: Always check whether r < R, r = R or r > R.
Exam-style question • Q3
Question: JEE Main calculation/concept MCQ: For Uniform Non-Conducting Solid Sphere, spot the most common wrong formula.
- A. Do not use E = 0 inside as if it were a conducting sphere.
- B. Do not use E = 0 inside as if it were a conducting sphere.
- C. A cylinder or off-centre sphere does not match spherical symmetry.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. Do not use E = 0 inside as if it were a conducting sphere.
Solution: Solid insulator: linear inside, point-charge outside.
Exam-style question • Q4
Question: JEE Main calculation/concept MCQ: For Infinite Thin Non-Conducting Sheet, decide the field direction for positive charge.
- A. Away from the sheet on both sides.
- B. Do not use σ/ε0 for a non-conducting sheet.
- C. A sphere or cylinder not crossing the sheet symmetrically is unhelpful.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. Away from the sheet on both sides.
Solution: Straight lines perpendicular to the sheet.
Exam-style question • Q5
Question: JEE Main calculation/concept MCQ: For Infinite Thick Sheet / Slab, decide the field direction for negative charge.
- A. Toward the central plane on both sides.
- B. Do not use σ/2ε0 unless converting the whole slab to an equivalent sheet outside.
- C. A spherical surface does not match planar symmetry.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. Toward the central plane on both sides.
Solution: Straight perpendicular lines away from the midplane for positive ρ.
Exam-style question • Q6
Question: JEE Main calculation/concept MCQ: For Infinite Line Charge, find the charge enclosed expression.
- A. λL.
- B. Q/(4πε0r2) is for point or spherical symmetry, not an infinite line.
- C. A sphere centred on the line is poor because E is not constant over it.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. λL.
Solution: q = λL.
Exam-style question • Q7
Question: JEE Main calculation/concept MCQ: For Hollow Conducting Sphere, choose the active flux area.
- A. A spherical Gaussian surface gives E(4πr2) where E is non-zero.
- B. Do not use ρr/3ε0; that is for a uniformly charged solid insulator.
- C. A non-concentric surface does not exploit spherical symmetry.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. A spherical Gaussian surface gives E(4πr2) where E is non-zero.
Solution: The outside active area is 4πr2.
Exam-style question • Q8
Question: JEE Main calculation/concept MCQ: For Thin Spherical Shell, explain the symmetry requirement.
- A. All points on a sphere around the centre are equivalent.
- B. Do not use the solid sphere inside formula.
- C. An off-centre Gaussian sphere cannot make E constant.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. All points on a sphere around the centre are equivalent.
Solution: Spherical symmetry makes E constant on a concentric sphere.
Exam-style question • Q9
Question: JEE Main calculation/concept MCQ: For Conducting Plane Sheet, rank two points in the field.
- A. All nearby outside points have equal magnitude in the ideal model.
- B. Do not use σ/2ε0 for a conducting surface.
- C. A symmetric pillbox with two outside faces would describe a non-conducting sheet, not one conducting surface.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. All nearby outside points have equal magnitude in the ideal model.
Solution: For an ideal infinite conducting plane, outside E is independent of distance.
Exam-style question • Q10
Question: JEE Main calculation/concept MCQ: For Two Parallel Infinite Sheets, recognize the boundary behavior.
- A. Each sheet creates a normal field jump of σ/ε0.
- B. Do not use σ/2ε0 for the final between-field of equal opposite sheets.
- C. Using only qenclosed for both plates without direction loses superposition information.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. Each sheet creates a normal field jump of σ/ε0.
Solution: Crossing a sheet changes the normal field by σ/ε0.
Exam-style question • Q11
Question: JEE Main calculation/concept MCQ: For Conducting Solid Sphere, separate flux from local field.
- A. A Gaussian surface inside has zero flux because it encloses no excess charge.
- B. Do not use the non-conducting inside formula Qr/(4πε0R3).
- C. An off-centre sphere ruins the constant-E symmetry.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. A Gaussian surface inside has zero flux because it encloses no excess charge.
Solution: A Gaussian surface inside has zero flux because it encloses no excess charge.
Exam-style question • Q12
Question: JEE Main calculation/concept MCQ: For Uniform Non-Conducting Solid Sphere, state when the ideal formula fails.
- A. Non-uniform density changes qenclosed and the graph.
- B. Do not use E = 0 inside as if it were a conducting sphere.
- C. A cylinder or off-centre sphere does not match spherical symmetry.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. Non-uniform density changes qenclosed and the graph.
Solution: Charge is uniformly distributed throughout the volume.
Exam-style question • Q13
Question: JEE Main calculation/concept MCQ: For Infinite Thin Non-Conducting Sheet, remember the centre or midplane result.
- A. No special centre; magnitude is constant on both sides.
- B. Do not use σ/ε0 for a non-conducting sheet.
- C. A sphere or cylinder not crossing the sheet symmetrically is unhelpful.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. No special centre; magnitude is constant on both sides.
Solution: No special centre; magnitude is constant on both sides.
Exam-style question • Q14
Question: JEE Main calculation/concept MCQ: For Infinite Thick Sheet / Slab, use proportionality for a ratio.
- A. Inside E ∝ x; outside E ∝ a.
- B. Do not use σ/2ε0 unless converting the whole slab to an equivalent sheet outside.
- C. A spherical surface does not match planar symmetry.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. Inside E ∝ x; outside E ∝ a.
Solution: Inside E ∝ x; outside E is constant.
Exam-style question • Q15
Question: JEE Main calculation/concept MCQ: For Infinite Line Charge, find a density from a measured field.
- A. λ can be found from λ = 2πε0rE.
- B. Q/(4πε0r2) is for point or spherical symmetry, not an infinite line.
- C. A sphere centred on the line is poor because E is not constant over it.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. λ can be found from λ = 2πε0rE.
Solution: λ/(ε0r) gives N C-1.
Exam-style question • Q16
Question: JEE Main calculation/concept MCQ: For Hollow Conducting Sphere, identify the correct graph clue.
- A. Zero inside and inverse-square outside.
- B. Do not use ρr/3ε0; that is for a uniformly charged solid insulator.
- C. A non-concentric surface does not exploit spherical symmetry.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. Zero inside and inverse-square outside.
Solution: If the graph rises linearly inside, it is not a hollow conducting sphere.
Exam-style question • Q17
Question: JEE Main calculation/concept MCQ: For Thin Spherical Shell, select the formula that matches the stated region.
- A. E = 0 inside; outside E = Q/(4πε0r2).
- B. Do not use the solid sphere inside formula.
- C. An off-centre Gaussian sphere cannot make E constant.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. E = 0 inside; outside E = Q/(4πε0r2).
Solution: Check whether r is less or greater than R.
Exam-style question • Q18
Question: JEE Main calculation/concept MCQ: For Conducting Plane Sheet, spot the most common wrong formula.
- A. Do not use σ/2ε0 for a conducting surface.
- B. Do not use σ/2ε0 for a conducting surface.
- C. A symmetric pillbox with two outside faces would describe a non-conducting sheet, not one conducting surface.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. Do not use σ/2ε0 for a conducting surface.
Solution: Conducting surface: one active face, so σ/ε0.
Exam-style question • Q19
Question: JEE Main calculation/concept MCQ: For Two Parallel Infinite Sheets, decide the field direction for positive charge.
- A. From the positive sheet toward the negative sheet between equal opposite sheets.
- B. Do not use σ/2ε0 for the final between-field of equal opposite sheets.
- C. Using only qenclosed for both plates without direction loses superposition information.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. From the positive sheet toward the negative sheet between equal opposite sheets.
Solution: Between equal opposite sheets, straight lines go from + to -.
Exam-style question • Q20
Question: JEE Main calculation/concept MCQ: For Conducting Solid Sphere, decide the field direction for negative charge.
- A. Radially inward.
- B. Do not use the non-conducting inside formula Qr/(4πε0R3).
- C. An off-centre sphere ruins the constant-E symmetry.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. Radially inward.
Solution: Radial outside and absent inside the metal.
Exam-style question • Q21
Question: JEE Main calculation/concept MCQ: For Uniform Non-Conducting Solid Sphere, find the charge enclosed expression.
- A. ρ(4/3)πr3 inside; Q outside.
- B. Do not use E = 0 inside as if it were a conducting sphere.
- C. A cylinder or off-centre sphere does not match spherical symmetry.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. ρ(4/3)πr3 inside; Q outside.
Solution: ρ = Q/[(4/3)πR3].
Exam-style question • Q22
Question: JEE Main calculation/concept MCQ: For Infinite Thin Non-Conducting Sheet, choose the active flux area.
- A. Two flat faces contribute: total flux = 2EA.
- B. Do not use σ/ε0 for a non-conducting sheet.
- C. A sphere or cylinder not crossing the sheet symmetrically is unhelpful.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. Two flat faces contribute: total flux = 2EA.
Solution: The active area is A on each of two faces.
Exam-style question • Q23
Question: JEE Main calculation/concept MCQ: For Infinite Thick Sheet / Slab, explain the symmetry requirement.
- A. The central plane is a mirror plane where E = 0.
- B. Do not use σ/2ε0 unless converting the whole slab to an equivalent sheet outside.
- C. A spherical surface does not match planar symmetry.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. The central plane is a mirror plane where E = 0.
Solution: Planar symmetry makes E perpendicular to the slab faces and equal on symmetric pillbox faces.
Exam-style question • Q24
Question: JEE Main calculation/concept MCQ: For Infinite Line Charge, rank two points in the field.
- A. The smaller r has the larger E.
- B. Q/(4πε0r2) is for point or spherical symmetry, not an infinite line.
- C. A sphere centred on the line is poor because E is not constant over it.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. The smaller r has the larger E.
Solution: E is inversely proportional to r.
Exam-style question • Q25
Question: JEE Main calculation/concept MCQ: For Hollow Conducting Sphere, recognize the boundary behavior.
- A. Field is zero inside the conductor and non-zero just outside if Q is present.
- B. Do not use ρr/3ε0; that is for a uniformly charged solid insulator.
- C. A non-concentric surface does not exploit spherical symmetry.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. Field is zero inside the conductor and non-zero just outside if Q is present.
Solution: The normal field jump is caused by surface charge.
Exam-style question • Q26
Question: JEE Main calculation/concept MCQ: For Thin Spherical Shell, separate flux from local field.
- A. Zero field inside follows from zero enclosed charge plus symmetry.
- B. Do not use the solid sphere inside formula.
- C. An off-centre Gaussian sphere cannot make E constant.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. Zero field inside follows from zero enclosed charge plus symmetry.
Solution: Zero field inside follows from zero enclosed charge plus symmetry.
Exam-style question • Q27
Question: JEE Main calculation/concept MCQ: For Conducting Plane Sheet, state when the ideal formula fails.
- A. A finite conductor has edge effects.
- B. Do not use σ/2ε0 for a conducting surface.
- C. A symmetric pillbox with two outside faces would describe a non-conducting sheet, not one conducting surface.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. A finite conductor has edge effects.
Solution: Electrostatic equilibrium in a conductor.
Exam-style question • Q28
Question: JEE Main calculation/concept MCQ: For Two Parallel Infinite Sheets, remember the centre or midplane result.
- A. At the midpoint between equal opposite sheets, E is the same as everywhere between.
- B. Do not use σ/2ε0 for the final between-field of equal opposite sheets.
- C. Using only qenclosed for both plates without direction loses superposition information.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. At the midpoint between equal opposite sheets, E is the same as everywhere between.
Solution: At the midpoint between equal opposite sheets, E is the same as everywhere between.
Exam-style question • Q29
Question: JEE Main calculation/concept MCQ: For Conducting Solid Sphere, use proportionality for a ratio.
- A. Outside E ∝ Q/r2.
- B. Do not use the non-conducting inside formula Qr/(4πε0R3).
- C. An off-centre sphere ruins the constant-E symmetry.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. Outside E ∝ Q/r2.
Solution: Outside E varies as 1/r2; inside it is zero.
Exam-style question • Q30
Question: JEE Main calculation/concept MCQ: For Uniform Non-Conducting Solid Sphere, find a density from a measured field.
- A. ρ, Q or R can be found from inside or surface data.
- B. Do not use E = 0 inside as if it were a conducting sphere.
- C. A cylinder or off-centre sphere does not match spherical symmetry.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. ρ, Q or R can be found from inside or surface data.
Solution: ρr/ε0 gives N C-1.
Exam-style question • Q31
Question: JEE Main calculation/concept MCQ: For Infinite Thin Non-Conducting Sheet, identify the correct graph clue.
- A. A horizontal line.
- B. Do not use σ/ε0 for a non-conducting sheet.
- C. A sphere or cylinder not crossing the sheet symmetrically is unhelpful.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. A horizontal line.
Solution: A distance-dependent graph is not an ideal infinite sheet.
Exam-style question • Q32
Question: JEE Main calculation/concept MCQ: For Infinite Thick Sheet / Slab, select the formula that matches the stated region.
- A. Inside E = ρx/ε0; outside E = ρa/ε0.
- B. Do not use σ/2ε0 unless converting the whole slab to an equivalent sheet outside.
- C. A spherical surface does not match planar symmetry.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. Inside E = ρx/ε0; outside E = ρa/ε0.
Solution: Check |x| < a or |x| ≥ a.
Exam-style question • Q33
Question: JEE Main calculation/concept MCQ: For Infinite Line Charge, spot the most common wrong formula.
- A. Q/(4πε0r2) is for point or spherical symmetry, not an infinite line.
- B. Q/(4πε0r2) is for point or spherical symmetry, not an infinite line.
- C. A sphere centred on the line is poor because E is not constant over it.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. Q/(4πε0r2) is for point or spherical symmetry, not an infinite line.
Solution: Line charge means cylinder and 1/r.
Exam-style question • Q34
Question: JEE Main calculation/concept MCQ: For Hollow Conducting Sphere, decide the field direction for positive charge.
- A. Radially outward outside the shell.
- B. Do not use ρr/3ε0; that is for a uniformly charged solid insulator.
- C. A non-concentric surface does not exploit spherical symmetry.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. Radially outward outside the shell.
Solution: No field lines in the empty cavity; radial lines outside.
Exam-style question • Q35
Question: JEE Main calculation/concept MCQ: For Thin Spherical Shell, decide the field direction for negative charge.
- A. Radially inward outside the shell.
- B. Do not use the solid sphere inside formula.
- C. An off-centre Gaussian sphere cannot make E constant.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. Radially inward outside the shell.
Solution: No lines inside; radial lines outside.
Exam-style question • Q36
Question: JEE Main calculation/concept MCQ: For Conducting Plane Sheet, find the charge enclosed expression.
- A. σA on the surface patch.
- B. Do not use σ/2ε0 for a conducting surface.
- C. A symmetric pillbox with two outside faces would describe a non-conducting sheet, not one conducting surface.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. σA on the surface patch.
Solution: q = σA on the selected surface patch.
Exam-style question • Q37
Question: JEE Main calculation/concept MCQ: For Two Parallel Infinite Sheets, choose the active flux area.
- A. For each non-conducting sheet, two faces give 2EA.
- B. Do not use σ/2ε0 for the final between-field of equal opposite sheets.
- C. Using only qenclosed for both plates without direction loses superposition information.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. For each non-conducting sheet, two faces give 2EA.
Solution: For each single sheet, the two pillbox faces are active.
Exam-style question • Q38
Question: JEE Main calculation/concept MCQ: For Conducting Solid Sphere, explain the symmetry requirement.
- A. All surface charge is uniformly distributed over a spherical outer surface.
- B. Do not use the non-conducting inside formula Qr/(4πε0R3).
- C. An off-centre sphere ruins the constant-E symmetry.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. All surface charge is uniformly distributed over a spherical outer surface.
Solution: Spherical symmetry makes E constant on a concentric sphere.
Exam-style question • Q39
Question: JEE Main calculation/concept MCQ: For Uniform Non-Conducting Solid Sphere, rank two points in the field.
- A. Inside, larger r gives larger E; outside, larger r gives smaller E.
- B. Do not use E = 0 inside as if it were a conducting sphere.
- C. A cylinder or off-centre sphere does not match spherical symmetry.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. Inside, larger r gives larger E; outside, larger r gives smaller E.
Solution: Inside E ∝ r; outside E ∝ 1/r2.
Exam-style question • Q40
Question: JEE Main calculation/concept MCQ: For Infinite Thin Non-Conducting Sheet, recognize the boundary behavior.
- A. The field jump across the sheet is σ/ε0.
- B. Do not use σ/ε0 for a non-conducting sheet.
- C. A sphere or cylinder not crossing the sheet symmetrically is unhelpful.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. The field jump across the sheet is σ/ε0.
Solution: Across a charged sheet the normal component changes by σ/ε0.
Exam-style question • Q41
Question: JEE Main calculation/concept MCQ: For Infinite Thick Sheet / Slab, separate flux from local field.
- A. Inside qenclosed uses thickness 2x, not 2a.
- B. Do not use σ/2ε0 unless converting the whole slab to an equivalent sheet outside.
- C. A spherical surface does not match planar symmetry.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. Inside qenclosed uses thickness 2x, not 2a.
Solution: Inside qenclosed uses thickness 2x, not 2a.
Exam-style question • Q42
Question: JEE Main calculation/concept MCQ: For Infinite Line Charge, state when the ideal formula fails.
- A. A finite rod does not have the same cylindrical symmetry near its ends.
- B. Q/(4πε0r2) is for point or spherical symmetry, not an infinite line.
- C. A sphere centred on the line is poor because E is not constant over it.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. A finite rod does not have the same cylindrical symmetry near its ends.
Solution: The line must be effectively infinite and uniformly charged.
Exam-style question • Q43
Question: JEE Main calculation/concept MCQ: For Hollow Conducting Sphere, remember the centre or midplane result.
- A. E = 0 at the centre of the empty cavity.
- B. Do not use ρr/3ε0; that is for a uniformly charged solid insulator.
- C. A non-concentric surface does not exploit spherical symmetry.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. E = 0 at the centre of the empty cavity.
Solution: E = 0 at the centre of the empty cavity.
Exam-style question • Q44
Question: JEE Main calculation/concept MCQ: For Thin Spherical Shell, use proportionality for a ratio.
- A. Outside E ∝ 1/r2.
- B. Do not use the solid sphere inside formula.
- C. An off-centre Gaussian sphere cannot make E constant.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. Outside E ∝ 1/r2.
Solution: Inside E = 0; outside E ∝ 1/r2.
Exam-style question • Q45
Question: JEE Main calculation/concept MCQ: For Conducting Plane Sheet, find a density from a measured field.
- A. σ can be found from σ = ε0E.
- B. Do not use σ/2ε0 for a conducting surface.
- C. A symmetric pillbox with two outside faces would describe a non-conducting sheet, not one conducting surface.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. σ can be found from σ = ε0E.
Solution: σ/ε0 gives N C-1.
Exam-style question • Q46
Question: JEE Main calculation/concept MCQ: For Two Parallel Infinite Sheets, identify the correct graph clue.
- A. Zero-constant-zero step graph.
- B. Do not use σ/2ε0 for the final between-field of equal opposite sheets.
- C. Using only qenclosed for both plates without direction loses superposition information.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. Zero-constant-zero step graph.
Solution: If outside is not zero for equal opposite sheets, a sign has been missed.
Exam-style question • Q47
Question: JEE Main calculation/concept MCQ: For Conducting Solid Sphere, select the formula that matches the stated region.
- A. E = 0 inside; E = Q/(4πε0r2) outside.
- B. Do not use the non-conducting inside formula Qr/(4πε0R3).
- C. An off-centre sphere ruins the constant-E symmetry.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. E = 0 inside; E = Q/(4πε0r2) outside.
Solution: Always check whether r < R, r = R or r > R.
Exam-style question • Q48
Question: JEE Main calculation/concept MCQ: For Uniform Non-Conducting Solid Sphere, spot the most common wrong formula.
- A. Do not use E = 0 inside as if it were a conducting sphere.
- B. Do not use E = 0 inside as if it were a conducting sphere.
- C. A cylinder or off-centre sphere does not match spherical symmetry.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. Do not use E = 0 inside as if it were a conducting sphere.
Solution: Solid insulator: linear inside, point-charge outside.
Exam-style question • Q49
Question: JEE Main calculation/concept MCQ: For Infinite Thin Non-Conducting Sheet, decide the field direction for positive charge.
- A. Away from the sheet on both sides.
- B. Do not use σ/ε0 for a non-conducting sheet.
- C. A sphere or cylinder not crossing the sheet symmetrically is unhelpful.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. Away from the sheet on both sides.
Solution: Straight lines perpendicular to the sheet.
Exam-style question • Q50
Question: JEE Main calculation/concept MCQ: For Infinite Thick Sheet / Slab, decide the field direction for negative charge.
- A. Toward the central plane on both sides.
- B. Do not use σ/2ε0 unless converting the whole slab to an equivalent sheet outside.
- C. A spherical surface does not match planar symmetry.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. Toward the central plane on both sides.
Solution: Straight perpendicular lines away from the midplane for positive ρ.
Exam-style question • Q51
Question: JEE Main calculation/concept MCQ: For Infinite Line Charge, find the charge enclosed expression.
- A. λL.
- B. Q/(4πε0r2) is for point or spherical symmetry, not an infinite line.
- C. A sphere centred on the line is poor because E is not constant over it.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. λL.
Solution: q = λL.
Exam-style question • Q52
Question: JEE Main calculation/concept MCQ: For Hollow Conducting Sphere, choose the active flux area.
- A. A spherical Gaussian surface gives E(4πr2) where E is non-zero.
- B. Do not use ρr/3ε0; that is for a uniformly charged solid insulator.
- C. A non-concentric surface does not exploit spherical symmetry.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. A spherical Gaussian surface gives E(4πr2) where E is non-zero.
Solution: The outside active area is 4πr2.
Exam-style question • Q53
Question: JEE Main calculation/concept MCQ: For Thin Spherical Shell, explain the symmetry requirement.
- A. All points on a sphere around the centre are equivalent.
- B. Do not use the solid sphere inside formula.
- C. An off-centre Gaussian sphere cannot make E constant.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. All points on a sphere around the centre are equivalent.
Solution: Spherical symmetry makes E constant on a concentric sphere.
Exam-style question • Q54
Question: JEE Main calculation/concept MCQ: For Conducting Plane Sheet, rank two points in the field.
- A. All nearby outside points have equal magnitude in the ideal model.
- B. Do not use σ/2ε0 for a conducting surface.
- C. A symmetric pillbox with two outside faces would describe a non-conducting sheet, not one conducting surface.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. All nearby outside points have equal magnitude in the ideal model.
Solution: For an ideal infinite conducting plane, outside E is independent of distance.
Exam-style question • Q55
Question: JEE Main calculation/concept MCQ: For Two Parallel Infinite Sheets, recognize the boundary behavior.
- A. Each sheet creates a normal field jump of σ/ε0.
- B. Do not use σ/2ε0 for the final between-field of equal opposite sheets.
- C. Using only qenclosed for both plates without direction loses superposition information.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. Each sheet creates a normal field jump of σ/ε0.
Solution: Crossing a sheet changes the normal field by σ/ε0.
Exam-style question • Q56
Question: JEE Main calculation/concept MCQ: For Conducting Solid Sphere, separate flux from local field.
- A. A Gaussian surface inside has zero flux because it encloses no excess charge.
- B. Do not use the non-conducting inside formula Qr/(4πε0R3).
- C. An off-centre sphere ruins the constant-E symmetry.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. A Gaussian surface inside has zero flux because it encloses no excess charge.
Solution: A Gaussian surface inside has zero flux because it encloses no excess charge.
Exam-style question • Q57
Question: JEE Main calculation/concept MCQ: For Uniform Non-Conducting Solid Sphere, state when the ideal formula fails.
- A. Non-uniform density changes qenclosed and the graph.
- B. Do not use E = 0 inside as if it were a conducting sphere.
- C. A cylinder or off-centre sphere does not match spherical symmetry.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. Non-uniform density changes qenclosed and the graph.
Solution: Charge is uniformly distributed throughout the volume.
Exam-style question • Q58
Question: JEE Main calculation/concept MCQ: For Infinite Thin Non-Conducting Sheet, remember the centre or midplane result.
- A. No special centre; magnitude is constant on both sides.
- B. Do not use σ/ε0 for a non-conducting sheet.
- C. A sphere or cylinder not crossing the sheet symmetrically is unhelpful.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. No special centre; magnitude is constant on both sides.
Solution: No special centre; magnitude is constant on both sides.
Exam-style question • Q59
Question: JEE Main calculation/concept MCQ: For Infinite Thick Sheet / Slab, use proportionality for a ratio.
- A. Inside E ∝ x; outside E ∝ a.
- B. Do not use σ/2ε0 unless converting the whole slab to an equivalent sheet outside.
- C. A spherical surface does not match planar symmetry.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. Inside E ∝ x; outside E ∝ a.
Solution: Inside E ∝ x; outside E is constant.
Exam-style question • Q60
Question: JEE Main calculation/concept MCQ: For Infinite Line Charge, find a density from a measured field.
- A. λ can be found from λ = 2πε0rE.
- B. Q/(4πε0r2) is for point or spherical symmetry, not an infinite line.
- C. A sphere centred on the line is poor because E is not constant over it.
- D. The result cannot be inferred from symmetry
Show Answer
Correct Answer: A. λ can be found from λ = 2πε0rE.
Solution: λ/(ε0r) gives N C-1.