UNIT-1 PHYSICS CATEGORIZED-PYQ -

📚

UNIT-1 PHYSICS CATEGORIZED-PYQ

Physics

⏱️ 30 Minutes

❓ 28 Questions

📅 2022, 2023, 2024 &2025

✍️ Long Answers

📌 Answers are locked once submitted — results and explanations appear at the end.

UNIT-1 PHYSICS CATEGORIZED-PYQ

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QUESTION 1 OF 28

An infinitely long wire is uniformly charged with linear charge density λ and placed in air. The electric field at a distance r from the wire is: (PYQ 2022 Shift 1)

Answer Review

1 An infinitely long wire is uniformly charged with linear charge density λ and placed in air. The electric field at a distance r from the wire is: (PYQ 2022 Shift 1)

A λ/(4πε₀r)

B λ/(4πε₀r²)

C λ/(2ε₀)

D λ/(2πε₀r)

💡 Short Explanation

�� Use Gauss's law for infinite line charge. �� Choose a cylindrical Gaussian surface. �� Electric field varies as 1/r.

✅ Correct Answer Explanation

(Detailed) → For an infinitely long charged wire, choose a cylindrical Gaussian surface of radius r and length L. → Total flux = E(2πrL) → Charge enclosed = λL → By Gauss's law: E(2πrL) = λL/ε₀ → E = λ/(2πε₀r)

❓ Why Other Options Are Incorrect

A) Incorrect factor; resembles point-charge constant.
B) Wrong inverse-square dependence.
C) Missing distance dependence.

🎯 Strategy

Used

�� Method used: Formula Substitution
�� Identify infinite line charge.
�� Use cylindrical symmetry.
�� Apply Gauss's law directly.

🧠 Memory Trick

📚 → Class 12, Physics, Chapter Electric Charges and Fields: Gauss's Law Application

2 Two point charges (−q) and (+4q) are placed at a separation r. Where should a third charge be placed so that the system is in equilibrium? (PYQ 2022 Shift 1)

A At distance r from −q on the outer side of −q

B At distance r from +4q on the outer side of +4q

C At distance r/2 from −q between the charges

D At distance r/4 from +4q between the charges

💡 Short Explanation

�� Net electric field must be zero. �� Between opposite charges, fields are in the same direction. �� Neutral point lies outside the smaller charge.

✅ Correct Answer Explanation

(Detailed) → Charges are −q and +4q. → Between unlike charges, electric fields add, so neutral point cannot lie between them. → Neutral point lies outside the smaller charge, i.e., outside −q. → Let distance from −q be x. → kq/x² = 4kq/(r + x)² → r + x = 2x → x = r

❓ Why Other Options Are Incorrect

B) Neutral point is outside the smaller charge, not the larger charge.
C) Between unlike charges, fields add and do not cancel.
D) No equilibrium point exists between the charges.

🎯 Strategy

Used

�� Method used: Conceptual Recall
�� Reject region between unlike charges.
�� Locate point outside smaller charge.
�� Use electric field equality.

🧠 Memory Trick

📚 → Class 12, Physics, Chapter Electric Charges and Fields: Electric Field Due to Point Charges

3 The variation of electric field with distance from the centre of a charged conducting spherical shell of radius R is represented by which graph? (PYQ 2022 Shift 1)

💡 Short Explanation

�� Inside a conducting shell, E = 0. �� Outside the shell, it behaves like a point charge. �� Electric field decreases as 1/r² outside.

✅ Correct Answer Explanation

(Detailed) → For r < R, electric field inside a conducting spherical shell is zero. → For r ≥ R, the shell behaves as if its entire charge is concentrated at the centre. → Therefore: E = 0 for r < R E = kQ/r² for r ≥ R → Hence, the correct graph is zero inside and decreases as 1/r² outside.

❓ Why Other Options Are Incorrect

A) Shows non-zero field inside the conductor.
C) Does not show correct 1/r² outside variation.
D) Shows incorrect continuous variation inside the shell.

🎯 Strategy

Used

�� Method used: Diagram Analysis
�� Check field inside conductor first.
�� Then check outside variation.
�� Select graph with E = 0 inside and 1/r² outside.

🧠 Memory Trick

📚 → Class 12, Physics, Chapter Electric Charges and Fields: Conductors in Electrostatic Equilibrium

4 A conducting sphere is charged. The electric field at a distance 20 cm from the centre is 1.2 × 10³ N/C and is directed radially inward. The net charge on the sphere is: (PYQ 2022 Shift 1)

A 4.5 × 10⁹ C

B −4.5 × 10⁻⁹ C

C 1.7 × 10⁹ C

D −5.3 × 10⁻⁹ C

💡 Short Explanation

�� Use E = kQ/r². �� Field is inward, so charge is negative. �� Convert 20 cm into metre.

✅ Correct Answer Explanation

(Detailed) → Given: E = 1.2 × 10³ N/C r = 20 cm = 0.20 m k = 9 × 10⁹ Nm²/C² → E = kQ/r² → Q = Er²/k → Q = (1.2 × 10³ × 0.20²)/(9 × 10⁹) → Q = 48/(9 × 10⁹) → Q ≈ 5.3 × 10⁻⁹ C → Since the field is radially inward, charge is negative. → Q = −5.3 × 10⁻⁹ C

❓ Why Other Options Are Incorrect

A) Wrong sign and wrong magnitude.
B) Sign is correct but magnitude is not accurate.
C) Wrong sign and wrong magnitude.

🎯 Strategy

Used

�� Method used: Formula Substitution
�� Convert cm into metre.
�� Use E = kQ/r².
�� Use field direction to decide sign.

🧠 Memory Trick

📚 → Class 12, Physics, Chapter Electric Charges and Fields: Electric Field Due to Charged Sphere

5 The expression for torque acting on an electric dipole with dipole moment p in a uniform electric field E is: (PYQ 2022 Shift 1)

A τ⃗ = P⃗ · E⃗

B τ⃗ = P⃗/E⃗

C τ⃗ = E⃗/P⃗

D τ⃗ = P⃗ × E⃗

💡 Short Explanation

�� Torque is a vector quantity. �� It is given by cross product. �� Magnitude is τ = pE sinθ.

✅ Correct Answer Explanation

(Detailed) → Torque acting on an electric dipole in a uniform electric field is: τ = pE sinθ → In vector form: τ⃗ = p⃗ × E⃗ → Therefore, the correct expression is: τ⃗ = P⃗ × E⃗

❓ Why Other Options Are Incorrect

A) Dot product gives scalar, not torque vector.
B) Vector division is not valid.
C) Vector division is not a valid torque relation.

🎯 Strategy

Used

�� Method used: Formula Recall
�� Torque uses cross product.
�� Check vector nature of answer.
�� Select p⃗ × E⃗.

🧠 Memory Trick

📚 → Class 12, Physics, Chapter Electric Charges and Fields: Electric Dipole in Electric Field

6 Two point charges +4 μC and −2 μC are placed at a certain distance in air. They experience an attractive force F. If they are brought in contact with each other and then separated to the same distance, the new repulsive force between them will be: (PYQ 2023 Shift 1)

A 8F

B 4F

C F/8

D F/4

💡 Short Explanation

�� Total charge after contact = +2 μC. �� Each identical body gets +1 μC. �� Force is proportional to product of charges.

✅ Correct Answer Explanation

(Detailed) → Initial charges: +4 μC and −2 μC → Initial force magnitude: F ∝ |(+4)(−2)| = 8 → After contact: Total charge = +4 − 2 = +2 μC → Each sphere gets: +2/2 = +1 μC → Final force: F′ ∝ (1)(1) = 1 → F′/F = 1/8 → F′ = F/8

❓ Why Other Options Are Incorrect

A) Wrongly assumes force increases.
B) Charges become smaller after contact, so force cannot become 4F.
D) Does not match the charge-product ratio.

🎯 Strategy

Used

�� Method used: Ratio Method
�� Use conservation of charge.
�� Share charge equally after contact.
�� Compare charge products before and after contact.

🧠 Memory Trick

📚 → Class 12, Physics, Chapter Electric Charges and Fields: Coulomb's Law and Charge Sharing

7 A spherical conducting shell of inner radius r₁ and outer radius r₂ has a charge Q. A charge q is kept at the centre of the shell. The surface charge density on outer surface of the shell is: (PYQ 2023 Shift 1)

A (Q + q)/(4πr₂²)

B (Q − q)/(4πr₂²)

C (Q + q)/[4π(r₂² − r₁²)]

D (Q − q)/[4π(r₂² − r₁²)]

💡 Short Explanation

�� Charge −q is induced on inner surface. �� Outer surface carries Q + q. �� Surface area of outer sphere is 4πr₂².

✅ Correct Answer Explanation

(Detailed) → Charge q is placed at the centre of the shell. → To make electric field zero inside conducting material, charge −q is induced on the inner surface. → Total charge on shell is Q. → Therefore, outer surface charge = Q + q. → Surface charge density: σ = Charge/Area → σ = (Q + q)/(4πr₂²)

❓ Why Other Options Are Incorrect

B) Uses Q − q instead of Q + q.
C) Uses wrong surface area.
D) Has both wrong charge and wrong area.

🎯 Strategy

Used

�� Method used: Charge Distribution
�� Apply electrostatic shielding.
�� Inner surface gets −q.
�� Outer surface carries Q + q.

🧠 Memory Trick

📚 → Class 12, Physics, Chapter Electric Charges and Fields: Conductors in Electrostatic Equilibrium

8 An infinite line charge produces an electric field of 9 × 10⁶ N/C at a distance of 20 cm. The linear charge density will be: (PYQ 2023 Shift 1)

A 10⁻³ C/m

B 10⁻⁴ C/m

C 10⁻² C/m

D 10⁻⁵ C/m

💡 Short Explanation

�� Electric field of line charge is E = λ/(2πε₀r). �� Rearrange to find λ. �� Substitute r = 0.20 m.

✅ Correct Answer Explanation

(Detailed) → Formula: E = λ/(2πε₀r) → λ = E(2πε₀r) → Given: E = 9 × 10⁶ N/C r = 20 cm = 0.20 m → λ = 9 × 10⁶ × 2π × 8.85 × 10⁻¹² × 0.20 → λ ≈ 10⁻⁴ C/m

❓ Why Other Options Are Incorrect

A) Too large.
C) Much larger than the calculated value.
D) Too small.

🎯 Strategy

Used

�� Method used: Formula Substitution
�� Use line charge field formula.
�� Rearrange for λ.
�� Convert cm into metre.

🧠 Memory Trick

📚 → Class 12, Physics, Chapter Electric Charges and Fields: Electric Field Due to Infinite Line Charge

9 A vertical electric field of magnitude 4.9 × 10⁵ N C⁻¹ just prevents a water droplet of mass 0.1 g from falling. What is the charge on the droplet?
(g = 9.8 m s⁻²) (PYQ 2023 Shift 1)

A 2.0 × 10⁻⁹ C

B 2.0 × 10⁻⁶ C

C 2.0 × 10⁵ C

D 2.0 × 10⁶ C

💡 Short Explanation

�� Droplet is just prevented from falling. �� Electric force balances weight. �� Use qE = mg.

✅ Correct Answer Explanation

(Detailed) → Given: m = 0.1 g = 0.1 × 10⁻³ kg E = 4.9 × 10⁵ N/C g = 9.8 m/s² → At equilibrium: qE = mg → q = mg/E → q = (0.1 × 10⁻³ × 9.8)/(4.9 × 10⁵) → q = 2.0 × 10⁻⁹ C

❓ Why Other Options Are Incorrect

B) Too large by a factor of 10³.
C) Physically unrealistic for a water droplet.
D) Extremely large and incorrect.

🎯 Strategy

Used

�� Method used: Force Balance
�� Convert gram into kilogram.
�� Equate electric force and weight.
�� Solve q = mg/E.

🧠 Memory Trick

📚 → Class 12, Physics, Chapter Electric Charges and Fields: Force on Charge in Electric Field

10 The electric charges are distributed in a small volume. The flux of the electric field through a sphere of radius 10 cm surrounding the total charge is 20 Vm. The flux over a concentric sphere of radius 20 cm will be: (PYQ 2023 Shift 2)

A 20 Vm

B 25 Vm

C 40 Vm

D 200 Vm

💡 Short Explanation

�� Electric flux depends only on enclosed charge. �� Radius of Gaussian surface does not affect total flux. �� Use Gauss's law.

✅ Correct Answer Explanation

(Detailed) → By Gauss's law: Φ = Q/ε₀ → Both concentric spheres enclose the same charge. → Hence, total electric flux remains unchanged. → Therefore, flux = 20 Vm.

❓ Why Other Options Are Incorrect

B) Flux does not change to 25 Vm with radius.
C) Flux does not double when radius doubles.
D) Flux does not scale with area or radius in this case.

🎯 Strategy

Used

�� Method used: Conceptual Recall
�� Use Gauss's law.
�� Check enclosed charge.
�� Ignore change in radius.

🧠 Memory Trick

📚 → Class 12, Physics, Chapter Electric Charges and Fields: Gauss's Law

11 A conducting sphere of radius R is given a charge Q. What is the electric field and the electric potential at the centre of the sphere? (PYQ 2023 Shift 2)

A Q/4πϵ₀R² and Q/4πϵ₀R

B Q/4πϵ₀R² and zero

C Zero and Q/4πϵ₀R

D Both electric field and electric potential are zero

💡 Short Explanation

�� Electric field inside a conductor is zero. �� Potential inside a conductor is constant. �� Potential at centre equals surface potential.

✅ Correct Answer Explanation

(Detailed) → For a charged conducting sphere, charge resides on the outer surface. → Inside the conducting sphere: E = 0 → Potential is same everywhere inside the conductor. → Surface potential: V = Q/4πϵ₀R → Therefore, at the centre: Electric field = 0 Electric potential = Q/4πϵ₀R

❓ Why Other Options Are Incorrect

A) Electric field at the centre is zero, not Q/4πϵ₀R².
B) Electric field is wrong and potential is not zero.
D) Electric field is zero, but potential is not zero.

🎯 Strategy

Used

�� Method used: Conceptual Recall
�� Use properties of conductor in electrostatic equilibrium.
�� Field inside conductor is zero.
�� Potential remains constant inside.

🧠 Memory Trick

📚 → Class 12, Physics, Chapter Electric Charges and Fields: Conductors in Electrostatic Equilibrium

12 A β-particle is moving perpendicular to the uniform electric field created by a potential difference of 6×10³ V across two parallel metal plates separated by a vertical distance of 6 cm. What is the acceleration of the particle between the plates?
(Mass of electron = 9.1×10⁻³¹ kg) (PYQ 2023 Shift 2)

A 1.76×10¹⁴ m s⁻²

B 1.76×10¹⁶ m s⁻²

C 1.8×10⁻¹⁶ m s⁻²

D 1.8×10⁻¹² m s⁻²

💡 Short Explanation

�� Use E = V/d. �� Force on charge is F = qE. �� Acceleration is a = F/m.

✅ Correct Answer Explanation

(Detailed) → Given: V = 6×10³ V d = 6 cm = 0.06 m → Electric field: E = V/d → E = 6×10³/0.06 → E = 1×10⁵ V/m → A β-particle is an electron. → q = 1.6×10⁻¹⁹ C → Acceleration: a = eE/m → a = (1.6×10⁻¹⁹ × 1×10⁵)/(9.1×10⁻³¹) → a ≈ 1.76×10¹⁶ m s⁻²

❓ Why Other Options Are Incorrect

A) Incorrect power of 10.
C) Wrong magnitude and physically impossible value.
D) Much smaller than the correct acceleration.

🎯 Strategy

Used

�� Method used: Formula Substitution
�� Convert cm into metre.
�� First find electric field.
�� Then use a = qE/m.

🧠 Memory Trick

📚 → Class 12, Physics, Chapter Electric Charges and Fields: Motion of Charge in Electric Field

13 Two protons P and Q are placed between two charged plates A and B as shown. Plate A is positively charged and plate B is negatively charged. (PYQ 2023 Shift 2)

A P experiences more force as it is closer to the positively charged plate

B Q experiences more force as it is closer to the negatively charged plate

C Both experience the same force towards the right-hand side

D Both experience the same force towards the left-hand side

💡 Short Explanation

�� Electric field between parallel plates is uniform. �� Force on charge is F = qE. �� Protons move along the electric field direction.

✅ Correct Answer Explanation

(Detailed) → Plate A is positively charged and plate B is negatively charged. → Electric field direction is from positive plate to negative plate. → Therefore, field is towards the right-hand side. → Both P and Q are protons, so both have the same charge q. → Electric field E is uniform between the plates. → Force: F = qE → Hence, both protons experience the same force towards the right-hand side.

❓ Why Other Options Are Incorrect

A) In a uniform electric field, force does not depend on closeness to the plate.
B) Force does not become larger near the negative plate in a uniform field.
D) Direction is wrong because a proton moves along the electric field.

🎯 Strategy

Used

�� Method used: Diagram Analysis
�� Identify positive and negative plates.
�� Electric field goes from + to −.
�� Use F = qE for both protons.

🧠 Memory Trick

📚 → Class 12, Physics, Chapter Electric Charges and Fields: Uniform Electric Field

14 Choose the correct statements from the following: (PYQ 2023 Shift 3)
(A) The total charge in any isolated system remains constant.
(B) When some charge is transferred to a conductor, it stays at the same place without getting disturbed over the entire surface.
(C) One Coulomb of negative charge is the total charge of 6.25 × 10¹⁸ electrons.
(D) Electric field is a scalar field.
(E) Potential due to a point charge depends on position independent of external electric field. Permanent dipole means that the dipole moment P exists irrespective of external electric field E.
Choose the correct answer from the options given below:

A (B), (C) and (E) only

B (A), (D) and (E) only

C (B) and (D) only

D (A), (C) and (E) only

💡 Short Explanation

�� Charge is conserved in an isolated system. �� 1 C charge corresponds to 6.25 × 10¹⁸ electrons. �� Electric field is a vector, not scalar.

✅ Correct Answer Explanation

(Detailed) → (A) is correct because total charge in an isolated system remains constant. → (B) is incorrect because charge given to a conductor spreads over its outer surface. → (C) is correct because: Number of electrons = 1/(1.6 × 10⁻¹⁹) = 6.25 × 10¹⁸ → (D) is incorrect because electric field has magnitude and direction. → (E) is correct because potential due to point charge depends on position, and permanent dipole has dipole moment even without external electric field.

❓ Why Other Options Are Incorrect

A) Includes incorrect statement (B).
B) Includes incorrect statement (D).
C) Both (B) and (D) are incorrect.

🎯 Strategy

Used

�� Method used: Statement Verification
�� Check every statement individually.
�� Remove statement saying charge stays fixed on conductor.
�� Remove statement saying electric field is scalar.

🧠 Memory Trick

📚 → Class 12, Physics, Chapter Electric Charges and Fields: Conservation of Charge and Electric Field

15 Choose the correct alternative from the following: (PYQ 2023 Shift 3)
(1) Gauss law is true for any open surface.
(2) Gauss law includes the use of all charges enclosed by the surface for calculation of electric flux through the surface.
(3) Gauss law can be used to calculate the magnetic field due to steady current.
(4) Gauss law is not based on the inverse square dependence of distance contained in Coulomb's law.

A 1 only

B 2 only

C 3 only

D 4 only

💡 Short Explanation

�� Gauss law applies to closed surfaces. �� Electric flux depends on enclosed charge. �� Magnetic field due to steady current is found using Ampere's law.

✅ Correct Answer Explanation

(Detailed) → Gauss law states: Φ = q_enclosed/ε₀ → It is applicable for closed Gaussian surfaces. → Statement (2) is correct because electric flux through a closed surface depends on total enclosed charge. → Therefore, only statement (2) is correct.

❓ Why Other Options Are Incorrect

A) Gauss law is not for any open surface.
C) Magnetic field due to steady current is calculated using Ampere's circuital law.
D) Gauss law is connected with inverse-square nature of Coulomb's law.

🎯 Strategy

Used

�� Method used: Statement Verification
�� Check whether surface is open or closed.
�� Check whether the law is for electric field or magnetic field.
�� Select only the enclosed-charge statement.

🧠 Memory Trick

📚 → Class 12, Physics, Chapter Electric Charges and Fields: Gauss's Law

16 Choose the correct answer from the following: If free charged particles are collinear and are in equilibrium, then: (PYQ 2023 Shift 3)

A All the charged particles have equal charge.

B All the charged particles have the same sign and the distances between two consecutive charged particles are same.

C All the charged particles have the same sign.

D All the charged particles cannot have the same sign.

💡 Short Explanation

�� Same sign charges only repel each other. �� Equilibrium needs balanced forces. �� Both attraction and repulsion are needed for net zero force.

✅ Correct Answer Explanation

(Detailed) → If all charges have the same sign, every charge repels every other charge. → The outermost charges experience net outward force. → For free collinear charged particles to remain in equilibrium, net force on every particle must be zero. → This is not possible if all charges have the same sign. → Hence, all charged particles cannot have the same sign.

❓ Why Other Options Are Incorrect

A) Equal charge is not a necessary condition for equilibrium.
B) Same sign charges cannot remain in equilibrium only by repulsion.
C) If all charges have the same sign, forces cannot balance properly.

🎯 Strategy

Used

�� Method used: Logical Elimination
�� Check nature of electrostatic force.
�� Same signs repel.
�� Equilibrium requires force cancellation.

🧠 Memory Trick

📚 → Class 12, Physics, Chapter Electric Charges and Fields: Electrostatic Force and Equilibrium

17 The electric field intensity due to an infinite thin plane sheet of surface charge density σ is: (PYQ 2023 Shift 3)

A σ/ε₀

B σ/(2ε₀)

C −σ/ε₀

D −2σ/ε₀

💡 Short Explanation

�� Use Gauss's law for infinite plane sheet. �� Electric field is uniform. �� Single sheet gives E = σ/(2ε₀).

✅ Correct Answer Explanation

(Detailed) → For an infinite thin plane sheet of charge, choose a cylindrical Gaussian pillbox. → Flux passes through both flat faces. → By Gauss's law: 2EA = σA/ε₀ → E = σ/(2ε₀)

❓ Why Other Options Are Incorrect

A) σ/ε₀ is the field between two oppositely charged plates, not a single sheet.
C) Magnitude and sign are incorrect for general field intensity.
D) Magnitude is incorrect.

🎯 Strategy

Used

�� Method used: Formula Recall
�� Identify it as a single infinite sheet.
�� Use Gaussian pillbox.
�� Remember factor 2 in denominator.

🧠 Memory Trick

📚 → Class 12, Physics, Chapter Electric Charges and Fields: Field Due to Infinite Plane Sheet

18 Two charged particles, placed at a distance d apart in vacuum, exert a force F on each other. Now, each of the charges is doubled. To keep the force unchanged, the distance between the charges should be changed to ______. (PYQ 2024 Shift 1)

A 4d

B 2d

C d

D d/2

💡 Short Explanation

�� Use Coulomb's law. �� Force is directly proportional to product of charges. �� Force is inversely proportional to square of distance.

✅ Correct Answer Explanation

(Detailed) → Coulomb's law: F = kq₁q₂/d² → When both charges are doubled: F′ = k(2q₁)(2q₂)/r² → F′ = 4kq₁q₂/r² → For force to remain unchanged: 4/r² = 1/d² → r² = 4d² → r = 2d

❓ Why Other Options Are Incorrect

A) 4d makes the force too small.
C) d gives force 4 times the original force.
D) d/2 gives force 16 times the original force.

🎯 Strategy

Used

�� Method used: Ratio Method
�� Track change in charge product.
�� Charge product becomes 4 times.
�� Distance must become 2 times.

🧠 Memory Trick

📚 → Class 12, Physics, Chapter Electric Charges and Fields: Coulomb's Law

19 Two large plane parallel sheets shown in the figure have equal but opposite surface charge densities +σ and −σ. A point charge q placed at points P₁, P₂ and P₃ experiences forces F⃗₁, F⃗₂ and F⃗₃ respectively. Then choose the correct answer from the options given below. (PYQ
2024 Shift 1)

A F⃗₁ = 0, F⃗₂ = 0, F⃗₃ = 0

B F⃗₁ = 0, F⃗₂ ≠ 0, F⃗₃ = 0

C F⃗₁ ≠ 0, F⃗₂ ≠ 0, F⃗₃ ≠ 0

D F⃗₁ = 0, F⃗₃ ≠ 0, F⃗₂ = 0

💡 Short Explanation

�� Outside oppositely charged parallel sheets, fields cancel. �� Between the sheets, fields add. �� Force exists only where electric field is non-zero.

✅ Correct Answer Explanation

(Detailed) → Electric field due to one infinite sheet is σ/2ε₀. → Outside the two oppositely charged sheets, electric fields are equal and opposite. → Therefore, net field outside = 0. → Between the sheets, both fields act in the same direction. → Therefore, net field between sheets ≠ 0. → Since F⃗ = qE⃗: F⃗₁ = 0, F⃗₂ ≠ 0, F⃗₃ = 0

❓ Why Other Options Are Incorrect

A) It ignores the non-zero field between the plates.
C) It wrongly assumes field is non-zero everywhere.
D) It gives the wrong force distribution.

🎯 Strategy

Used

�� Method used: Diagram Analysis
�� Use superposition of fields.
�� Outside fields cancel.
�� Between plates fields add.

🧠 Memory Trick

📚 → Class 12, Physics, Chapter Electric Charges and Fields: Electric Field Due to Infinite Plane Sheet

20 The transfer of integral number of ____________ is one of the evidence of quantization of electric charge. (PYQ 2024 Shift 1)

A photons

B nuclei

C electrons

D neutrons

💡 Short Explanation

�� Electric charge is quantized. �� Charge exists in integral multiples of e. �� Electron transfer shows this quantization.

✅ Correct Answer Explanation

(Detailed) → Quantization of charge means: q = ±ne → Here, n is an integer and e is the fundamental electronic charge. → Charging of a body occurs due to transfer of electrons. → Since electrons are transferred as whole particles, charge changes in integral multiples of e. → Therefore, transfer of electrons is evidence of quantization of charge.

❓ Why Other Options Are Incorrect

A) Photons are quanta of electromagnetic radiation, not carriers of electric charge.
B) Nuclei are not generally transferred in ordinary charging.
D) Neutrons are electrically neutral.

🎯 Strategy

Used

�� Method used: Conceptual Recall
�� Recall q = ne.
�� Identify particle responsible for ordinary charge transfer.
�� Select electrons.

🧠 Memory Trick

📚 → Class 12, Physics, Chapter Electric Charges and Fields: Quantization of Charge

21 A copper ball of density 8.0 g/cc and 1 cm in diameter is immersed in oil of density 0.8 g/cc. The charge on the ball if it remains just suspended in oil in an electric field of intensity 600π V/m acting in the upward direction is _______________. (PYQ 2024 Shift 1)

A 2 × 10⁻⁶ C

B 2 × 10⁻⁵ C

C 1 × 10⁻⁵ C

D 1 × 10⁻⁶ C

💡 Short Explanation

�� Electric force balances effective weight. �� Effective weight = weight − buoyant force. �� Use qE = (ρ_copper − ρ_oil)Vg.

✅ Correct Answer Explanation

(Detailed) → Diameter = 1 cm → Radius = 0.5 cm = 5 × 10⁻³ m → Volume of sphere: V = (4/3)πr³ → V = (4/3)π(5 × 10⁻³)³ → Density difference: ρ = 8.0 − 0.8 = 7.2 g/cc = 7200 kg/m³ → Effective weight: W = ρVg → W = 7200 × (4/3)π(5 × 10⁻³)³ × 9.8 → W = 0.01176π N → Electric field: E = 600π V/m → For suspension: qE = W → q = W/E → q = 0.01176π/600π → q = 1.96 × 10⁻⁵ C → q ≈ 2 × 10⁻⁵ C

❓ Why Other Options Are Incorrect

A) It is smaller by a factor of 10.
C) It is nearly half of the correct value.
D) It is too small.

🎯 Strategy

Used

�� Method used: Force Balance
�� Find effective weight in oil.
�� Balance it with electric force.
�� Solve q = W/E.

🧠 Memory Trick

📚 → Class 12, Physics, Chapter Electric Charges and Fields: Force on Charge in Electric Field

22 For an electric dipole in a non-uniform electric field with dipole moment parallel to direction of the field, the force F and torque τ on the dipole respectively are ____________. (PYQ 2024 Shift 1)

A F = 0, τ = 0

B F ≠ 0, τ = 0

C F = 0, τ ≠ 0

D F ≠ 0, τ ≠ 0

💡 Short Explanation

�� Non-uniform electric field can exert net force on dipole. �� Torque depends on angle between p⃗ and E⃗. �� Here p⃗ is parallel to E⃗, so torque is zero.

✅ Correct Answer Explanation

(Detailed) → Torque on an electric dipole: τ = pE sinθ → Dipole moment is parallel to electric field. → Therefore: θ = 0° → τ = pE sin0° → τ = 0 → Since the electric field is non-uniform, field strength is different at the two ends of the dipole. → Hence, forces on +q and −q are unequal. → Therefore, net force is non-zero. → So, F ≠ 0 and τ = 0.

❓ Why Other Options Are Incorrect

A) Force is not zero in a non-uniform field.
C) Torque is zero because dipole moment is parallel to field.
D) Torque is not non-zero for parallel alignment.

🎯 Strategy

Used

�� Method used: Conceptual Recall
�� Use τ = pE sinθ.
�� Put θ = 0°.
�� Remember non-uniform field gives net force.

🧠 Memory Trick

📚 → Class 12, Physics, Chapter Electric Charges and Fields: Electric Dipole in Non-uniform Electric Field

23 An object A is charged by rubbing with an object B. In the process 10⁹ electrons are transferred from A to B. Charge developed on A will be:
(Take electronic charge e = 1.602 × 10⁻¹⁹ C) (PYQ 2024 Shift 2)

A 10⁹ C

B 10⁻⁹ C

C 1.6 × 10⁻¹⁰ C

D −1.6 × 10⁻¹⁰ C

💡 Short Explanation

�� Charge q = ne. �� Electrons are transferred from A to B. �� A loses electrons, so A becomes positively charged.

✅ Correct Answer Explanation

(Detailed) → Number of electrons transferred: n = 10⁹ → Electronic charge: e = 1.602 × 10⁻¹⁹ C → Charge developed: q = ne → q = 10⁹ × 1.602 × 10⁻¹⁹ → q = 1.602 × 10⁻¹⁰ C → Since A loses electrons, charge on A is positive. → q ≈ +1.6 × 10⁻¹⁰ C

❓ Why Other Options Are Incorrect

A) It ignores the value of electronic charge.
B) It has incorrect power of 10.
D) Sign is wrong because A loses electrons and becomes positive.

🎯 Strategy

Used

�� Method used: Formula Substitution
�� Use q = ne.
�� Track direction of electron transfer.
�� Losing electrons means positive charge.

🧠 Memory Trick

📚 → Class 12, Physics, Chapter Electric Charges and Fields: Quantization of Charge

24 If the protons and electrons are the only basic charges in the universe, all observable charges have to be integral multiples of e. Thus, if an object contains x electrons and y protons, the net charge on the object will be: (PYQ 2025 Shift 1)

A (x + y)e

B (x + y)/e

C (x − y)/e

D (y − x)e

💡 Short Explanation

�� Electron has charge −e. �� Proton has charge +e. �� Net charge = proton charge + electron charge.

✅ Correct Answer Explanation

(Detailed) → Charge of one proton = +e → Charge of y protons = +ye → Charge of one electron = −e → Charge of x electrons = −xe → Net charge: q = ye − xe → q = (y − x)e

❓ Why Other Options Are Incorrect

A) It ignores the negative charge of electrons.
B) It has wrong expression and wrong unit.
C) It has wrong sign and wrong unit.

🎯 Strategy

Used

�� Method used: Algebraic Charge Addition
�� Add charges with signs.
�� Proton contributes positive charge.
�� Electron contributes negative charge.

🧠 Memory Trick

📚 → Class 12, Physics, Chapter Electric Charges and Fields: Quantization of Charge

25 A charge of magnitude 3 × 10⁻⁷ C is located at a distance of 0.09 m from a point P. Obtain the work done in bringing a charge of 2 × 10⁻⁹ C from infinity to the point P. (PYQ 2025 Shift 1)

A 1.6 × 10⁻⁴ J

B 6 × 10⁻² J

C 6 × 10⁻⁵ J

D 6 × 10⁵ J

💡 Short Explanation

�� Work done equals increase in electrostatic potential energy. �� Use W = kQq/r. �� Substitute values carefully.

✅ Correct Answer Explanation

(Detailed) → Work done in bringing charge from infinity: W = kQq/r → Given: Q = 3 × 10⁻⁷ C q = 2 × 10⁻⁹ C r = 0.09 m k = 9 × 10⁹ Nm²/C² → W = (9 × 10⁹ × 3 × 10⁻⁷ × 2 × 10⁻⁹)/0.09 → W = (54 × 10⁻⁷)/0.09 → W = 600 × 10⁻⁷ → W = 6 × 10⁻⁵ J

❓ Why Other Options Are Incorrect

A) It results from calculation error.
B) It has power of 10 error.
D) It is unrealistically large for such small charges.

🎯 Strategy

Used

�� Method used: Formula Substitution
�� Use W = qV.
�� Substitute V = kQ/r.
�� Track powers of 10 carefully.

🧠 Memory Trick

📚 → Class 12, Physics, Chapter Electric Charges and Fields: Electrostatic Potential Energy

26 If the net flux through a cube is 1.05 Nm² C⁻¹, what will be the total charge inside the cube?
(Given: Permittivity of free space = 8.85 × 10⁻¹² C² N⁻¹ m⁻²) (PYQ 2025 Shift 1)

A 9.29 × 10⁻¹¹ C

B 9.27 × 10⁻¹⁰ C

C 9.22 × 10⁻⁶ C

D 9.29 × 10⁻¹² C

💡 Short Explanation

�� Use Gauss's law. �� Φ = Q/ε₀. �� Therefore, Q = ε₀Φ. �� Use Gauss's law. �� Q = ε₀Φ. �� Multiplication gives 9.29 × 10⁻¹² C.

✅ Correct Answer Explanation

(Detailed) → By Gauss's law: Φ = Q/ε₀ → Therefore: Q = ε₀Φ → Given: ε₀ = 8.85 × 10⁻¹² Φ = 1.05 → Q = 8.85 × 10⁻¹² × 1.05 → Q = 9.2925 × 10⁻¹² C → Q ≈ 9.29 × 10⁻¹² C

❓ Why Other Options Are Incorrect

A) It is 10 times greater than the correct value.
B) It has wrong power of 10.
C) It is much larger than the correct value.

🎯 Strategy

Used

�� Method used: Formula Substitution
�� Use Gauss's law.
�� Rearrange to Q = ε₀Φ.
�� Track exponent correctly.

🧠 Memory Trick

📚 → Class 12, Physics, Chapter Electric Charges and Fields: Gauss's Law

27 The electric potential due to an electric dipole: (PYQ 2025 Shift 1)
(A) depends on r
(B) depends on angle
(C) falls as 1/r²
(D) independent of separation
Choose the correct answer from the options given below:

A (A), (B) and (D) only

B (A), (B) and (C) only

C (A), (B), (C) and (D)

D (B), (C) and (D) only

💡 Short Explanation

�� Dipole potential depends on distance r. �� It also depends on angle θ. �� It falls as 1/r².

✅ Correct Answer Explanation

(Detailed) → Electric potential due to a short dipole is: V = (1/4πε₀)(p cosθ/r²) → (A) Correct: It depends on r. → (B) Correct: It depends on angle θ. → (C) Correct: It falls as 1/r². → (D) Incorrect: Dipole moment p = q × 2a, so it depends on charge separation. → Correct statements are (A), (B) and (C) only.

❓ Why Other Options Are Incorrect

A) It includes incorrect statement (D).
C) It includes all statements, including incorrect statement (D).
D) It misses correct statement (A) and includes incorrect statement (D).

🎯 Strategy

Used

�� Method used: Statement Verification
�� Write dipole potential formula.
�� Check r dependence.
�� Check angular and separation dependence.

🧠 Memory Trick

📚 → Class 12, Physics, Chapter Electric Charges and Fields: Electric Dipole Potential

28 Two point charges placed a distance d apart in vacuum exert a force of magnitude F on each other. One of the two charges is doubled. To keep the magnitude of force same the separation between the charges should be changed to: (PYQ 2025 Shift 1)

A 2d

B d/2

C √2d

D d/√2

💡 Short Explanation

�� Coulomb force F ∝ q₁q₂/r². �� One charge is doubled, so force becomes double. �� Distance must increase by √2.

✅ Correct Answer Explanation

(Detailed) → Initial force: F = kq₁q₂/d² → One charge is doubled. → New force should remain same: F = k(2q₁)q₂/r² → Equate both: kq₁q₂/d² = 2kq₁q₂/r² → 1/d² = 2/r² → r² = 2d² → r = √2d

❓ Why Other Options Are Incorrect

A) 2d increases distance too much and force becomes smaller.
B) d/2 decreases distance and force increases greatly.
D) d/√2 decreases distance and force becomes larger.

🎯 Strategy

Used

�� Method used: Ratio Method
�� Track change in charge product.
�� One charge doubled gives factor 2.
�� Distance must become √2 times.

🧠 Memory Trick

📚 → Class 12, Physics, Chapter Electric Charges and Fields: Coulomb's Law

UNIT-1 PHYSICS CATEGORIZED-PYQ

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