NEET Physics: Electrostatics and Current Electricity · Chapter 1

Electric Charges and Fields

What to remember

Coulomb's law: F = k q₁q₂/r² with k = 1/(4πε₀) = 9 × 10⁹ N m² C⁻²; the force is along the line joining the charges, and forces from many charges add as vectors (superposition).
Charge is conserved, additive and quantised: q = ne, with e = 1.6 × 10⁻¹⁹ C.
Gauss's law: the net flux through a closed surface = q_enclosed/ε₀. It gives E = λ/(2πε₀r) for a long wire, E = σ/(2ε₀) for an infinite sheet, and for a thin shell E = 0 inside and kq/r² outside.
A dipole has moment p = q × 2a (from –q to +q). In a uniform field the torque is τ = pE sin θ, the potential energy is U = –pE cos θ, and the net force is zero.

Electric charge, conductors and insulators

Charge is a basic property of matter. There are two kinds: positive and negative. Like charges repel and unlike charges attract. The SI unit of charge is the coulomb (C). Rubbing a glass rod with silk makes the rod positive; rubbing an ebonite rod with fur makes the rod negative. Only electrons move in this process, so the object that loses electrons becomes positive.

Conductors (metals, the human body, Earth) allow charge to flow through them easily because they have free electrons. Insulators (glass, plastic, rubber, dry wood) do not allow charge to flow. Charge given to an insulator stays where it is placed. Charge given to a conductor spreads over its outer surface.

Charging by induction. Bring a charged rod near an uncharged metal sphere on an insulating stand, without touching it. The near side of the sphere gets the opposite charge and the far side gets the same charge as the rod. Earth the sphere with a wire (the far-side charge flows to Earth), remove the earth wire, and then remove the rod. The sphere keeps a charge opposite to that of the rod. The rod itself loses no charge in induction.

Charging by conduction (contact). A charged body touches a neutral conductor and the two share charge. Both end up with charge of the same sign.

Basic properties of charge

Property	Meaning	Example
Additivity	Total charge is the algebraic sum of the charges	+5 μC and –2 μC together make +3 μC
Conservation	Total charge of an isolated system never changes	Charge is created only in equal and opposite pairs
Quantisation	Charge is an integer multiple of e	q = ±ne; the smallest free charge is 1.6 × 10⁻¹⁹ C

Quantisation shows that the charge on a body can never be 0.5e. Macroscopic charges are so large compared with e that the steps look smooth. One microcoulomb contains about 6 × 10¹² electrons' worth of charge.

Coulomb's law and superposition

For two point charges q₁ and q₂ at distance r in vacuum:

F = (1/4πε₀) q₁q₂/r² = k q₁q₂/r²

Here ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻² is the permittivity of free space and k = 9 × 10⁹ N m² C⁻². The force is attractive for opposite signs and repulsive for like signs. The forces on the two charges are equal and opposite (Newton's third law). In a medium of dielectric constant K the force becomes F/K.

Vector form. The force on charge q₂ due to q₁ is F₂₁ = k q₁q₂ r̂₁₂/r², where r̂₁₂ is the unit vector pointing from q₁ to q₂.

Superposition. The force on one charge due to many charges is the vector sum of the forces from each charge taken alone. The presence of other charges does not change the pair force.

Electric field

The electric field at a point is the force per unit positive test charge: E = F/q₀, in N C⁻¹ or V m⁻¹. The test charge must be very small so that it does not disturb the source charges.

Point charge Q: E = kQ/r², directed away from Q if Q is positive and towards Q if Q is negative.
System of charges: E is the vector sum of the fields from each charge.
Force on a charge q in a field E: F = qE.

Electric field lines are a picture of the field.

Property	Statement
Direction	Start on positive charge, end on negative charge (or go to infinity)
Tangent	The tangent at a point gives the direction of E there
Density	Closer lines mean a stronger field
Crossing	Two lines never cross, because E has one direction at a point
Shape	Lines never form closed loops in electrostatics
Conductor	Lines meet a conductor surface at right angles

Electric dipole

A dipole is a pair of equal and opposite charges +q and –q separated by a small distance 2a. The dipole moment is the vector p = q × 2a, pointing from –q to +q; its SI unit is C m. A dipole is neutral overall.

Position	Field magnitude (for r much greater than a)	Direction
On the axis	E = 2kp/r³ = p/(2πε₀r³)	Along p
On the equatorial line	E = kp/r³ = p/(4πε₀r³)	Opposite to p

On the axis the field is exactly twice the equatorial field at the same distance. Dipole field falls as 1/r³, faster than the 1/r² of a point charge.

Dipole in a uniform field E.

Net force = 0, because the two forces are equal and opposite.
Torque: τ = p × E, of magnitude pE sin θ, where θ is the angle between p and E. The torque tries to rotate p towards E.
Potential energy: U = –pE cos θ. It is minimum (–pE) when θ = 0 (stable equilibrium) and maximum (+pE) when θ = 180° (unstable equilibrium).
Work done in turning the dipole from θ₁ to θ₂: W = pE (cos θ₁ – cos θ₂).

In a non-uniform field a dipole also feels a net force.

Continuous charge distribution

When charge is spread over a line, surface or volume, we use densities:

Distribution	Symbol	Definition	SI unit
Linear	λ	charge per unit length	C m⁻¹
Surface	σ	charge per unit area	C m⁻²
Volume	ρ	charge per unit volume	C m⁻³

A small element dq = λ dl (or σ dS, or ρ dV) is treated as a point charge, and the fields of all elements are added by integration.

Electric flux and Gauss's law

Electric flux through a small area vector ΔS is Δφ = E · ΔS = E ΔS cos θ. Its unit is N m² C⁻¹. Flux is maximum when the field is perpendicular to the surface and zero when the field is parallel to it. Outward flux is taken as positive.

Gauss's law: the total flux through any closed surface is φ = q_enclosed/ε₀. Charges outside the surface contribute zero net flux, although they affect the field at the surface. The surface used is an imaginary Gaussian surface and is chosen with the symmetry of the charge.

Applications of Gauss's law

Charge distribution	Field	Notes
Infinitely long straight wire (λ per unit length)	E = λ/(2πε₀r)	Radial; falls as 1/r; cylindrical Gaussian surface
Infinite plane sheet (σ per unit area)	E = σ/(2ε₀)	Same value at every distance; points away from a positive sheet
Thin spherical shell, radius R, total charge q	Outside: E = kq/r² (r ≥ R)	As if all charge is at the centre
Thin spherical shell	Inside: E = 0	No enclosed charge
Thin shell, on the surface	E = kq/R²	Value just outside the shell

For a thin charged shell the field is zero inside and drops from kq/R² to zero as the point moves just inside the surface. Two large parallel sheets with equal and opposite charge density σ have field σ/ε₀ between them and zero outside.

Constants and formula summary

Quantity	Formula	Units
Coulomb force	F = kq₁q₂/r²	N
Field of a point charge	E = kQ/r²	N C⁻¹
Axial dipole field	E = 2kp/r³	N C⁻¹
Equatorial dipole field	E = kp/r³	N C⁻¹
Torque on dipole	τ = pE sin θ	N m
Dipole energy	U = –pE cos θ	J
Gauss's law	φ = q/ε₀	N m² C⁻¹
Wire	E = λ/(2πε₀r)	N C⁻¹
Sheet	E = σ/(2ε₀)	N C⁻¹

Constants: k = 9 × 10⁹ N m² C⁻², e = 1.6 × 10⁻¹⁹ C, ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻², mₑ = 9.1 × 10⁻³¹ kg.

Worked examples

Example 1 (Coulomb force). Two point charges of 2 μC and 3 μC are 3 cm apart in air. Find the force.

F = 9 × 10⁹ × (2 × 10⁻⁶)(3 × 10⁻⁶)/(3 × 10⁻²)² = 9 × 10⁹ × 6 × 10⁻¹²/(9 × 10⁻⁴) = 60 N, repulsive.

Example 2 (dipole torque). A dipole has charges ±2 μC separated by 1 cm. It is placed at 30° to a field of 10⁵ N C⁻¹.

p = 2 × 10⁻⁶ × 10⁻² = 2 × 10⁻⁸ C m. τ = pE sin 30° = 2 × 10⁻⁸ × 10⁵ × 0.5 = 10⁻³ N m.

Example 3 (Gauss's law, sheet). An infinite sheet has σ = 1.77 × 10⁻⁸ C m⁻². Find E.

E = σ/(2ε₀) = 1.77 × 10⁻⁸/(2 × 8.85 × 10⁻¹²) = 1000 N C⁻¹.

Example 4 (flux). A closed surface encloses a net charge of 17.7 nC. The flux is q/ε₀ = 17.7 × 10⁻⁹/8.85 × 10⁻¹² = 2000 N m² C⁻¹.

Example 5 (field of two charges). Charges +4 μC and –4 μC are 0.6 m apart. At the midpoint each charge gives a field in the same direction (towards the negative charge). Each field is 9 × 10⁹ × 4 × 10⁻⁶/(0.3)² = 4 × 10⁵ N C⁻¹, so the net field is 8 × 10⁵ N C⁻¹.

NEET traps

1. Axial and equatorial dipole fields. Axial is 2kp/r³ along p; equatorial is kp/r³ opposite to p. Do not swap them.
2. Force in a uniform field on a dipole is zero, but torque is not. In a non-uniform field there is also a net force.
3. Dipole stable position. θ = 0 is stable (U minimum); θ = 180° is unstable (U maximum). Torque is zero at both.
4. Sheet versus shell. The field of an infinite sheet does not depend on distance; the field of a shell outside falls as 1/r² and inside is zero.
5. Plate-to-plate field. Two oppositely charged parallel plates give σ/ε₀ between them, not σ/2ε₀.
6. Flux depends only on enclosed charge, but the field at the surface depends on all charges, inside and outside.
7. Charging by induction leaves the rod's charge unchanged and gives the body the opposite sign. Charging by conduction gives the same sign.
8. Field lines do not cross and do not form closed loops in electrostatics. They also never start or stop in empty space.
9. Quantisation: a body can have charge 3.2 × 10⁻¹⁹ C (2e) but not 2.4 × 10⁻¹⁹ C, because 2.4/1.6 = 1.5 is not an integer.
10. Medium effect. In a medium of dielectric constant K the force is divided by K; the value of k in vacuum is not changed.

One-liners

1. The SI unit of charge is the coulomb.
2. The charge of an electron is –1.6 × 10⁻¹⁹ C.
3. Charge is conserved in every process.
4. Coulomb force between two charges is along the line joining them.
5. Electric field is force per unit positive test charge.
6. SI unit of electric field is N C⁻¹, which equals V m⁻¹.
7. Electric field lines are never closed loops in electrostatics.
8. Dipole moment points from the negative to the positive charge.
9. The SI unit of dipole moment is C m.
10. Net force on a dipole in a uniform field is zero.
11. The torque on a dipole is maximum at θ = 90°.
12. The unit of electric flux is N m² C⁻¹.
13. The flux through a closed surface with no net charge inside is zero.
14. The field inside a uniformly charged thin spherical shell is zero.
15. The field of an infinite sheet is independent of distance from the sheet.

Practice questions

Two point charges of 4 μC and 5 μC are placed 0.2 m apart in vacuum. Taking k = 9 × 10⁹ N m² C⁻², the force between them is
1. 4.5 N
2. 9 N
3. 45 N
4. 0.45 N
Answer
A. 4.5 N
F = 9 × 10⁹ × (4 × 10⁻⁶)(5 × 10⁻⁶)/(0.2)² = 9 × 10⁹ × 20 × 10⁻¹²/0.04 = 4.5 N.
The force between two point charges is F. If each charge is doubled and the distance between them is halved, the new force is
1. 8 F
2. 2 F
3. 16 F
4. 4 F
Answer
C. 16 F
F ∝ q₁q₂/r². Charges ×2 each gives ×4; r halved gives ×4. Total ×16.
A body has a charge of 3.2 μC. The number of excess electrons on it (e = 1.6 × 10⁻¹⁹ C) is
1. 2 × 10¹²
2. 2 × 10¹³
3. 5 × 10¹³
4. 2 × 10¹⁹
Answer
B. 2 × 10¹³
n = q/e = 3.2 × 10⁻⁶/1.6 × 10⁻¹⁹ = 2 × 10¹³.
The electric field at 0.3 m from a point charge of 8 nC in vacuum is (k = 9 × 10⁹ N m² C⁻²)
1. 240 N C⁻¹
2. 80 N C⁻¹
3. 2400 N C⁻¹
4. 800 N C⁻¹
Answer
D. 800 N C⁻¹
E = kq/r² = 9 × 10⁹ × 8 × 10⁻⁹/0.09 = 72/0.09 = 800 N C⁻¹.
The force on an electron placed in a uniform electric field of 2 × 10⁴ N C⁻¹ is (e = 1.6 × 10⁻¹⁹ C)
1. 3.2 × 10⁻²³ N
2. 1.6 × 10⁻¹⁵ N
3. 3.2 × 10⁻¹⁵ N
4. 3.2 × 10⁻¹³ N
Answer
C. 3.2 × 10⁻¹⁵ N
F = eE = 1.6 × 10⁻¹⁹ × 2 × 10⁴ = 3.2 × 10⁻¹⁵ N.
A dipole of moment 3 × 10⁻⁸ C m is in air. The electric field on its axis at a distance 0.3 m (much larger than the dipole size) is (k = 9 × 10⁹ N m² C⁻²)
1. 4 × 10⁴ N C⁻¹
2. 2 × 10⁴ N C⁻¹
3. 6 × 10³ N C⁻¹
4. 1 × 10⁴ N C⁻¹
Answer
B. 2 × 10⁴ N C⁻¹
E = 2kp/r³ = 2 × 9 × 10⁹ × 3 × 10⁻⁸/(0.027) = 540/0.027 = 2 × 10⁴ N C⁻¹. (1 × 10⁴ would be the equatorial value.)
The maximum torque on a dipole of moment 5 × 10⁻⁹ C m in a uniform field of 4 × 10⁴ N C⁻¹ is
1. 1 × 10⁻⁴ N m
2. 2 × 10⁻⁵ N m
3. 2 × 10⁻³ N m
4. 2 × 10⁻⁴ N m
Answer
D. 2 × 10⁻⁴ N m
τ_max = pE (at θ = 90°) = 5 × 10⁻⁹ × 4 × 10⁴ = 2 × 10⁻⁴ N m.
A dipole of moment 2 × 10⁻⁸ C m lies along a uniform field of 10⁵ N C⁻¹. The work needed to turn it through 180° is
1. 4 × 10⁻³ J
2. zero
3. 8 × 10⁻³ J
4. 2 × 10⁻³ J
Answer
A. 4 × 10⁻³ J
W = pE(cos 0 – cos 180°) = 2pE = 2 × 2 × 10⁻⁸ × 10⁵ = 4 × 10⁻³ J.
A closed surface encloses a net charge of 26.55 nC. Taking ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻², the electric flux through the surface is
1. 3000 N m² C⁻¹
2. 1500 N m² C⁻¹
3. 239 N m² C⁻¹
4. 6000 N m² C⁻¹
Answer
A. 3000 N m² C⁻¹
φ = q/ε₀ = 26.55 × 10⁻⁹/8.85 × 10⁻¹² = 3000 N m² C⁻¹.
An infinitely long straight wire carries a uniform linear charge density 2 × 10⁻⁸ C m⁻¹. The field at 0.4 m from the wire is (k = 9 × 10⁹ N m² C⁻²)
1. 1800 N C⁻¹
2. 900 N C⁻¹
3. 450 N C⁻¹
4. 2250 N C⁻¹
Answer
B. 900 N C⁻¹
E = λ/(2πε₀r) = 2kλ/r = 2 × 9 × 10⁹ × 2 × 10⁻⁸/0.4 = 360/0.4 = 900 N C⁻¹.
A thin spherical shell of radius 0.1 m carries a charge of 5 μC. The electric field at 0.3 m from its centre is (k = 9 × 10⁹ N m² C⁻²)
1. 1.5 × 10⁵ N C⁻¹
2. zero
3. 5 × 10⁵ N C⁻¹
4. 1.5 × 10⁶ N C⁻¹
Answer
C. 5 × 10⁵ N C⁻¹
Outside the shell E = kq/r² = 9 × 10⁹ × 5 × 10⁻⁶/0.09 = 5 × 10⁵ N C⁻¹.
Which of the following cannot be the charge on a body? (e = 1.6 × 10⁻¹⁹ C)
1. 6.4 × 10⁻¹⁹ C
2. 4.8 × 10⁻¹⁹ C
3. 8.0 × 10⁻¹⁹ C
4. 2.4 × 10⁻¹⁹ C
Answer
D. 2.4 × 10⁻¹⁹ C
Charge must be an integer multiple of e. 2.4/1.6 = 1.5 is not an integer; the others are 3e, 4e and 5e.
An infinite plane sheet produces a field of 500 N C⁻¹ in air. Its surface charge density (ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻²) is
1. 8.85 × 10⁻¹² C m⁻²
2. 8.85 × 10⁻⁹ C m⁻²
3. 4.4 × 10⁻⁹ C m⁻²
4. 1.77 × 10⁻⁸ C m⁻²
Answer
B. 8.85 × 10⁻⁹ C m⁻²
E = σ/2ε₀, so σ = 2ε₀E = 2 × 8.85 × 10⁻¹² × 500 = 8.85 × 10⁻⁹ C m⁻².
A dipole placed in a uniform electric field is in stable equilibrium when the angle between its moment p and the field E is
1. 0°
2. 90°
3. 45°
4. 180°
Answer
A. 0°
U = –pE cos θ is minimum at θ = 0°, so equilibrium there is stable.
On the equatorial line of an electric dipole, the electric field is directed
1. along the dipole moment
2. perpendicular to the dipole moment
3. opposite to the dipole moment
4. radially away from the centre
Answer
C. opposite to the dipole moment
On the equatorial line the net field is kp/r³ and points opposite to p.
A point charge is at the centre of a sphere. If the radius of the sphere is doubled, the electric flux through the sphere
1. becomes half
2. becomes four times
3. becomes double
4. remains the same
Answer
D. remains the same
Flux = q/ε₀ depends only on the enclosed charge, not on the size of the closed surface.
Given below are two statements: Statement I: The net force on an electric dipole placed in a uniform electric field is zero. Statement II: The torque on a dipole in a uniform electric field is always zero. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
Forces on +q and –q are equal and opposite, so net force is zero. Torque pE sin θ is zero only when θ is 0° or 180°.
Given below are two statements: Statement I: Two electric field lines never cross each other. Statement II: Electrostatic field lines can form closed loops. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
A crossing would give two directions of E at one point. Electrostatic field lines start on positive charge and end on negative charge; they do not form closed loops.
Given below are two statements: Statement I: The electric field inside a uniformly charged thin spherical shell is zero. Statement II: Outside the shell the field is the same as that of a point charge at the centre carrying the total charge. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
A. Both Statement I and Statement II are correct
Gauss's law gives zero enclosed charge inside and kq/r² outside, as if the charge were at the centre.
Given below are two statements: Statement I: In charging by induction, the body gets a charge of the same sign as the charging rod. Statement II: The charging rod loses some of its charge during induction. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
B. Both Statement I and Statement II are incorrect
Induction gives the body the opposite sign, and the rod never touches the body, so its charge is unchanged.
Given below are two statements: Statement I: At the same far distance, the field of a dipole on its axis is half the field on its equatorial line. Statement II: The equatorial field of a dipole is directed opposite to the dipole moment. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
Axial field is 2kp/r³ (twice the equatorial kp/r³), so Statement I is wrong. Statement II is correct.
Which of the following statements are correct? 1. The SI unit of electric field is N C⁻¹. 2. The SI unit of electric flux is N m² C⁻¹. 3. The SI unit of dipole moment is C m⁻¹.
1. 1 and 2 only
2. 2 and 3 only
3. 1 and 3 only
4. 1, 2 and 3
Answer
A. 1 and 2 only
E has unit N C⁻¹ and flux has N m² C⁻¹. Dipole moment is charge × distance, so its unit is C m, not C m⁻¹.
Which of the following statements are correct? 1. Charge is quantised. 2. A single positive charge can be created alone in an isolated system. 3. The total charge of an isolated system is conserved.
1. 1 and 2 only
2. 2 and 3 only
3. 1 and 3 only
4. 1, 2 and 3
Answer
C. 1 and 3 only
Charge is quantised and conserved. Charge is created only in equal and opposite pairs, so statement 2 is false.
Which of the following statements are correct? 1. The flux through a closed surface depends on charges placed outside it. 2. Gauss's law holds for any closed surface. 3. The flux is zero if the net enclosed charge is zero.
1. 1 and 2 only
2. 2 and 3 only
3. 1 and 3 only
4. 1, 2 and 3
Answer
B. 2 and 3 only
Flux depends only on enclosed charge. Outside charges give equal inflow and outflow.
Given below are two statements: Statement I: The force between two charges in a medium of dielectric constant K is K times the force in vacuum. Statement II: The electric field of a point charge falls as 1/r² in vacuum. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
In a medium the force is divided by K, so Statement I is wrong. E = kq/r² is correct.
Given below are two statements: Statement I: The torque on a dipole in a uniform field is maximum when the dipole moment is perpendicular to the field. Statement II: The potential energy of a dipole is maximum when its moment is opposite to the field. Choose the correct option.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
A. Both Statement I and Statement II are correct
τ = pE sin θ is maximum at 90°. U = –pE cos θ is maximum (+pE) at θ = 180°.
Assertion (A): The electric field inside a charged thin conducting spherical shell is zero. Reason (R): No charge is enclosed by a Gaussian sphere drawn inside the shell, and the field has spherical symmetry.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
Gauss's law with zero enclosed charge and spherical symmetry gives E = 0 inside.
Assertion (A): The charge on any body is always an integral multiple of the electronic charge. Reason (R): Electric charge is conserved in an isolated system.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
B. Both A and R are true but R is not the correct explanation of A
Both statements are true. Conservation does not explain why charge comes in whole multiples of e.
Assertion (A): The net force on a dipole in a uniform electric field is zero. Reason (R): The torque on a dipole in a uniform electric field is always zero.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
C. A is true but R is false
A is true because the two forces cancel. R is false since torque is pE sin θ.
Assertion (A): A body is charged by induction by touching it with a charged rod. Reason (R): A body charged by induction gets a charge opposite to that of the rod.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
D. A is false but R is true
Induction does not need contact, so A is false. R is true.
Assertion (A): A dipole has zero torque when its moment is parallel to the field. Reason (R): Equilibrium at θ = 0° is stable and at θ = 180° is unstable.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
B. Both A and R are true but R is not the correct explanation of A
Both true, but zero torque follows from sin θ = 0, not from the stability of equilibrium.
Assertion (A): The net flux through a closed surface is zero if no charge is inside it. Reason (R): Gauss's law states that the flux through a closed surface equals the enclosed charge divided by ε₀.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
With q_enclosed = 0, φ = q/ε₀ = 0. R is the law that explains A.
Match List I with List II. List I: a. Dipole moment b. Electric flux c. Linear charge density d. Surface charge density. List II: i. N m² C⁻¹ ii. C m iii. C m⁻¹ iv. C m⁻².
1. a-i, b-ii, c-iii, d-iv
2. a-iii, b-ii, c-i, d-iv
3. a-ii, b-i, c-iv, d-iii
4. a-ii, b-i, c-iii, d-iv
Answer
D. a-ii, b-i, c-iii, d-iv
Dipole moment: C m. Flux: N m² C⁻¹. λ: C m⁻¹. σ: C m⁻².
Match List I with List II. List I: a. Infinite straight wire b. Infinite plane sheet c. Thin shell, outside d. Thin shell, inside. List II: i. σ/2ε₀ ii. λ/2πε₀r iii. kq/r² iv. zero.
1. a-iii, b-i, c-ii, d-iv
2. a-i, b-ii, c-iii, d-iv
3. a-ii, b-i, c-iii, d-iv
4. a-ii, b-i, c-iv, d-iii
Answer
C. a-ii, b-i, c-iii, d-iv
Wire λ/2πε₀r; sheet σ/2ε₀; shell outside kq/r²; shell inside zero.
Match List I with List II. List I: a. Axial field of a dipole b. Equatorial field of a dipole c. Torque on a dipole at 90° d. Minimum potential energy of a dipole. List II: i. –pE ii. pE iii. kp/r³ iv. 2kp/r³.
1. a-iv, b-iii, c-i, d-ii
2. a-iv, b-iii, c-ii, d-i
3. a-iii, b-iv, c-ii, d-i
4. a-ii, b-iii, c-iv, d-i
Answer
B. a-iv, b-iii, c-ii, d-i
Axial 2kp/r³; equatorial kp/r³; τ at 90° = pE; U_min = –pE.
Match List I with List II. List I: a. Charging by conduction b. Charging by induction c. Glass rod rubbed with silk d. Ebonite rod rubbed with fur. List II: i. Glass rod becomes positive ii. Ebonite rod becomes negative iii. Body gets the same sign as the charged body iv. Body gets the opposite sign.
1. a-iv, b-iii, c-i, d-ii
2. a-i, b-ii, c-iii, d-iv
3. a-iii, b-iv, c-ii, d-i
4. a-iii, b-iv, c-i, d-ii
Answer
D. a-iii, b-iv, c-i, d-ii
Conduction gives the same sign; induction the opposite. Glass becomes positive with silk; ebonite becomes negative with fur.
Match List I with List II (k = 9 × 10⁹ N m² C⁻², ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻²). List I: a. Force between two 1 μC charges 0.1 m apart b. Field at 0.3 m from a 3 μC charge c. Torque on a dipole of 10⁻⁹ C m at 90° in 10⁴ N C⁻¹ d. Flux through a closed surface enclosing 8.85 nC. List II: i. 10⁻⁵ N m ii. 10³ N m² C⁻¹ iii. 3 × 10⁵ N C⁻¹ iv. 0.9 N.
1. a-i, b-iii, c-iv, d-ii
2. a-iv, b-iii, c-i, d-ii
3. a-iv, b-iii, c-ii, d-i
4. a-iii, b-iv, c-i, d-ii
Answer
B. a-iv, b-iii, c-i, d-ii
a: 9 × 10⁹ × 10⁻¹²/0.01 = 0.9 N. b: 9 × 10⁹ × 3 × 10⁻⁶/0.09 = 3 × 10⁵. c: 10⁻⁹ × 10⁴ = 10⁻⁵ N m. d: 8.85 × 10⁻⁹/8.85 × 10⁻¹² = 10³.
Match List I with List II. List I: a. Field lines of an isolated positive charge b. Field lines of an isolated negative charge c. Field lines of a dipole d. Field lines of a uniform field. List II: i. Parallel and equally spaced ii. Radially inward iii. Radially outward iv. Begin on the positive charge and end on the negative charge.
1. a-iii, b-ii, c-i, d-iv
2. a-iv, b-ii, c-iii, d-i
3. a-ii, b-iii, c-iv, d-i
4. a-iii, b-ii, c-iv, d-i
Answer
D. a-iii, b-ii, c-iv, d-i
Positive charge: lines outward; negative: inward; dipole: from + to –; uniform field: parallel, equal spacing.
Two identical small metal spheres carry +6 μC and –2 μC. They are touched together and then placed 0.3 m apart. With k = 9 × 10⁹ N m² C⁻², the force between them is
1. 0.4 N, repulsive
2. 0.4 N, attractive
3. 1.2 N, attractive
4. 1.2 N, repulsive
Answer
A. 0.4 N, repulsive
Each sphere gets (6 – 2)/2 = +2 μC. F = 9 × 10⁹ × 4 × 10⁻¹²/0.09 = 0.4 N, repulsive.
Charges +q and +4q are 0.3 m apart. The point on the line joining them where the net electric field is zero is at a distance from the charge +q of
1. 0.15 m
2. 0.2 m
3. 0.1 m
4. 0.05 m
Answer
C. 0.1 m
kq/x² = 4kq/(0.3 – x)². So 0.3 – x = 2x and x = 0.1 m.
An electron is placed in a uniform field of 910 N C⁻¹. Taking e = 1.6 × 10⁻¹⁹ C and mₑ = 9.1 × 10⁻³¹ kg, its acceleration is
1. 1.6 × 10¹⁵ m s⁻²
2. 1.6 × 10¹³ m s⁻²
3. 1.6 × 10¹⁶ m s⁻²
4. 1.6 × 10¹⁴ m s⁻²
Answer
D. 1.6 × 10¹⁴ m s⁻²
a = eE/m = 1.6 × 10⁻¹⁹ × 910/9.1 × 10⁻³¹ = 1.6 × 10¹⁴ m s⁻².
A dipole of moment 2 × 10⁻⁸ C m is at 60° to a uniform field of 5 × 10⁴ N C⁻¹. The work done by the field in turning it to align with the field is
1. 5 × 10⁻⁴ J
2. 1 × 10⁻³ J
3. 1.5 × 10⁻³ J
4. 8.7 × 10⁻⁴ J
Answer
A. 5 × 10⁻⁴ J
Work by field = U_i – U_f = –pE cos 60° + pE = pE/2 = 10⁻³/2 = 5 × 10⁻⁴ J.
A point charge of 5.31 μC is at the centre of a cube. With ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻², the flux through one face of the cube is
1. 6 × 10⁵ N m² C⁻¹
2. 1.5 × 10⁵ N m² C⁻¹
3. 1 × 10⁵ N m² C⁻¹
4. 3 × 10⁵ N m² C⁻¹
Answer
C. 1 × 10⁵ N m² C⁻¹
Total flux = 5.31 × 10⁻⁶/8.85 × 10⁻¹² = 6 × 10⁵. By symmetry one face gets one sixth, 1 × 10⁵.
A thin spherical shell of radius 0.1 m carries 4 μC. The field just outside its surface is (k = 9 × 10⁹ N m² C⁻²)
1. zero
2. 3.6 × 10⁶ N C⁻¹
3. 1.8 × 10⁶ N C⁻¹
4. 3.6 × 10⁵ N C⁻¹
Answer
B. 3.6 × 10⁶ N C⁻¹
E = kq/R² = 9 × 10⁹ × 4 × 10⁻⁶/0.01 = 3.6 × 10⁶ N C⁻¹.
Two large parallel plates carry equal and opposite uniform surface charge densities of magnitude 8.85 × 10⁻⁸ C m⁻². The field between the plates is (ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻²)
1. 2 × 10⁴ N C⁻¹
2. zero
3. 1 × 10⁴ N C⁻¹
4. 5 × 10³ N C⁻¹
Answer
C. 1 × 10⁴ N C⁻¹
Fields of the two sheets add between the plates: σ/ε₀ = 8.85 × 10⁻⁸/8.85 × 10⁻¹² = 1 × 10⁴ N C⁻¹.

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1.The unit of electric charge is?
2.Who invented the electric bulb?
3.What is the unit of electric resistance?
4.How many players are in a Kabaddi team on the field?
5.The Arjuna Award is given in which field?
6.The Pulitzer Prize is associated with which field?
7.The 'SATH-E' project is associated with which of the following fields? (RRB NTPC CBT-I (30/12/2020) Shift 1)
8.The magnetic field inside a long straight solenoid carrying current is _____________.
9.The method of file organization in which data records in a file are arranged in a specified order according to a key field is known as the
10.The Hornbill Festival, also known as 'the Festival of Festivals', is celebrated in which Indian state? (RRB ALP CBT 2 (Electrician)(22/01/2019) Shift 3)

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