Class 12 · Physics · Part 1

Chapter 1: Electric Charges and Fields

Revision notes, practice numericals, a question bank, assertion and reason questions and answers to the NCERT questions. Tap a group to open it; opening one closes the others.

1.1Electric chargeCharging by rubbing, two kinds of charge, the electroscope

Charging by rubbing

When a glass rod is rubbed with silk, or a plastic comb is rubbed with dry hair, the objects begin to attract small bits of paper. They are said to be electrified or charged. The word comes from the Greek elektron for amber, which shows this effect when rubbed.

Two kinds of charge

There are only two kinds of charge. Benjamin Franklin named them positive and negative.
Like charges repel and unlike charges attract each other.
A glass rod rubbed with silk becomes positive and the silk becomes negative. A plastic rod rubbed with fur becomes negative.
Electrification by rubbing happens because electrons are transferred from one body to the other. The body that loses electrons becomes positive and the body that gains them becomes negative.
A body with equal positive and negative charge is neutral. Neutral does not mean there is no charge in it; it means the total is zero.

Like charges repel each other and unlike charges attract each other. The arrows show the force on each body.

Gold-leaf electroscope

An electroscope detects charge. When a charged object touches the knob, the charge spreads to the metal rod and the two gold leaves. The leaves carry like charge, so they repel and spread apart. The more charge, the wider the leaves open.

Gold-leaf electroscope. When a charged object touches the knob, charge flows to the leaves and they diverge; a bigger spread means more charge.

Unit of charge

The SI unit of charge is the coulomb (C).
One coulomb is the charge that flows in one second through a wire carrying a current of one ampere: 1 C = 1 A × 1 s.
Common smaller units: 1 mC = 10⁻³ C, 1 μC = 10⁻⁶ C, 1 nC = 10⁻⁹ C.

1.2Conductors and insulatorsHow charge moves, earthing and charging by induction

Conductors and insulators

Conductors such as metals, the human body and the earth allow charge to move through them easily, because some electrons are free to move.
Insulators such as glass, plastic, rubber and dry wood do not allow charge to flow easily. Charge placed on them stays where it is.
When a charged body is connected to the earth by a conductor, its extra charge flows into the earth. This is called earthing (grounding).

Charging by conduction

When a charged conducting body touches an uncharged one, charge is shared between them. Both end up with charge of the same sign. Two identical spheres share the total charge equally.

Charging by induction

Bring a charged rod near (but not touching) a neutral metal sphere on an insulating stand. The free electrons of the sphere shift: the near side gets the opposite charge and the far side gets the same charge as the rod.
Touch the far side with a finger (earth it). The same-sign charge flows to the earth.
Remove the finger first and then the rod. The sphere is left with charge opposite to that of the rod.
Here the rod loses no charge, and the sphere acquires charge of the opposite sign.

Why charged bodies attract paper

A charged comb induces opposite charge on the near side of a bit of paper. The attraction to the near, opposite charge is stronger than the repulsion from the far, same charge, so the paper is pulled towards the comb.

1.3Basic properties of electric chargeAdditivity, conservation and quantisation

Additivity of charge

Charge is a scalar. The total charge of a system is the algebraic sum of the charges in it, with their signs. For example, +5 μC and −3 μC together make +2 μC.

Conservation of charge

The total charge of an isolated system never changes. Charge can be transferred from one body to another but it is neither created nor destroyed.
In rubbing, the charge gained by one body equals the charge lost by the other.
Even in processes such as the creation of an electron–positron pair, the total charge stays the same.

Quantisation of charge

Charge always comes in whole-number multiples of a basic unit, the charge of an electron: q = ± n e, where n = 1, 2, 3, ….
The elementary charge is e = 1.6 × 10⁻¹⁹ C. An electron has charge −e and a proton has charge +e.
Charge was first measured by Millikan in his oil-drop experiment.
In everyday situations we deal with charges of millions of e, so the steps are too small to notice and charge appears to change smoothly. At the microscopic level the quantisation is important.
To find how many electrons make a charge: n = q / e.

1.4Coulomb's lawThe force between two point charges

Statement

Coulomb's law: the force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. It acts along the line joining the charges.
In magnitude: F = k q₁ q₂ / r²
The constant is k = 1 / (4π ε₀) = 9 × 10⁹ N m² C⁻², where ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻² is the permittivity of free space.
A point charge is a charged body whose size is very small compared with the distance to other bodies.

Coulomb force between two point charges. The two forces are equal in magnitude and opposite in direction, along the line joining the charges.

Direction and vector form

If q₁q₂ > 0 (same sign) the force is repulsive; if q₁q₂ < 0 (opposite sign) it is attractive.
The forces on the two charges are equal and opposite (Newton's third law): F₁₂ = −F₂₁.
In vector form, the force on charge 2 due to charge 1 is F₂₁ = k q₁q₂ r̂₂₁ / r², where r̂₂₁ is the unit vector from 1 to 2.

Coulomb force and gravity

Both forces follow an inverse-square law and act along the line joining the bodies.
Gravity is always attractive, but the electric force can attract or repel.
The electric force is far stronger. For an electron and a proton, F_electric / F_gravity ≈ 2.4 × 10³⁹.
Charge does not depend on speed, and the electric force between charges does not depend on whether other charges are nearby (see superposition).

1.5Forces between multiple chargesThe principle of superposition

Principle of superposition

The force on a charge due to several other charges is the vector sum of the forces due to each of them, found as if the others were absent.
For a charge q₁ with other charges q₂, q₃, …: F₁ = F₁₂ + F₁₃ + …
Each force is found with Coulomb's law. The forces must be added as vectors, using components.

Principle of superposition: the force on q₁ is the vector sum of F₁₂ and F₁₃, each found as if the other charge were absent.

How to solve

Draw the charges and show the direction of each force (away for like charges, towards for unlike charges).
Find each magnitude with F = k q₁q₂ / r².
Resolve into x and y components, add them, and then find the resultant: F = √(Fₓ² + F_y²).
On a straight line, simply add the forces with a sign for direction.

1.6Electric fieldField of a point charge and of a system

Definition

A charge Q changes the space around it. The charge sets up an electric field, and any other charge in the region feels a force because of it.
The electric field at a point is the force per unit positive test charge: E = F / q₀, taking a very small q₀ so that it does not disturb the source charges.
SI unit: newton per coulomb (N C⁻¹). It is a vector quantity.
A charge q placed in a field E feels the force F = qE. The force is along E for a positive charge and opposite to E for a negative charge.

Field of a point charge

At distance r from a point charge Q: E = k Q / r² = Q / (4π ε₀ r²).
The direction is radially outward for a positive charge and radially inward for a negative charge.
The field depends only on the distance, so it is the same size at all points of a sphere around the charge.

The electric field of a point charge is radial. The lines spread out as we go away from the charge, so the field becomes weaker.

Field due to a system of charges

By superposition, the field at a point is the vector sum of the fields due to the individual charges: E = E₁ + E₂ + E₃ + …
The field of a point charge is a vector field: it has a size and a direction at every point of space.
The field exists whether or not a test charge is placed there.

1.7Electric field linesPicturing the field

What field lines show

A field line is a curve whose tangent at any point gives the direction of the field at that point.
The closeness of lines shows the strength of the field: crowded lines mean a strong field, widely spaced lines mean a weak field.
Field lines are a way to picture the field; they are imaginary, not real threads.

Field lines of two equal and opposite charges (a dipole). Lines start on +q and end on −q and never cross.

Properties of field lines

Field lines start on a positive charge and end on a negative charge (or go to infinity). They do not start or stop in the middle of empty space.
Two field lines never cross: if they did, the field would have two directions at the crossing point.
The lines of a static electric field do not form closed loops, unlike magnetic field lines.
The lines are continuous curves without breaks.
The tangent to a line at a point gives the direction of the force on a positive charge placed there.
Where the lines are closer the field is stronger. Around a point charge the lines spread out, so the field gets weaker with distance.

1.8Electric fluxHow much field passes through a surface

Definition

For a small flat area ΔS in a field E, the electric flux is Δφ = E ΔS cos θ, where θ is the angle between E and the normal to the surface.
Flux is a scalar. Its SI unit is N m² C⁻¹ (or V m).
It is largest when the surface faces the field (θ = 0°) and zero when the field lies along the surface (θ = 90°).
For a closed surface the normal is taken to point outward. Field lines going out give positive flux; lines coming in give negative flux.
The total flux through a closed surface is the sum of the flux through all its small parts.

The flux through a tilted area element depends on the angle θ between E and the normal n̂. It is largest at θ = 0 and zero at θ = 90°.

Flux through a closed surface

If as many field lines enter a closed surface as leave it, the net flux is zero. This is the case when the surface contains no net charge. If more lines leave than enter, the surface encloses a net positive charge.

1.9Electric dipoleDipole moment, its field and the torque on it

Dipole and dipole moment

An electric dipole is a pair of equal and opposite charges +q and −q separated by a small distance 2a.
The dipole moment is p = q × 2a. It is a vector directed from −q to +q. Its SI unit is the coulomb metre (C m).
The total charge of a dipole is zero, but its field is not zero.
Many molecules are dipoles. Polar molecules such as H₂O and HCl have a permanent dipole moment; non-polar molecules such as O₂ and CO₂ do not.

Field of a dipole

On the axis (the line through both charges) at a distance r from the centre, for r ≫ a: E = 2 k p / r³, in the direction of p.
On the equatorial line (the perpendicular bisector) at distance r, for r ≫ a: E = k p / r³, in the direction opposite to p.
So the axial field is twice the equatorial field at the same distance.
A dipole field falls as 1/r³, faster than the 1/r² of a single charge, because the fields of the two charges partly cancel.

Dipole field at large distance: along the axis E points the same way as p (E = 2p / 4πε₀r³); on the equatorial line E points opposite to p (E = p / 4πε₀r³ in magnitude).

Dipole in a uniform field

The forces +qE and −qE on the two charges are equal and opposite, so the net force is zero.
They act at different points, so they form a couple. The torque is τ = p × E, with magnitude τ = p E sin θ.
The torque turns the dipole so that p lines up with E. It is greatest at θ = 90° (τ = pE) and zero at θ = 0° or 180°.
In a non-uniform field there is also a net force on the dipole.

In a uniform field the two forces are equal and opposite, so the net force is zero, but they act at different points and produce a torque that turns p towards E.

1.10Continuous charge distributionLine, surface and volume charge densities

Charge densities

When charge is spread over a body, we describe it by a density.
Linear charge density λ = Δq / Δl (unit C m⁻¹), for charge on a wire.
Surface charge density σ = Δq / ΔS (unit C m⁻²), for charge on a surface.
Volume charge density ρ = Δq / ΔV (unit C m⁻³), for charge in a volume.

Field of a continuous distribution

Divide the body into small elements of charge Δq, treat each as a point charge, and add their fields as vectors. When the elements are made very small, the sum becomes an integral. For a uniformly charged body, q = λ l, q = σ A or q = ρ V.

1.11Gauss's lawFlux and enclosed charge

Statement

Gauss's law: the total electric flux through a closed surface equals 1/ε₀ times the net charge enclosed by it: φ = q_enclosed / ε₀.
The closed surface used is called a Gaussian surface. It is an imaginary surface chosen for convenience.
Only the charge inside contributes to the flux. A charge outside sends as many lines in as out, so its net flux is zero. But E at a point on the surface is due to all charges, inside and outside.
The flux does not depend on the shape or the size of the Gaussian surface, only on the enclosed charge.

A sphere of radius r around a point charge. The field is the same size at every point of the sphere and is along the outward normal, so the flux is simply E × 4πr².

Gauss's law and Coulomb's law

For a point charge q at the centre of a sphere of radius r, E is the same everywhere on the sphere and is along the outward normal. So E × 4π r² = q / ε₀, which gives E = q / (4π ε₀ r²), exactly Coulomb's law.

Choosing a Gaussian surface

Gauss's law is easy to use when the charge distribution is symmetric. Choose a surface on which E has a constant magnitude and is perpendicular to the surface (or parallel, where it gives no flux).

1.12Applications of Gauss's lawWire, sheet and spherical shell

Infinitely long straight charged wire

Take a coaxial cylinder of radius r and length l. The field is radial, so the flux goes only through the curved surface of area 2π r l.
E × 2π r l = λ l / ε₀, so E = λ / (2π ε₀ r), directed away from the wire if λ > 0.
The field falls as 1/r.

Gaussian cylinder around an infinitely long charged wire. Only the curved surface (area 2πrl) contributes to the flux; the two ends do not.

Infinite thin plane sheet

Take a small box straddling the sheet. The flux passes through the two faces parallel to the sheet, each of area A.
2 E A = σ A / ε₀, so E = σ / (2 ε₀), directed away from a positively charged sheet.
The field is uniform: it does not depend on the distance from the sheet.

Gaussian box through an infinite charged sheet. Only the two faces parallel to the sheet carry flux, so 2EA = σA / ε₀ and E = σ / 2ε₀.

Uniformly charged thin spherical shell

Outside (r > R): the shell acts as if all its charge q were at the centre: E = q / (4π ε₀ r²).
On the surface (r = R): E = q / (4π ε₀ R²).
Inside (r < R): a Gaussian sphere encloses no charge, so E = 0.
The surface charge density is σ = q / (4π R²).

Solid circle: the charged shell of radius R. Dashed circle: the Gaussian sphere through P. Outside, the enclosed charge is q; inside it is zero.

Quick referenceFormulas at a glanceRevise before the exam

Key results

Quantisation: q = n e, e = 1.6 × 10⁻¹⁹ C
Coulomb's law: F = k q₁q₂ / r², k = 9 × 10⁹ N m² C⁻²
Field of a point charge: E = k Q / r²; force F = qE
Flux: φ = E S cos θ; Gauss's law: φ = q / ε₀
Dipole moment: p = 2 a q; axial E = 2kp / r³; equatorial E = kp / r³
Torque on a dipole: τ = p E sin θ
Infinite wire: E = λ / (2π ε₀ r); infinite sheet: E = σ / (2 ε₀)
Thin shell: E = kq / r² outside, 0 inside

Common mistakes

Adding forces or fields as numbers instead of as vectors.
Using the distance in cm in k q₁q₂ / r²: always convert to metres.
Forgetting that Gauss's law counts only the enclosed charge for flux.
Writing the dipole field direction wrongly: along p on the axis and opposite to p on the equator.

Try each problem first, then open the solution to check your working.

1.1Coulomb's law and chargeQuantisation, force and sharing of charge · 8 problems

Q1Counting electrons

How many electrons must be removed from a neutral body to give it a charge of +4.8 μC?

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Given: q = 4.8 μC = 4.8 × 10⁻⁶ C; e = 1.6 × 10⁻¹⁹ C

Use: q = n e

n = q / e = (4.8 × 10⁻⁶) / (1.6 × 10⁻¹⁹) = 3 × 10¹³

Answer: 3 × 10¹³ electrons

Q2Force between two charges

Two point charges of +4 μC and +6 μC are 20 cm apart in air. Find the force between them.

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Given: q₁ = 4 × 10⁻⁶ C; q₂ = 6 × 10⁻⁶ C; r = 0.2 m

Use: F = k q₁q₂ / r²

F = (9 × 10⁹ × 4 × 10⁻⁶ × 6 × 10⁻⁶) / (0.2)² = 0.216 / 0.04 = 5.4 N

Both charges are positive, so the force is repulsive.

Answer: 5.4 N, repulsive

Q3Finding the distance

Two equal charges of 2 μC repel each other with a force of 0.9 N. How far apart are they?

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Given: q₁ = q₂ = 2 × 10⁻⁶ C; F = 0.9 N

Use: r² = k q₁q₂ / F

r² = (9 × 10⁹ × 4 × 10⁻¹²) / 0.9 = 0.036 / 0.9 = 0.04 m²

r = √0.04 = 0.2 m

Answer: r = 0.2 m (20 cm)

Q4Touching spheres

Two identical small metal spheres carry +8 μC and −2 μC and are 10 cm apart. (a) Find the force between them. (b) They are touched together and placed 10 cm apart again. Find the new force.

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Given: q₁ = 8 μC; q₂ = −2 μC; r = 0.1 m

Use: F = k q₁q₂ / r²; identical spheres share total charge equally

(a) F = (9 × 10⁹ × 8 × 10⁻⁶ × 2 × 10⁻⁶) / (0.1)² = 0.144 / 0.01 = 14.4 N (attractive)

(b) Total charge = 8 − 2 = +6 μC, so each sphere has +3 μC.

F′ = (9 × 10⁹ × 3 × 10⁻⁶ × 3 × 10⁻⁶) / (0.1)² = 0.081 / 0.01 = 8.1 N (repulsive)

Answer: (a) 14.4 N attractive; (b) 8.1 N repulsive

Q5Point of zero force

Charges +4 μC and +9 μC are 50 cm apart. At what point on the line joining them would a third charge feel no net force?

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Given: q₁ = 4 μC; q₂ = 9 μC; d = 0.5 m

Use: The point lies between the charges, where the two forces are equal and opposite: q₁ / x² = q₂ / (d − x)²

Let x be the distance from the 4 μC charge. Then 4 / x² = 9 / (0.5 − x)²

Taking square roots: 2 / x = 3 / (0.5 − x), so 2(0.5 − x) = 3x

1 − 2x = 3x, so x = 0.2 m

Answer: 0.2 m from the 4 μC charge (0.3 m from the 9 μC charge)

Q6Electric force versus gravity

Compare the electric repulsion with the gravitational attraction between two electrons. (G = 6.67 × 10⁻¹¹ N m² kg⁻², mₑ = 9.11 × 10⁻³¹ kg)

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Given: e = 1.6 × 10⁻¹⁹ C; mₑ = 9.11 × 10⁻³¹ kg

Use: F_e / F_g = k e² / (G mₑ²) (the distance cancels)

Numerator: 9 × 10⁹ × (1.6 × 10⁻¹⁹)² = 2.30 × 10⁻²⁸

Denominator: 6.67 × 10⁻¹¹ × (9.11 × 10⁻³¹)² = 5.54 × 10⁻⁷¹

Ratio = 2.30 × 10⁻²⁸ / 5.54 × 10⁻⁷¹ ≈ 4.2 × 10⁴²

Answer: F_e / F_g ≈ 4.2 × 10⁴²; gravity is completely negligible

Q7Changing charge and distance

Two point charges exert a force of 12 N on each other. What is the force if one charge is doubled and the distance is halved?

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Given: F₁ = 12 N; q₁′ = 2q₁; r′ = r / 2

Use: F ∝ q₁q₂ / r²

F₂ / F₁ = (2) × (1 / (½)²) = 2 × 4 = 8

F₂ = 8 × 12 = 96 N

Answer: 96 N

Q8Sharing charge among three spheres

Three identical metal spheres A, B and C stand on insulating stands. A carries 8 μC, and B and C are neutral. A touches B and is removed. Then A touches C and is removed. Find the final charge on each.

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Given: Q_A = 8 μC; Q_B = Q_C = 0

Use: Identical spheres share their total charge equally when they touch

A touches B: each has 8 / 2 = 4 μC. So A = 4 μC, B = 4 μC.

A (4 μC) touches C (0): each has 2 μC. So A = 2 μC, C = 2 μC.

Check: 2 + 4 + 2 = 8 μC, the original total (charge is conserved).

Answer: A = 2 μC, B = 4 μC, C = 2 μC

1.2Superposition of forcesVector addition of Coulomb forces · 4 problems

Q9Three charges in a line

Charges +2 μC, −3 μC and +4 μC lie in a line as shown. The first two are 10 cm apart and the last two are 20 cm apart. Find the net force on the −3 μC charge.

Three charges in a line.

Show solutionHide solution

Given: q₁ = +2 μC; q₂ = −3 μC; q₃ = +4 μC; r₁₂ = 0.1 m; r₂₃ = 0.2 m

Use: F = k q q′ / r², then add the forces with signs

Force from +2 μC (attractive, towards it): F = (9 × 10⁹ × 2 × 10⁻⁶ × 3 × 10⁻⁶) / (0.1)² = 5.4 N, towards the left.

Force from +4 μC (attractive, towards it): F = (9 × 10⁹ × 4 × 10⁻⁶ × 3 × 10⁻⁶) / (0.2)² = 2.7 N, towards the right.

Net force = 5.4 − 2.7 = 2.7 N, towards the left.

Answer: 2.7 N towards the +2 μC charge

Q10Charges on the axes

A +1 μC charge is at the origin, a +3 μC charge at x = 0.3 m and a +4 μC charge at y = 0.4 m. Find the force on the +1 μC charge.

Charges placed on the axes.

Show solutionHide solution

Given: q = 1 μC at the origin; 3 μC at (0.3 m, 0); 4 μC at (0, 0.4 m)

Use: Add the two forces as vectors (they are at right angles)

Force from 3 μC: F₁ = (9 × 10⁹ × 1 × 10⁻⁶ × 3 × 10⁻⁶) / (0.3)² = 0.3 N, repulsive, so along −x.

Force from 4 μC: F₂ = (9 × 10⁹ × 1 × 10⁻⁶ × 4 × 10⁻⁶) / (0.4)² = 0.225 N, repulsive, so along −y.

F = √(0.3² + 0.225²) = √0.140625 = 0.375 N

tan α = 0.225 / 0.3 = 0.75, so α ≈ 37° from the −x axis, towards −y.

Answer: 0.375 N, at 37° below the −x axis

Q11Charges at the corners of a square

Four equal charges of +2 μC are at the corners of a square of side 10 cm. Find the force on the charge at one corner.

Four equal charges at the corners of a square.

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Given: q = 2 × 10⁻⁶ C; side = 0.1 m; diagonal = 0.1√2 m

Use: Two neighbours at distance a, one opposite corner at distance a√2

Force from each neighbour: F = (9 × 10⁹ × 4 × 10⁻¹²) / (0.1)² = 3.6 N. The two are at 90°, so their resultant is 3.6√2 = 5.09 N along the diagonal, away from the centre.

Force from the opposite corner: r² = 0.02 m², so F = 0.036 / 0.02 = 1.8 N, also along the diagonal, away from the centre.

Total = 5.09 + 1.8 ≈ 6.9 N

Answer: ≈ 6.9 N, along the diagonal away from the centre

Q12Equilateral triangle

Three charges of +1 μC each are at the corners of an equilateral triangle of side 10 cm. Find the force on any one of them.

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Given: q = 1 × 10⁻⁶ C; a = 0.1 m

Use: Two equal forces F at 60° to each other: R = 2F cos 30° = F√3

F = (9 × 10⁹ × 10⁻¹²) / (0.1)² = 0.9 N from each of the other two.

R = 0.9 × √3 = 1.56 N, directed away from the centre of the triangle.

Answer: ≈ 1.56 N, directed away from the centre

1.3Electric fieldPoint charges and systems of charges · 6 problems

Q13Field of a point charge

Find the electric field at a distance of 30 cm from a point charge of +5 μC.

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Given: Q = 5 × 10⁻⁶ C; r = 0.3 m

Use: E = k Q / r²

E = (9 × 10⁹ × 5 × 10⁻⁶) / (0.3)² = 4.5 × 10⁴ / 0.09 = 5 × 10⁵ N/C

Answer: 5 × 10⁵ N/C, directed away from the charge

Q14Midpoint of a pair

Charges +4 μC and −4 μC are 10 cm apart. Find the field at the midpoint of the line joining them.

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Given: q = ±4 μC; distance of the midpoint from each = 0.05 m

Use: Both fields point the same way at the midpoint: away from + and towards −

Each field: E = (9 × 10⁹ × 4 × 10⁻⁶) / (0.05)² = 3.6 × 10⁴ / 0.0025 = 1.44 × 10⁷ N/C

Total E = 2 × 1.44 × 10⁷ = 2.88 × 10⁷ N/C

Answer: 2.88 × 10⁷ N/C, directed from the +4 μC towards the −4 μC charge

Q15An electron in a field

An electron is placed in a uniform field of 2000 N/C. Find the force on it and its acceleration. (mₑ = 9.11 × 10⁻³¹ kg)

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Given: E = 2000 N/C; e = 1.6 × 10⁻¹⁹ C

Use: F = qE and a = F / m

F = 1.6 × 10⁻¹⁹ × 2000 = 3.2 × 10⁻¹⁶ N

a = 3.2 × 10⁻¹⁶ / 9.11 × 10⁻³¹ = 3.51 × 10¹⁴ m/s²

The force is opposite to the field, because the charge is negative.

Answer: F = 3.2 × 10⁻¹⁶ N opposite to E; a ≈ 3.51 × 10¹⁴ m/s²

Q16Finding the source charge

At a distance of 2 m from a point charge the field is 900 N/C. Find the charge and the field at 6 m.

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Given: E₁ = 900 N/C; r₁ = 2 m; r₂ = 6 m

Use: E = kQ / r² so Q = E r² / k

Q = (900 × 2²) / (9 × 10⁹) = 3600 / 9 × 10⁹ = 4 × 10⁻⁷ C

E ∝ 1/r², so E₂ = 900 × (2/6)² = 900 / 9 = 100 N/C

Answer: Q = 4 × 10⁻⁷ C (0.4 μC); E at 6 m = 100 N/C

Q17Where the field is zero

Charges +4 μC and +1 μC are 30 cm apart. At what point on the line joining them is the electric field zero?

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Given: q₁ = 4 μC; q₂ = 1 μC; d = 0.3 m

Use: Between like charges the fields oppose: k q₁ / x² = k q₂ / (d − x)²

4 / x² = 1 / (0.3 − x)²

Square roots: 2 / x = 1 / (0.3 − x), so 2(0.3 − x) = x

0.6 = 3x, so x = 0.2 m

Answer: 0.2 m from the 4 μC charge (0.1 m from the 1 μC charge)

Q18Field at the vertex of a triangle

Charges +2 μC and −2 μC are at two corners of an equilateral triangle of side 10 cm. Find the field at the third corner.

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Given: q = ±2 μC; r = 0.1 m for each

Use: Find each field, then add as vectors

Each field: E = (9 × 10⁹ × 2 × 10⁻⁶) / (0.1)² = 1.8 × 10⁶ N/C.

The field of +2 μC points away from it and the field of −2 μC points towards it. These two directions make an angle of 120° with each other.

For two equal vectors at 120°, the resultant equals each vector. So E = 1.8 × 10⁶ N/C, parallel to the line from +2 μC to −2 μC.

Answer: 1.8 × 10⁶ N/C, parallel to the line from the + charge to the − charge

1.4Electric dipoleDipole moment, field and torque · 5 problems

Q19Dipole moment

A dipole has charges +5 μC and −5 μC separated by 2 cm. Find its dipole moment.

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Given: q = 5 × 10⁻⁶ C; 2a = 0.02 m

Use: p = q × 2a

p = 5 × 10⁻⁶ × 0.02 = 1 × 10⁻⁷ C m

Answer: 1 × 10⁻⁷ C m, directed from the negative to the positive charge

Q20Axial and equatorial fields

A short dipole has p = 5 × 10⁻⁸ C m. Find its field at a point 0.2 m from the centre (a) on the axis and (b) on the equatorial line.

Point P on the axis of a short dipole.

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Given: p = 5 × 10⁻⁸ C m; r = 0.2 m

Use: E_axis = 2kp / r³; E_eq = kp / r³ (for r much larger than the dipole size)

kp / r³ = (9 × 10⁹ × 5 × 10⁻⁸) / (0.2)³ = 450 / 0.008 = 5.625 × 10⁴ N/C

(a) E_axis = 2 × 5.625 × 10⁴ = 1.125 × 10⁵ N/C, in the direction of p.

(b) E_eq = 5.625 × 10⁴ N/C ≈ 5.6 × 10⁴ N/C, opposite to p.

Answer: (a) 1.125 × 10⁵ N/C along p; (b) ≈ 5.6 × 10⁴ N/C opposite to p

Q21Torque on a dipole

A dipole of moment 2 × 10⁻⁸ C m is in a uniform field of 3 × 10⁴ N/C and makes an angle of 30° with the field. Find the torque and the net force on it.

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Given: p = 2 × 10⁻⁸ C m; E = 3 × 10⁴ N/C; θ = 30°

Use: τ = p E sin θ

τ = 2 × 10⁻⁸ × 3 × 10⁴ × sin 30° = 6 × 10⁻⁴ × 0.5 = 3 × 10⁻⁴ N m

The forces on the two charges are equal and opposite, so the net force is zero.

Answer: τ = 3 × 10⁻⁴ N m; net force = 0

Q22Maximum torque

A dipole has charges ±2 μC separated by 1 cm. It is placed in a field of 10⁵ N/C. Find the maximum torque on it and the angle at which it occurs.

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Given: q = 2 × 10⁻⁶ C; 2a = 0.01 m; E = 10⁵ N/C

Use: p = q × 2a; τ_max = pE at 90°

p = 2 × 10⁻⁶ × 0.01 = 2 × 10⁻⁸ C m

τ_max = 2 × 10⁻⁸ × 10⁵ = 2 × 10⁻³ N m

Answer: 2 × 10⁻³ N m, when the dipole is at 90° to the field

Q23Finding the angle

A dipole of moment 4 × 10⁻⁹ C m in a field of 5 × 10⁴ N/C experiences a torque of 10⁻⁴ N m. Find the angle between p and E.

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Given: p = 4 × 10⁻⁹ C m; E = 5 × 10⁴ N/C; τ = 10⁻⁴ N m

Use: sin θ = τ / (pE)

pE = 4 × 10⁻⁹ × 5 × 10⁴ = 2 × 10⁻⁴ N m

sin θ = 10⁻⁴ / 2 × 10⁻⁴ = 0.5

θ = 30° (or 150°)

Answer: θ = 30° (or equivalently 150°)

1.5Flux and Gauss's lawFlux, enclosed charge and symmetric distributions · 9 problems

Q24Flux through a flat area

A uniform field of 500 N/C passes through a flat area of 0.2 m². Find the flux when the normal to the area makes an angle of (a) 0°, (b) 60°, (c) 90° with the field.

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Given: E = 500 N/C; S = 0.2 m²

Use: φ = E S cos θ

ES = 500 × 0.2 = 100 N m²/C

(a) φ = 100 × cos 0° = 100

(b) φ = 100 × cos 60° = 50

Answer: (a) 100, (b) 50, (c) 0 (all in N m²/C)

Q25Charge at the centre of a cube

A charge of 6 nC is placed at the centre of a cube. Find the total flux and the flux through one face. (ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻²)

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Given: q = 6 × 10⁻⁹ C

Use: Gauss's law, φ = q / ε₀; by symmetry the six faces share the flux equally

φ = 6 × 10⁻⁹ / 8.85 × 10⁻¹² = 678 ≈ 6.8 × 10² N m²/C

Flux through one face = 678 / 6 = 113 N m²/C

Answer: Total ≈ 6.8 × 10² N m²/C; each face 113 N m²/C

Q26Net enclosed charge

A closed surface encloses charges of +5 μC and −2 μC. A −4 μC charge lies outside it. Find the net flux through the surface.

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Given: q₁ = +5 μC; q₂ = −2 μC inside; −4 μC outside

Use: Only the enclosed charge counts: φ = q_enc / ε₀

q_enc = 5 − 2 = +3 μC (the outside charge adds nothing)

φ = 3 × 10⁻⁶ / 8.85 × 10⁻¹² = 3.4 × 10⁵ N m²/C

Answer: 3.4 × 10⁵ N m²/C, outward

Q27Charge from flux

The net flux out of a closed surface is 1000 N m²/C. Find the charge enclosed.

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Given: φ = 1000 N m²/C

Use: q = ε₀ φ

q = 8.85 × 10⁻¹² × 1000 = 8.85 × 10⁻⁹ C

Answer: 8.85 × 10⁻⁹ C (about 8.85 nC), positive

Q28Long charged wire

A very long straight wire has a uniform linear charge density of 2 × 10⁻⁷ C/m. Find the field at 0.5 m from it.

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Given: λ = 2 × 10⁻⁷ C/m; r = 0.5 m

Use: E = λ / (2π ε₀ r) = 2kλ / r

E = (2 × 9 × 10⁹ × 2 × 10⁻⁷) / 0.5 = 3600 / 0.5 = 7200 N/C

Answer: 7200 N/C, radially away from the wire

Q29Infinite sheet

An infinite plane sheet carries a surface charge density of 5 μC/m². Find the field near it.

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Given: σ = 5 × 10⁻⁶ C/m²

Use: E = σ / (2 ε₀)

E = 5 × 10⁻⁶ / (2 × 8.85 × 10⁻¹²) = 5 × 10⁻⁶ / 1.77 × 10⁻¹¹ = 2.8 × 10⁵ N/C

It is the same at every distance from the sheet.

Answer: 2.8 × 10⁵ N/C, perpendicular to the sheet and pointing away

Q30A charged spherical shell

A thin spherical shell of radius 10 cm carries a charge of +4 μC spread uniformly. Find the field (a) 20 cm from the centre, (b) at the surface and (c) inside.

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Given: q = 4 × 10⁻⁶ C; R = 0.1 m

Use: Outside and on the surface: E = kq / r²; inside: Gauss's law with no enclosed charge

(a) r = 0.2 m: E = (9 × 10⁹ × 4 × 10⁻⁶) / (0.2)² = 3.6 × 10⁴ / 0.04 = 9 × 10⁵ N/C

(b) r = 0.1 m: E = 3.6 × 10⁴ / 0.01 = 3.6 × 10⁶ N/C

Answer: (a) 9 × 10⁵ N/C; (b) 3.6 × 10⁶ N/C; (c) 0

Q31Surface charge density of the shell

Find the surface charge density of the shell in the previous problem.

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Given: q = 4 × 10⁻⁶ C; R = 0.1 m

Use: σ = q / (4π R²)

Area = 4π × (0.1)² = 0.1257 m²

σ = 4 × 10⁻⁶ / 0.1257 = 3.2 × 10⁻⁵ C/m²

Answer: 3.2 × 10⁻⁵ C/m²

Q32Flux through a cylinder

A cylinder of length 2 m is coaxial with a long wire of λ = 2 × 10⁻⁷ C/m. Find the flux through the cylinder, whatever its radius.

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Given: λ = 2 × 10⁻⁷ C/m; l = 2 m

Use: Gauss's law: φ = λ l / ε₀

Enclosed charge = λ l = 4 × 10⁻⁷ C

φ = 4 × 10⁻⁷ / 8.85 × 10⁻¹² = 4.5 × 10⁴ N m²/C

Answer: 4.5 × 10⁴ N m²/C; it does not depend on the radius

Answers to every exercise question of the NCERT chapter. Try each one first, then tap Show answer.

1.1Exercises 1.1 to 1.8Coulomb's law, quantisation, field lines and fields · 8 questions

Q11.1 What is the force between two small charged spheres having charges of 2 × 10⁻⁷ C and 3 × 10⁻⁷ C placed 30 cm apart in air?

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F = k q₁q₂ / r² = (9 × 10⁹ × 2 × 10⁻⁷ × 3 × 10⁻⁷) / (0.3)² = 5.4 × 10⁻⁴ / 0.09

F = 6 × 10⁻³ N. Both charges are positive, so the force is repulsive.

Q21.2 The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge −0.8 μC in air is 0.2 N. (a) What is the distance between the two spheres? (b) What is the force on the second sphere due to the first?

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(a) r² = k q₁q₂ / F = (9 × 10⁹ × 0.4 × 10⁻⁶ × 0.8 × 10⁻⁶) / 0.2 = 2.88 × 10⁻³ / 0.2 = 1.44 × 10⁻² m²

r = 0.12 m (12 cm).

(b) By Newton's third law the force on the second sphere has the same size, 0.2 N, and the opposite direction. The charges are of opposite signs, so the force is attractive: each sphere is pulled towards the other.

Q31.3 Check that the ratio ke²/(G mₑ mₚ) is dimensionless. Look up a table of physical constants and determine the value of this ratio. What does the ratio signify?

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Dimensions. From Coulomb's law, k e² is a force × distance², so its unit is N m². From Newton's law of gravitation, G mₑ mₚ is also a force × distance², unit N m² (G is in N m² kg⁻² and the masses in kg). The two have the same unit, so the ratio has no dimensions.

Value. With e = 1.6 × 10⁻¹⁹ C, G = 6.67 × 10⁻¹¹ N m² kg⁻², mₑ = 9.11 × 10⁻³¹ kg and mₚ = 1.67 × 10⁻²⁷ kg:

ke² = 9 × 10⁹ × (1.6 × 10⁻¹⁹)² = 2.3 × 10⁻²⁸ N m²; G mₑ mₚ = 1.0 × 10⁻⁶⁷ N m²

Ratio ≈ 2.3 × 10³⁹ (the NCERT solution quotes about 2.4 × 10³⁹, using slightly different rounding of the constants).

Meaning. It is the ratio of the electrostatic force to the gravitational force between an electron and a proton. It does not depend on the distance. The electric force is about 10³⁹ times stronger.

Q41.4 (a) Explain the meaning of the statement 'electric charge of a body is quantised'. (b) Why can one ignore quantisation of electric charge when dealing with macroscopic, i.e. large-scale, charges?

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(a) The charge on a body is always an integral multiple of a basic unit of charge, the electron charge: q = ± n e with n = 0, 1, 2, … and e = 1.6 × 10⁻¹⁹ C. Fractions of e cannot be transferred, so a body cannot have, for example, 2.5e.

(b) In large-scale situations the charges involved are of the order of μC or mC, which contain about 10¹³ to 10¹⁶ electrons. One electron charge is so tiny compared with this that the steps are not noticeable and the charge appears to vary continuously.

Q51.5 When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge.

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Before rubbing, both bodies are neutral, so the total charge is zero. When they are rubbed, electrons pass from one body to the other. The body that loses electrons becomes positive and the body that gains them becomes negative, and the two charges are equal in size.

The total charge after rubbing is again (+q) + (−q) = 0. Charge is only transferred, not created or destroyed, which agrees with the law of conservation of charge.

Q61.6 Four point charges q_A = 2 μC, q_B = −5 μC, q_C = 2 μC and q_D = −5 μC are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1 μC placed at the centre of the square?

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The centre O is the same distance from every corner (half the diagonal), so charges at opposite corners act in opposite directions along the same line.

A and C are both +2 μC and lie on opposite ends of a diagonal. Their forces on the 1 μC charge are equal in size and opposite in direction, so they cancel.

B and D are both −5 μC and lie on the other diagonal. Their forces are also equal and opposite and cancel.

The net force on the charge at the centre is zero.

Q71.7 (a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not? (b) Explain why two field lines never cross each other at any point.

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(a) A field line shows the direction of the force on a positive test charge that moves from point to point. A charge moving in the field would feel a force at every point, so the path it follows is continuous. A break would mean there is no field (no force) at some points of the path, which is not possible in empty space between charges.

(b) If two lines crossed, there would be two tangents, and therefore two different directions of the field, at the point of crossing. A positive charge placed there would have to move in two directions at once. The field at a point has only one direction, so the lines cannot cross.

Q81.8 Two point charges q_A = 3 μC and q_B = −3 μC are located 20 cm apart in vacuum. (a) What is the electric field at the midpoint O of the line AB joining the two charges? (b) If a negative test charge of magnitude 1.5 × 10⁻⁹ C is placed at this point, what is the force experienced by the test charge?

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(a) O is 0.1 m from each charge. Field due to A: E = kq/r² = (9 × 10⁹ × 3 × 10⁻⁶) / (0.1)² = 2.7 × 10⁶ N/C, directed away from A (towards B).

Field due to B: the same size, 2.7 × 10⁶ N/C, directed towards B (the charge is negative).

The two fields point the same way, so E = 2.7 × 10⁶ + 2.7 × 10⁶ = 5.4 × 10⁶ N/C, directed from A to B.

(b) F = qE = 1.5 × 10⁻⁹ × 5.4 × 10⁶ = 8.1 × 10⁻³ N. The test charge is negative, so the force is opposite to the field, that is directed towards A.

1.2Exercises 1.9 to 1.15Dipoles, charging, forces and flux · 7 questions

Q91.9 A system has two charges q_A = 2.5 × 10⁻⁷ C and q_B = −2.5 × 10⁻⁷ C located at points A: (0, 0, −15 cm) and B: (0, 0, +15 cm), respectively. What are the total charge and electric dipole moment of the system?

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Total charge = 2.5 × 10⁻⁷ + (−2.5 × 10⁻⁷) = 0.

The charges are equal and opposite, so the system is a dipole. The separation is AB = 30 cm = 0.3 m, so p = q × 2a = 2.5 × 10⁻⁷ × 0.3 = 7.5 × 10⁻⁸ C m.

The dipole moment points from the negative charge (at B, +z side) to the positive charge (at A, −z side), that is along the negative z-axis.

Q101.10 An electric dipole with dipole moment 4 × 10⁻⁹ C m is aligned at 30° with the direction of a uniform electric field of magnitude 5 × 10⁴ N C⁻¹. Calculate the magnitude of the torque acting on the dipole.

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τ = p E sin θ = 4 × 10⁻⁹ × 5 × 10⁴ × sin 30° = 2 × 10⁻⁴ × 0.5

τ = 1 × 10⁻⁴ N m

Q111.11 A polythene piece rubbed with wool is found to have a negative charge of 3 × 10⁻⁷ C. (a) Estimate the number of electrons transferred (from which to which?). (b) Is there a transfer of mass from wool to polythene?

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(a) n = q / e = (3 × 10⁻⁷) / (1.6 × 10⁻¹⁹) = 1.9 × 10¹² electrons. The polythene is negative, so it gained electrons: they were transferred from the wool to the polythene.

(b) Yes, but the mass is extremely small. Mass transferred = n × mₑ = 1.9 × 10¹² × 9.1 × 10⁻³¹ kg ≈ 1.7 × 10⁻¹⁸ kg, which is negligible.

Q121.12 (a) Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5 × 10⁻⁷ C? The radii of A and B are negligible compared with the distance of separation. (b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?

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(a) F = kq² / r² = (9 × 10⁹ × (6.5 × 10⁻⁷)²) / (0.5)² = (9 × 10⁹ × 4.225 × 10⁻¹³) / 0.25 = 1.5 × 10⁻² N (repulsive).

(b) The charges are doubled so q₁q₂ becomes 4 times. The distance is halved so 1/r² becomes 4 times. The force is 4 × 4 = 16 times: F′ = 16 × 1.52 × 10⁻² = 0.24 N.

Q131.13 Figure 1.30 shows tracks of three charged particles in a uniform electrostatic field (the upper plate is positive and the lower plate is negative, so the field points downward). Give the signs of the three charges. Which particle has the highest charge to mass ratio?

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The field points from the upper (positive) plate to the lower (negative) plate. A positive charge is pushed along the field (downwards) and a negative charge against it (upwards).

Particles 1 and 2 bend towards the upper, positive plate, so they are negatively charged. Particle 3 bends towards the lower, negative plate, so it is positively charged.

All the particles are in the same field, so the deflection is larger when q/m is larger (acceleration a = qE/m). Particle 3 is deflected the most, so it has the highest charge to mass ratio.

Q141.14 Consider a uniform electric field E = 3 × 10³ î N/C. (a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane? (b) What is the flux through the same square if the normal to its plane makes a 60° angle with the x-axis?

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Area S = (0.1 m)² = 0.01 m².

(a) A plane parallel to the yz plane has its normal along the x-axis, parallel to E, so θ = 0°: φ = ES cos 0° = 3 × 10³ × 0.01 = 30 N m²/C.

(b) θ = 60°: φ = ES cos 60° = 30 × 0.5 = 15 N m²/C.

Q151.15 What is the net flux of the uniform electric field of Exercise 1.14 through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes?

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The net flux is zero.

The field is along x. It goes through the two faces perpendicular to the x-axis: the flux is inward through one face (−ES) and outward through the opposite face (+ES), which cancel. For the other four faces the field is parallel to the surface, so the flux is zero. Also, there is no charge inside the cube, so by Gauss's law φ = 0.

1.3Exercises 1.16 to 1.23Gauss's law and symmetric charge distributions · 8 questions

Q161.16 Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is 8.0 × 10³ N m²/C. (a) What is the net charge inside the box? (b) If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box? Why or why not?

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(a) By Gauss's law, q = ε₀ φ = 8.85 × 10⁻¹² × 8.0 × 10³ = 7.1 × 10⁻⁸ C (about 0.07 μC).

(b) No. Zero net flux means only that the net (algebraic sum of) charge inside is zero. The box could contain equal positive and negative charges, for example a dipole, with no net charge.

Q171.17 A point charge +10 μC is a distance 5 cm directly above the centre of a square of side 10 cm. What is the magnitude of the electric flux through the square? (Hint: think of the square as one face of a cube with edge 10 cm.)

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Place the charge at the centre of a cube of edge 10 cm. The charge is 5 cm from the square face, which is the distance from the centre of the cube to the middle of a face. The square is then one face of that cube.

Total flux = q / ε₀. The six faces share it equally, so the flux through one face is q / 6ε₀ = 10 × 10⁻⁶ / (6 × 8.85 × 10⁻¹²) = 1.9 × 10⁵ N m²/C.

Q181.18 A point charge of 2.0 μC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?

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φ = q / ε₀ = (2.0 × 10⁻⁶) / (8.85 × 10⁻¹²) = 2.3 × 10⁵ N m²/C. The size of the cube does not matter, because only the enclosed charge counts.

Q191.19 A point charge causes an electric flux of −1.0 × 10³ N m²/C to pass through a spherical Gaussian surface of 10.0 cm radius centred on the charge. (a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface? (b) What is the value of the point charge?

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(a) The flux depends only on the enclosed charge, not on the radius. The charge is still inside, so the flux stays −1.0 × 10³ N m²/C.

(b) q = ε₀ φ = 8.85 × 10⁻¹² × (−1.0 × 10³) = −8.85 × 10⁻⁹ C (about −8.9 nC). The charge is negative because the flux is inward.

Q201.20 A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the centre of the sphere is 1.5 × 10³ N/C and points radially inward, what is the net charge on the sphere?

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Outside a charged sphere the field is that of a point charge at its centre: E = k|q| / r². The field points inward, so the charge is negative.

|q| = E r² / k = (1.5 × 10³ × (0.2)²) / (9 × 10⁹) = 60 / (9 × 10⁹) = 6.67 × 10⁻⁹ C

Net charge = −6.7 × 10⁻⁹ C (about −6.7 nC).

Q211.21 A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80.0 μC/m². (a) Find the charge on the sphere. (b) What is the total electric flux leaving the surface of the sphere?

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Radius R = 1.2 m. Surface area = 4πR² = 4π × 1.44 = 18.1 m².

(a) q = σ × area = 80 × 10⁻⁶ × 18.1 = 1.45 × 10⁻³ C.

(b) φ = q / ε₀ = 1.45 × 10⁻³ / 8.85 × 10⁻¹² = 1.6 × 10⁸ N m²/C.

Q221.22 An infinite line charge produces a field of 9 × 10⁴ N/C at a distance of 2 cm. Calculate the linear charge density.

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For a line charge, E = λ / (2π ε₀ r) = 2kλ / r, so λ = E r / (2k).

λ = (9 × 10⁴ × 0.02) / (2 × 9 × 10⁹) = 1800 / (1.8 × 10¹⁰) = 10⁻⁷ C/m.

Q231.23 Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 17.0 × 10⁻²² C/m². What is E: (a) in the outer region of the first plate, (b) in the outer region of the second plate, and (c) between the plates?

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Treat each plate as a large sheet of charge, with density +σ on one inner face and −σ on the other. Each sheet gives a field σ / 2ε₀, directed away from the positive sheet and towards the negative one.

(a) and (b) Outside the plates the two fields are equal and opposite, so E = 0 in both outer regions.

(c) Between the plates the two fields point the same way and add: E = σ / ε₀ = 17.0 × 10⁻²² / 8.85 × 10⁻¹² = 1.9 × 10⁻¹⁰ N/C, directed from the positive plate to the negative plate.