Class 12 Physics · NCERT Chapter 1
~33 min readElectric Charges and Fields Class 12 Notes
Complete, exam-ready notes on electric charge and its properties, Coulomb's law and superposition, electric field and field lines, charge distributions, electric flux and Gauss's law with its applications to sheets, lines and spheres — every NCERT topic with MCQs, 1/2/3-mark questions and solved numericals for CBSE, JEE and NEET.
State Gauss's law in electrostatics.
The net electric flux through any closed surface equals the net charge enclosed divided by ε₀: Φ_E = ∮ E·dA = q_enc/ε₀. Charges outside the surface contribute zero net flux, and for a sphere of a point charge Q at its centre the flux is Q/ε₀ regardless of the sphere's size.
Electric Charge and Its Properties
Electric charge
An intrinsic property of matter (protons and electrons) that produces electric force. The SI unit is the coulomb (C). A neutral body has equal numbers of protons and electrons; removing electrons makes it positively charged, adding electrons makes it negatively charged.
Electrostatics
The branch of physics that deals with charges at rest. A glass rod rubbed with silk transfers some electrons to the silk, leaving the rod positively charged and the silk equally negative.
- charge is a scalar; the net charge of a system is the algebraic sum of individual charges
- .
- charge always exists in integer multiples of the elementary charge,
- , with
- .
- the total charge of an isolated system is constant; charge can be transferred but never created or destroyed.
- charge does not depend on the speed of the charged body.
- charge cannot exist without mass, though mass can exist without charge.
Like repels, unlike attracts
MCQ Questions
1 Mark Questions
- 1State the SI unit of electric charge.
- 2How does a glass rod become positively charged when rubbed with silk?
- 3What is the magnitude of charge on an electron?
- 4Why is repulsion taken as a sure test of electrification but attraction is not?
2 Mark Questions
- 1State and explain three basic properties of electric charge.
- 2Why is charge quantised? Explain with q = ne.
- 3Does the mass of a positively charged body differ from its neutral mass? Explain.
- 4Distinguish between conductors and insulators with reference to charge flow.
3 Mark Questions
- 1How many electrons must be given to a neutral body so that it acquires a charge of 4.0 pC?
- 2Explain, with examples, that charge is an additive, conserved and quantised physical quantity.
- 3A body has an excess of 1.5 × 10⁷ electrons. Find the charge on the body.
Coulomb's Law and Superposition
Coulomb's law
The electrostatic 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 and obeys Newton's third law: the force on each charge is equal in magnitude and opposite in direction.
Here and is the permittivity of free space. is the force on charge 1 due to charge 2, directed along the vector from 2 to 1.
Relative permittivity (dielectric constant)
The ratio of the absolute permittivity of a medium to that of vacuum, . Equivalently, the ratio of the force between two charges in vacuum to the force between them at the same distance in the medium. For air or vacuum K = 1; for water K ≈ 80.
Principle of superposition
The net electrostatic force on a charge is the vector sum of the forces from every other charge taken one at a time: . The presence of other charges does not alter the force exerted by any single charge.
Coulomb vs gravitational force
MCQ Questions
1 Mark Questions
- 1State Coulomb's law in electrostatics.
- 2Define relative permittivity (dielectric constant) of a medium.
- 3Write the value of the Coulomb constant k and of ε₀.
- 4What does the principle of superposition of forces state?
2 Mark Questions
- 1Two identical balls with equal mass and charge are suspended by insulating strings of equal length. State how the electrostatic force and weight combine in equilibrium (Lami's theorem approach).
- 2Compare the electrostatic and gravitational forces between two protons.
- 3A charge Q is to be divided into two parts so that the force between them is maximum. Find the division.
- 4Why does the force between two charges decrease when an insulating medium replaces vacuum?
3 Mark Questions
- 1Calculate the Coulomb force between two α-particles (q = +3.2 × 10⁻¹⁹ C each) separated by 3.2 × 10⁻¹⁵ m.
- 2Three charges +Q, −Q and +q are placed at the vertices of an equilateral triangle. Using the principle of superposition, find the net force on +q.
- 3Two identical spheres P and Q each carry charge q. A third identical uncharged sphere R is touched first with P and then with Q. Find the final force between P and Q in terms of the initial force.
Electric Field Due to a Point Charge
Electric field
The region around a charge in which its electrical effects are observed. The electric field strength at a point is the force experienced per unit positive test charge placed there, . It is a vector, direction same as the force on a positive test charge, unit N/C (also V/m).
For a positive source charge the field points radially away; for a negative source it points radially towards the charge. A force acts on a charge q placed in the field, along the field for +q and opposite to it for −q.
Force on a charge in a field
If a charge q is placed in an electric field E, the electrostatic force on it is F = qE — in the direction of E for a positive charge and opposite to E for a negative charge.
Test charge must be small
MCQ Questions
1 Mark Questions
- 1Define electric field intensity at a point.
- 2Write the expression for the electric field due to a point charge.
- 3Give the SI unit of electric field.
- 4What is the direction of the force on a negative charge placed in an electric field?
2 Mark Questions
- 1Define electric field as the force per unit positive test charge and justify why the test charge must be infinitesimally small.
- 2Two point charges q₁ and q₂ are placed at two corners of a square. Using superposition of fields, express the field at the empty corner.
- 3A uniform electric field E is set up between two parallel plates. An electron enters symmetrically with speed v₀ and the plates have length l. Derive the angle of deviation of the electron using m, e and E.
3 Mark Questions
- 1Derive the expression E = q/(4πε₀r²) for the field of a point charge from Coulomb's law.
- 2An electron of charge e and mass m enters a uniform electric field E between two parallel plates of length l with speed v₀. Show that it emerges deviated by an angle θ = tan⁻¹(eEl/mv₀²).
- 3Calculate the electric field at the midpoint of the line joining two charges +q and −q separated by a distance d.
Charge Distributions: Density, Ring and Line
Charge densities
Linear charge density (C/m), surface charge density (C/m²), and volume charge density (C/m³).
For a continuous distribution, the field is the vector integral over all elements dq: , where depending on the geometry.
- (centre of the ring): E = 0, by symmetry.
- :
- — the ring behaves like a point charge.
- E is maximum at
- with
- .
- :
- — far away the line behaves like a point charge.
- Infinite line
- :
- — falls off as 1/x.
Ring field has only an axial component
MCQ Questions
1 Mark Questions
- 1Define linear, surface and volume charge densities and give their SI units.
- 2Write the expression for the electric field on the axis of a ring of charge.
- 3At what distance on the axis of a charged ring is the field maximum?
- 4Write the field of an infinitely long line charge.
2 Mark Questions
- 1Derive E = qx/[4πε₀(x² + R²)^{3/2}] for the electric field on the axis of a uniformly charged ring.
- 2Show that for x ≫ R the ring field becomes that of a point charge.
- 3Derive the electric field on the perpendicular bisector of a uniformly charged line of length 2a.
3 Mark Questions
- 1A charge Q is uniformly distributed on a circular ring of radius a. Derive the electric field at a point on its axis at distance x from the centre and hence obtain the value at the centre.
- 2Obtain the electric field at a distance r from an infinitely long uniformly charged line of linear charge density λ using proper integration, and show it behaves like a point charge for r ≫ a.
- 3A non-conducting rod of length L carries uniform positive charge density λ and total charge Q, lying along the x-axis. A point P is at distance x₀ from one end on the axis. Using superposition, express the magnitude of the electric field at P.
Electric Field Lines
Electric field lines
Imaginary smooth curves along which a free, isolated unit positive charge would move. A tangent at any point gives the direction of the electric field there.
- Field lines start at positive charges and end on negative charges; they extend from a lone positive charge to infinity and from infinity to a lone negative charge.
- No two field lines intersect — the field has a unique direction at every point.
- The tangent to a field line at a point gives the direction of the field.
- Line density indicates field strength: crowded lines mean a stronger field.
- Near a conductor field lines are perpendicular to the surface and do not enter the conductor.
- The number of lines leaving (or entering) a charge is proportional to the magnitude of the charge.
- Static-charge field lines never form closed loops and do not enter conductors.
Lines are not trajectories
MCQ Questions
1 Mark Questions
- 1What is an electric line of force?
- 2Why do two electric field lines never intersect?
- 3Where do electric field lines start and end?
- 4State how the number of field lines is related to the magnitude of charge.
2 Mark Questions
- 1Sketch the field lines of (a) a single positive charge, (b) a single negative charge, and (c) an electric dipole.
- 2Why is the line density of the field a measure of the field strength?
- 3Why do static field lines never form closed loops?
3 Mark Questions
- 1Two point charges q₁ and q₂ are shown with 6 field lines originating from q₁ and 18 lines ending on q₂. Deduce the signs and the ratio q₁/q₂.
- 2Describe the electric field line pattern for a system of two equal positive charges and for a dipole, and explain why the lines never cross.
- 3Explain with the help of field lines why the electric field is zero inside a closed conducting surface.
Electric Flux
Electric flux
The number of electric field lines that penetrate a surface. For a uniform field making an angle θ with the normal to a surface of area A, . It is a scalar with unit N·m²/C.
Flux is positive where field lines leave the surface and negative where they enter. For a closed surface the normal is chosen outward. The area element dA on a sphere subtends a solid angle , and the total solid angle around a point is 4π sr.
Area vector
MCQ Questions
1 Mark Questions
- 1Define electric flux through a surface.
- 2Write the unit and dimensional formula of electric flux.
- 3What is the flux through a surface placed parallel to a uniform electric field?
- 4What is a solid angle? What is its unit?
2 Mark Questions
- 1Derive the expression Φ_E = EA cosθ for the flux through a tilted surface in a uniform field.
- 2A non-uniform field E = 3.0x i + 4.0 j pierces the faces of a cube. Find the flux through (a) the right face at x = 3.0 m of area 4.0 m² and (b) the top face.
- 3Explain why the flux through a closed surface is positive for outward lines and negative for inward lines.
3 Mark Questions
- 1Consider the non-uniform field E = 3.0x i + 4.0 j newton per coulomb (x in metres) piercing the faces of a cube with faces at x = 1.0 m, x = 3.0 m and area 4.0 m². Calculate the net flux through the right, left and top faces and the net charge enclosed.
- 2Show that the flux of a point charge q through a spherical surface of radius r enclosing it is q/ε₀, independent of r.
- 3A small area ΔA on a sphere of radius r₁ subtends a solid angle ΔΩ at the centre. Express ΔA in terms of r₁ and ΔΩ, and find the total solid angle of a full sphere.
Gauss's Law
Gauss's law
The net electric flux through any closed surface equals the net charge enclosed by the surface divided by ε₀: . Charges outside the surface contribute zero net flux, and the result is independent of the shape of the Gaussian surface.
For a point charge Q at the centre of a sphere, E = Q/4πε₀r² on the surface and the flux is — independent of r. This is why Gauss's law is exact and powerful for symmetric distributions.
Problem-solving tactics for Gauss's law
MCQ Questions
1 Mark Questions
- 1State Gauss's law of electrostatics.
- 2Does the flux through a Gaussian surface depend on charges outside it?
- 3Write Gauss's law for a point charge at the centre of a sphere.
- 4Why is the flux through a closed surface independent of the distribution of the enclosed charge?
2 Mark Questions
- 1Prove that the flux of a point charge Q through a spherical Gaussian surface is Q/ε₀.
- 2Five charged lumps lie near a Gaussian surface: q₁ = q₄ = +3.1 nC, q₂ = q₅ = −5.9 nC, q₃ = −3.1 nC; q₃, q₁ and a neutral coin are inside. Compute the net flux.
- 3Explain how the shape of a Gaussian surface affects the total flux (or does not) and why.
3 Mark Questions
- 1Using Gauss's law, show that the net flux through a closed surface enclosing charges q₁ = +3.1 nC, q₂ = −5.9 nC and q₃ = −3.1 nC is −670 N·m²/C (ε₀ = 8.85 × 10⁻¹²).
- 2State the conditions under which Gauss's law may be used to easily find the electric field, and apply it to a uniformly charged thin spherical shell.
- 3A point charge is placed at the centre of a cube. Find the flux through one face in terms of the total flux.
Applications of Gauss's Law
- , uniform and independent of distance, on both sides of the sheet.
- outside the surface.
- radially outward, varying as 1/r.
- E = 0 inside; outside the shell
- .
- — field grows linearly with r inside.
- .
Discontinuity at the surface
MCQ Questions
1 Mark Questions
- 1Write the electric field due to an infinite non-conducting sheet of charge density σ.
- 2What is the electric field inside a uniformly charged spherical shell?
- 3Write the field inside a uniformly charged non-conducting solid sphere at distance r from the centre.
- 4State the field of an infinite line charge at distance r.
2 Mark Questions
- 1Using Gauss's law, derive E = σ/2ε₀ for an infinite non-conducting sheet using a cylindrical Gaussian surface (pillbox).
- 2Derive the electric field inside and outside a uniformly charged thin spherical shell by Gauss's law.
- 3Derive E = λ/2πε₀r for an infinitely long line charge using a coaxial cylindrical Gaussian surface.
3 Mark Questions
- 1Derive the electric field at a point inside and outside a uniformly charged non-conducting solid sphere of radius R and total charge Q by Gauss's law, and sketch E versus r.
- 2Use Gauss's law to find the field of an infinitely long rod of uniform linear charge density λ at a distance r, and verify it agrees with E = λ/2πε₀r.
- 3A thin spherical shell of radius a carries charge +Q uniformly on its surface. Obtain E at (i) r < a and (ii) r ≥ a by Gauss's law and comment on the discontinuity at r = a.
NCERT Exercise Questions
From the textbook
Exercises (NCERT Q1.1–Q1.16)
- 1.1) 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?
- 1.2) 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?
- 1.3) Check that the ratio ke²/Gmₑmₚ is dimensionless. Look up a table of physical constants and determine the value of this ratio. What does the ratio signify?
- 1.4) (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?
- 1.5) 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.
- 1.6) 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?
- 1.7) (a) An electrostatic field line is a continuous curve: a field line cannot have sudden breaks. Why? (b) Explain why two field lines never cross each other at any point.
- 1.8) 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?
- 1.9) 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?
- 1.10) 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. Calculate the magnitude of the torque acting on the dipole.
- 1.11) 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?
- 1.12) Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. (a) What is the mutual force of electrostatic repulsion if the charge on each is 6.5 × 10⁻⁷ C? (b) What is the force of repulsion if each sphere is charged double the above amount and the distance between them is halved?
- 1.13) Suppose the spheres A and B in Exercise 1.12 have identical sizes. A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with the second, and finally removed from both. What is the new force of repulsion between A and B?
- 1.14) The figure below shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge-to-mass ratio?
- 1.15) 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?
- 1.16) What is the net flux of the uniform electric field of Exercise 1.15 through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes?
Additional Exercises (NCERT Q1.17–Q1.34)
- 1.17) 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?
- 1.18) A point charge +10 µC is at a distance 5 cm directly above the centre of a square of side 10 cm, as shown in the figure. 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.)
- 1.19) 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?
- 1.20) 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?
- 1.21) 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?
- 1.22) 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?
- 1.23) An infinite line charge produces a field of 9 × 10⁴ N/C at a distance of 2 cm. Calculate the linear charge density.
- 1.24) 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 in (a) the outer region of the first plate, (b) the outer region of the second plate, and (c) between the plates?
- 1.25) An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 × 10⁴ N/C (Millikan's oil-drop experiment). The density of the oil is 1.26 g/cm³. Estimate the radius of the drop. (g = 9.81 m/s²; e = 1.60 × 10⁻¹⁹ C.)
- 1.26) Which among the curves shown in the figure cannot possibly represent electrostatic field lines? Give reasons for your answers.
- 1.27) In a certain region of space the electric field is along the z-direction throughout. The magnitude of E is not constant but increases uniformly along the positive z-direction at the rate of 10⁵ N/C per metre. What are the force and torque experienced by a system having a total dipole moment equal to 10⁻⁷ C m in the negative z-direction?
- 1.28) (a) A conductor A with a cavity is given a charge Q. Show that the entire charge must appear on the outer surface of the conductor. (b) Another conductor B with charge q is inserted into the cavity keeping B insulated from A. Show that the total charge on the outside surface of A is Q + q. (c) A sensitive instrument is to be shielded from the strong electrostatic fields in its environment. Suggest a possible way.
- 1.29) A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is (σ/2ε₀) n̂, where n̂ is the unit vector in the outward normal direction and σ is the surface charge density near the hole.
- 1.30) Obtain the formula for the electric field due to a long thin wire of uniform linear charge density λ without using Gauss's law. [Hint: use Coulomb's law directly and evaluate the necessary integral.]
- 1.31) A proton and a neutron are made of 'up' quarks (u, charge +2/3 e) and 'down' quarks (d, charge −1/3 e). Suggest a possible quark composition of a proton and of a neutron.
- 1.32) (a) Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (E = 0) of the configuration. Show that the equilibrium of the test charge is necessarily unstable. (b) Verify this result for the simple configuration of two charges of the same magnitude and sign placed a certain distance apart.
- 1.33) A particle of mass m and charge −q enters the region between two charged plates initially moving along the x-axis with speed vₓ. The length of each plate is L and a uniform electric field E is maintained between the plates. Show that the vertical deflection of the particle at the far edge of the plate is qEL²/(2mvₓ²). Compare this motion with the motion of a projectile in a gravitational field.
- 1.34) Suppose the particle in Exercise 1.33 is an electron projected with velocity vₓ = 2.0 × 10⁶ m/s. If E between the plates separated by 0.5 cm is 9.1 × 10² N/C, where will the electron strike the upper plate? (|e| = 1.6 × 10⁻¹⁹ C, mₑ = 9.1 × 10⁻³¹ kg.)
Answers at a glance
- 1Q1.1 — F = 6 × 10⁻³ N (repulsive).
- 2Q1.2 — (a) r = 0.12 m; (b) 0.2 N, equal and opposite (Newton's third law).
- 3Q1.3 — Dimensionless; ratio ≈ 2.3 × 10³⁹ (electric to gravitational force between a proton and an electron).
- 4Q1.4 — (a) Charge exists only as integral multiples of e; (b) macroscopic charges are huge compared with e, so the granularity is unobservable.
- 5Q1.5 — Rubbing only transfers electrons; equal positive and negative charges appear, so the total charge is conserved.
- 6Q1.6 — Net force = 0 N (the forces from the four corner charges cancel by symmetry).
- 7Q1.7 — (a) A charge moves continuously, so the field line cannot break; (b) crossing would give two directions of E at one point.
- 8Q1.8 — (a) E = 5.4 × 10⁶ N/C along OB; (b) F = 8.1 × 10⁻³ N along OA.
- 9Q1.9 — Total charge = 0; p = 7.5 × 10⁻⁸ C m along the +z-axis.
- 10Q1.10 — τ = pE sinθ = 10⁻⁴ N·m.
- 11Q1.11 — (a) n ≈ 1.87 × 10¹² electrons, transferred from wool to polythene; (b) yes, mass transferred ≈ 1.7 × 10⁻¹⁸ kg (negligible).
- 12Q1.12 — (a) F = 1.52 × 10⁻² N; (b) F = 0.243 N (16 times larger).
- 13Q1.13 — F ≈ 5.7 × 10⁻³ N (3/8 of the initial force; A carries q/2, B carries 3q/4).
- 14Q1.14 — Particles 1 and 2 are negative, particle 3 is positive; particle 3 has the highest charge-to-mass ratio.
- 15Q1.15 — (a) Φ = 30 N·m²/C; (b) Φ = 15 N·m²/C.
- 16Q1.16 — Net flux = 0 (uniform field: lines entering equal lines leaving).
- 17Q1.17 — (a) q = 0.07 µC; (b) no — zero net flux only means zero net enclosed charge.
- 18Q1.18 — Φ = 1.88 × 10⁵ N·m²/C.
- 19Q1.19 — Φ = 2.26 × 10⁵ N·m²/C.
- 20Q1.20 — (a) Same, −10³ N·m²/C (flux is independent of the Gaussian radius); (b) q = −8.85 nC.
- 21Q1.21 — q = 6.67 × 10⁻⁹ C = 6.67 nC (negative, radially inward).
- 22Q1.22 — (a) Q = 1.45 × 10⁻³ C; (b) Φ = 1.63 × 10⁸ N·m²/C.
- 23Q1.23 — λ = 10 µC/m.
- 24Q1.24 — E = 0 in both outer regions; between the plates E = σ/ε₀ = 1.92 × 10⁻¹⁰ N/C.
- 25Q1.25 — r ≈ 9.82 × 10⁻⁷ m.
- 26Q1.26 — Only (c) is valid; (a), (b), (d) and (e) cannot represent electrostatic field lines.
- 27Q1.27 — F = −10⁻² N (opposite to z); τ = 0.
- 28Q1.28 — (a) E = 0 inside the conductor compels q_enc = 0 on the inner surface; (b) outer surface charge = Q + q; (c) shield the instrument by enclosing it in a closed metal box (Faraday cage).
- 29Q1.29 — E_hole = σ/2ε₀ along n̂.
- 30Q1.30 — E = λ/2πε₀r.
- 31Q1.31 — Proton = uud; neutron = udd.
- 32Q1.32 — (a) The null point is an unstable equilibrium (displacement of the test charge grows); (b) the midpoint of two equal like charges is an unstable equilibrium.
- 33Q1.33 — y = qEL²/(2mvₓ²); identical to projectile motion under gravity.
- 34Q1.34 — The electron strikes the upper plate after travelling 1.6 cm.
CBSE Pattern Questions
How CBSE asks this chapter
MCQ Questions
1 Mark Questions
- 1State Coulomb's law in vector form.
- 2State Gauss's law in electrostatics.
- 3What is an electric dipole? Give its SI unit of dipole moment.
- 4Why is the electrostatic field inside a conductor zero in equilibrium?
2 Mark Questions
- 1Two charges of +3 μC and −3 μC are placed 30 cm apart in air. Find the magnitude of the force between them. (Ans: F = (1/4πε₀)q₁q₂/r² = 0.9 N)
- 2An electric dipole of moment 5 × 10⁻⁹ C m is placed in a uniform field of 2 × 10⁴ N/C. Find the maximum torque on it. (Ans: τ = pE = 10⁻⁴ N m)
- 3Find the electric field at a point 20 cm from a point charge of 4 μC. (Ans: E = (1/4πε₀)q/r² = 9 × 10⁵ N/C)
3 Mark Questions
- 1Derive the expression for the electric field due to an infinitely long thin straight wire of uniform linear charge density using Gauss's law. (Ans: E = λ/(2πε₀r))
- 2Derive the expression for the electric field due to a uniformly charged infinite plane sheet using Gauss's law. (Ans: E = σ/2ε₀)
- 3Derive the electric field at a point on the axial line of an electric dipole. (Ans: E = (1/4πε₀)2p/r³)
5 Mark Questions
- 1Derive the electric field of a uniformly charged spherical shell of radius R both (a) outside and (b) inside the shell using Gauss's law, and sketch the E versus r graph.
- 2Describe the electric dipole — define dipole moment, derive the field on the axial and equatorial lines, and find the torque and potential energy in a uniform external field. (Ans: τ = p × E, U = −p·E)
PDF: Assignments & Tests
Printable study material
Free PDF Downloads & Assignments
Suggested path: finish the two topic-wise assignments first, then attempt each timed 40-mark test in 90 minutes and check yourself against the answer key.
Every concept of the chapter, arranged topic by topic: electric charge and its properties, Coulomb's law, electric field and field lines, electric flux and Gauss's law.
Higher-difficulty second set — mixed multi-topic numericals and derivations covering the whole chapter.
Full board-pattern test: 1-mark MCQs/VSA, 2-mark, 3-mark and 5-mark long answers totalling 40 marks — to be done in 90 minutes, with an answer key.
Second full paper with a different question set — same 40-mark, 90-minute blueprint and answer key.
Quick Revision
Key formulas at a glance
Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.
Coulomb's law
k = 9 × 10⁹ N m²/C²; force in a medium = F_air/K.
Force in a medium
K = ε/ε₀ = relative permittivity (dielectric constant).
Superposition of forces
vector sum of individual Coulomb forces.
Electric field / force on charge
unit N/C or V/m.
Field of a point charge
radially outward for +q, inward for −q.
Field on ring axis
max at x = R/√2 = 2q/(3√3·4πε₀R²); E = 0 at centre.
Field of line charge (finite)
perpendicular bisector of a line of length 2a.
Infinite line charge
varies as 1/r, radially outward.
Infinite non-conducting sheet
uniform, distance-independent.
Infinite conducting sheet
outside the surface.
Electric flux
unit N·m²/C.
Gauss's law
independent of the Gaussian surface shape.
Solid conducting/insulating sphere (inside)
E ∝ r inside a uniformly charged sphere.
Sphere outside
acts like a point charge at the centre.
Exam Strategy
How this chapter is asked
High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.
- Charge is quantised (q = ±ne), conserved, scalar and invariant.
- Repulsion is a sure test of electrification; attraction alone is not.
- Coulomb's law works for point charges; net force uses superposition.
- Field of a point charge ∝ 1/r²; infinite line ∝ 1/r; infinite sheet uniform.
- Ring field E = kqx/(x²+R²)^{3/2}; E = 0 at centre, max at x = R/√2.
- Gauss's law: Φ_E = q_enc/ε₀ — use it for sheets, lines, shells and spheres.
- Inside a charged conducting shell E = 0; inside a solid non-conducting sphere E ∝ r.
- Discontinuity of the normal field across a surface is σ/ε₀.
- Field lines never cross, never enter conductors, and static lines never close back on themselves.
Practice Numericals
Step-by-step worked examples
Standard exam-level problems solved with clear conceptual breakdowns.
Coulomb force between two α-particles
Step-by-Step Solution
- 1F = (1/4πε₀)(q₁q₂/r²) = 9 × 10⁹ × (3.2 × 10⁻¹⁹)² ÷ (3.2 × 10⁻¹⁵)².
- 2Numerator: 9 × 10⁹ × 10.24 × 10⁻³⁸ = 9.216 × 10⁻²⁸; denominator: 10.24 × 10⁻³⁰.
- 3F ≈ 90 N, repulsive (like charges).
Final Answer
≈ 90 N (repulsive)
Charge acquired by adding electrons
Step-by-Step Solution
- 1Giving electrons makes the body negative: q = −4.0 × 10⁻¹² C.
- 2n = q/e = 4.0 × 10⁻¹² ÷ 1.6 × 10⁻¹⁹ = 2.5 × 10⁷ electrons.
- 3For 1.5 × 10⁷ excess electrons: q = ne = 1.5 × 10⁷ × 1.6 × 10⁻¹⁹ = 2.4 × 10⁻¹² C = 2.4 pC, negative.
Final Answer
2.5 × 10⁷ electrons; −2.4 pC
Sharing charge by contact
Step-by-Step Solution
- 1Touching R with P: charge divides equally, P becomes q/2, R becomes q/2.
- 2Touching R (now q/2) with Q: total on R + Q is q + q/2 = 3q/2, divided equally: Q = 3q/4.
- 3New force F′ = k(q/2)(3q/4)/r² = 3/8 × kq²/r² = 3F/8.
Final Answer
F′ = 3F/8
Electric field on the axis of a charged ring
Step-by-Step Solution
- 1Element dq on the ring produces dE = k dq/(a² + x²) at P on the axis.
- 2Perpendicular components of diametrically opposite elements cancel; only the axial component dE cosθ with cosθ = x/√(a² + x²) remains.
- 3E = k ∫ dq·x/(a² + x²)^{3/2} = kQx/(a² + x²)^{3/2}.
- 4At x = 0, E = 0; for x ≫ a, E ≈ kQ/x² (point charge limit).
Final Answer
E = kQx/(a² + x²)^{3/2}
Flux through the faces of a cube
Step-by-Step Solution
- 1Right face (x = 3.0 m): dA = dA i, so Φ = ∫3.0x dA = 3.0 × 3.0 × 4.0 = 36 N·m²/C.
- 2Left face (x = 1.0 m): dA = −dA i, so Φ = −3.0 × 1.0 × 4.0 = −12 N·m²/C.
- 3Top face: dA = dA j, only the 4.0 j term contributes: Φ = 4.0 × 4.0 = 16 N·m²/C.
Final Answer
36, −12 and 16 N·m²/C
Gauss's law with a charged shell
Step-by-Step Solution
- 1For r < a, the Gaussian sphere encloses no charge: q_enc = 0, so E = 0 inside.
- 2For r ≥ a, all charge is enclosed: q_enc = Q, so E(4πr²) = Q/ε₀ gives E = Q/4πε₀r².
- 3The field right outside the shell is E = Q/4πε₀a² = σ/ε₀; at the surface itself the field is discontinuous by σ/ε₀ between inside (0) and outside.
Final Answer
E = 0 (r < a); E = Q/4πε₀r² (r ≥ a)
FAQ
Frequently asked questions
Why is electric charge quantised?
Charge on a body comes from gaining or losing whole electrons, so it is always an integral multiple of the elementary charge e = 1.6 × 10⁻¹⁹ C: q = ±ne. In macroscopic charges the discreteness is invisible, but it is exact.
Why is the field zero at the centre of a uniformly charged ring?
By symmetry, diametrically opposite elements of the ring produce equal and opposite perpendicular field components at the centre, so they cancel and the net field is zero.
Why does an infinite sheet give a uniform field independent of distance?
The field lines from an infinite uniformly charged plane diverge only perpendicular to it; the flux through a pillbox of any height is the same, so E = σ/2ε₀ anywhere — no inverse-square fall-off.
Why is the electric field zero inside a charged conducting shell?
In electrostatic equilibrium, charges redistribute on the outer surface so that the internal field cancels. By Gauss's law a Gaussian surface inside the shell encloses zero net charge, giving E = 0.
Why do field lines never intersect?
The electric field has a unique direction at every point. If two lines crossed, there would be two field directions at the same point, which is impossible for a single well-defined E.
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