From individual charges to symmetry: calculate fields, interpret flux and predict motion.
JEE Main · JEE Advanced · NEET UG85 min study guide9 indexed sections
BY THE END, YOU CAN
Apply Coulomb superposition and charge conservation with conductor-equilibration conditions.
Use continuous distributions and limiting cases to interpret electric fields.
Distinguish flux from field and derive symmetric Gauss-law results.
Analyse dipole torque and energy and charged-particle motion in a uniform field.
BEFORE YOU START
Vector components and inverse-square forces.
Elementary integration, constant-acceleration kinematics and work–energy ideas.
Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.
CONNECT THE CONCEPT TO THE EXAM
Selected exam questions & worked solutions
These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.
Available in this lesson: JEE Advanced 2013. This is a limited lesson sample, not a complete exam-question bank.
1 sourced sample question available across the listed exams and years.
Independently derived and checked solution. Archived key finality is unverified; this is not presented as an official final answer.
Two insulating solid spheres partially overlap, with distinct centres A and B and uniform charge densities +ρ and −ρ. In the overlap, assess whether E vanishes, potential is constant, |E| is constant, or E has one fixed direction.
Reveal answer & worked solution
Answer: Constant field magnitude and direction only (independently derived; archived C,D).
For a point P inside a uniformly charged sphere, Gauss’s law gives E=ρr/(3ε₀), with r measured from that sphere’s centre.
Superpose the two fields: E=ρ(AP−BP)/(3ε₀)=ρAB/(3ε₀), where the symbols denote vectors.
AB is a fixed nonzero vector. Therefore magnitude and direction are constant, but E is not zero. Since E=−∇V, the potential is not constant across the overlap.
Source: JEE (Advanced) 2013, Paper 2, Physics Q2; independently derived solution; key finality unverified. Solution written for this lesson.
01 / UNDERSTAND & APPLY
1. Charge conservation and Coulomb’s interaction
Electric charge is a signed property of matter. Ordinary charging usually redistributes electrons rather than creating charge. An isolated collection conserves its total algebraic charge, although individual objects may gain or lose electrons. Quantisation means a freely observed ordinary body’s net charge changes in integer multiples of the elementary charge e; in macroscopic problems those steps are usually too small to resolve.
For stationary point charges in vacuum, Coulomb’s law gives force magnitude k|q₁q₂|/r² with k = 1/(4πε₀). Like signs repel and unlike signs attract along the line joining the charges. The vector on charge 2 due to charge 1 is kq₁q₂(r₂ − r₁)/|r₂ − r₁|³. Keeping the sign in this vector formula automatically handles attraction and repulsion.
For an extended distribution, add the contributions of small charge elements or use an appropriate symmetry argument. The point-charge formula does not automatically apply using centre-to-centre separation for every extended body. An isolated spherically symmetric distribution behaves externally like a central point charge, but nearby conductors can lose that symmetry through induced charge redistribution.
When conductors are connected, charges flow until their potentials are equal, not necessarily until their charges are equal. Identical isolated conducting spheres in symmetric surroundings share total charge equally. Widely separated spheres of radii R₁ and R₂ share charge in the ratio R₁:R₂ in the isolated-sphere approximation because their potentials are kQ/R. A nearby object or connecting-wire capacitance can modify that approximation.
ORIGINAL WORKED EXAMPLE
Identical spheres touch
Two identical small conducting spheres carry +8 nC and −2 nC. They are touched and then separated far enough for a point-charge approximation at 0.30 m. Find their final force with k = 9 × 10⁹.
Total charge is +6 nC and is conserved. Identical spheres each retain +3 nC after equilibration.
The final force is repulsive because both charges are positive.
Magnitude is 9 × 10⁹ × (3 × 10⁻⁹)²/(0.30)² = 9 × 10⁻⁷ N.
Answer 9 × 10⁻⁷ N, repulsive.
02 / UNDERSTAND & APPLY
2. A field assigns a force response to each point
The electric field at a point is force per unit positive test charge in the limit that the test charge does not significantly disturb the source configuration. Once E is known, a particle of signed charge q experiences F = qE. A negative charge accelerates opposite to the field direction; the field itself does not reverse merely because the test charge changes sign.
For a point source Q, E = kQ r̂/r², with r̂ directed from source to observation point. Superposition adds these fields vectorially. Draw each contribution at the observation point before combining magnitudes. Symmetry may cancel components, but it is not a licence to cancel vectors pointing in the same direction.
For two equal positive charges, their fields cancel at the midpoint. For equal opposite charges, the midpoint fields reinforce from positive toward negative. For unequal like charges, a zero-field point lies between them and nearer the smaller charge. For unlike unequal charges it lies outside the pair on the side of the smaller magnitude; equal opposite charges have no finite zero-field point along their axis.
Field lines are a visual convention: their tangent gives field direction and denser spacing indicates larger field magnitude when drawn consistently. They do not depict a particle’s trajectory, because a particle can enter with velocity in another direction. Electrostatic field lines cannot cross where the field is well-defined, since the field cannot have two directions at one point.
ORIGINAL WORKED EXAMPLE
A zero-field location
Charges +Q and +4Q lie at x = 0 and x = 3a. Locate the finite zero-field point.
Between the charges the fields oppose; outside they point the same way and cannot cancel.
Let the point be x from +Q. Equality of magnitudes gives kQ/x² = 4kQ/(3a − x)².
Positive distances give 3a − x = 2x, so x = a. It is nearer the smaller source as expected.
Answer x = a, between the charges.
03 / UNDERSTAND & APPLY
3. Integrate charge, then use symmetry
For a continuous distribution, write dq = λ dl for a line, σ dA for a surface or ρ dV for a volume. The field contribution is dE = k dq r̂/r². Express distance and direction using the integration variable before integrating. Charge density need not be uniform; when it varies, keep its position dependence inside the integral.
A uniformly charged ring is a good example of component cancellation. At an axial point distance x from its centre, opposite elements cancel transverse fields while axial components add. Every element has distance √(R² + x²), yielding Eₓ = kQx/(R² + x²)^(3/2). At the centre it is zero; far away it approaches the field of a point charge.
For a uniformly charged disk, regard each narrow annulus as a ring. On the positive axis of a positively charged disk, integrating rings gives E = σ[1 − x/√(x² + R²)]/(2ε₀). Far from the disk it approaches kQ/x², while very near a large disk it approaches σ/(2ε₀). These limits provide a strong check on both geometry and prefactors.
Use symmetry before calculus: decide which components survive, then integrate only those. An incorrect scalar addition of all magnitudes would predict a nonzero field at the centre of a uniformly charged ring. A zero vector field there does not imply that the potential is zero; potential is a different, scalar quantity.
Eaxis,ring = kQx/(R² + x²)^(3/2)
Uniform ring of radius R, centre at x = 0; the signed axial component is shown.
04 / UNDERSTAND & APPLY
4. Flux measures a normal component through a surface
Electric flux through a small oriented area is dΦ = E·dA. The area vector points normal to the surface. For a uniform field through a flat area, Φ = EA cosθ, where θ is the angle between field and area normal, not between field and the surface plane. Tilting a surface reduces its projected area perpendicular to the field.
For a closed surface, choose outward area normals. Field leaving contributes positive flux and field entering negative flux. Flux is a scalar signed integral, not the field magnitude. A uniform field can have zero net flux through a closed box even though the field is nonzero at every point on the box.
A closed Gaussian surface is an imaginary mathematical surface, not necessarily a material boundary. Charges outside it can contribute local field and both entering and leaving flux, but their total net flux through the closed surface is zero. The net enclosed charge controls total flux; it does not by itself determine the field at each point.
Φ = ∫E·dA; for uniform E and a plane, Φ = EA cosθ
Area orientation and whether the surface is open or closed must be specified.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Perspective schematic of the plane and vectors; angle and lengths are illustrative, not a calibrated 3 D projection. Reversing the chosen normal reverses the signed flux.
ORIGINAL WORKED EXAMPLE
Angle to the plane is not angle to its normal
A 0.020 m² flat surface lies in a uniform 300 N C⁻¹ field. The field makes 30° with the surface plane and points through the chosen positive side. Find flux.
The angle with the normal is 60°, not 30°.
Flux is EA cos60° = 300 × 0.020 × 0.5 = 3 N m² C⁻¹.
Answer +3 N m² C⁻¹.
05 / UNDERSTAND & APPLY
5. Gauss’s law is exact; extracting E requires symmetry
Gauss’s law states that total electric flux through any closed surface equals enclosed charge divided by ε₀. This is exact for electrostatic fields regardless of surface shape. Its usefulness for calculating a field depends on symmetry: the field must have a known direction and a constant magnitude on the contributing portions of a well-chosen surface.
For a spherical charge distribution, use a concentric sphere. Radial field is constant in magnitude on it, so E(4πr²) = Qenclosed/ε₀. Outside a uniformly charged sphere the result is kQ/r². Inside a uniformly charged nonconducting solid sphere of radius R, enclosed charge is Qr³/R³, giving E = kQr/R³. Its field increases linearly from zero at the centre to the surface value.
For an infinite uniform line charge, use a coaxial cylinder of radius r and length L. The curved surface contributes E(2πrL); the end caps contribute zero because their normals are parallel to the line while E is radial. Enclosed charge is λL, giving E = λ/(2πε₀r). The infinite-line idealisation requires observation scales small compared with the actual wire length and far from its ends.
For an infinite uniformly charged nonconducting sheet, use a pillbox straddling it. Equal flux leaves through the two flat faces, so 2EA = σA/ε₀ and E = σ/(2ε₀) on each side. A conducting surface has different boundary conditions: immediately outside an electrostatic conductor E = σ/ε₀, because the interior field is zero. Do not confuse a single nonconducting sheet with a conductor surface.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
The dashed surface is imaginary, not a physical conductor. The diagram applies outside the entire source distribution; the enclosed charge is Q. Circular cross-sections represent spheres.
ORIGINAL WORKED EXAMPLE
Inside a uniformly charged solid sphere
A nonconducting sphere of radius R holds uniform total charge Q. Compare the fields at r = R/2 and r = 2R.
Inside, only Qenclosed = Q(1/2)³ = Q/8 contributes to the Gaussian flux.
E(R/2) = k(Q/8)/(R²/4) = kQ/(2R²).
Outside, E(2R) = kQ/(4R²), so the inside point has twice the field magnitude of the outside point.
Answer E(R/2)/E(2R) = 2.
06 / UNDERSTAND & APPLY
6. Electrostatic equilibrium fixes what charges may do
In an ideal conductor at electrostatic equilibrium, mobile charges have stopped their macroscopic redistribution. A nonzero electric field inside the conducting material would drive them, so that interior field is zero. The conductor is an equipotential. Any excess charge resides on its surfaces, and the external field at a smooth surface is normal because a tangential component would move surface charge.
A Gaussian surface entirely inside the conducting material therefore encloses zero net charge. If an isolated cavity contains a charge q, its inner surface carries net induced charge −q, while the remaining charge appears on the outer surface according to total-charge conservation. Zero field in the metal does not imply zero field inside a cavity containing charge.
For an empty cavity in a conductor in electrostatic equilibrium, with no charges inside the cavity, external static fields are screened from the cavity. This statement relies on electrostatic conditions; time-varying electromagnetic shielding requires more care. Earthing lets charge flow to or from a large reservoir while maintaining the conductor at the reservoir potential; it does not mean the conductor must always be electrically neutral.
For a spherical conducting shell with no internal charge and spherically symmetric conditions, field is zero inside and kQ/r² outside. For a nonconducting shell with a uniformly fixed surface charge the same radial result follows from symmetry, but the physical reason is different. Distinguish mobile charges from fixed charge distributions when the surroundings are changed.
ORIGINAL WORKED EXAMPLE
A charged cavity in a neutral conductor
A +5 nC point charge is placed inside a cavity of an otherwise neutral isolated conductor. Find net charge on inner and outer surfaces.
The field in the conducting material is zero. A Gaussian surface lying in the metal around the cavity must enclose zero net charge.
Thus inner-surface charge is −5 nC.
The conductor remains neutral overall, so outer-surface charge is +5 nC. Its detailed distribution need not be uniform.
Answer Inner −5 nC; outer +5 nC.
07 / UNDERSTAND & APPLY
7. A dipole has zero net charge but a nonzero field
An electric dipole consists of charges −q and +q separated by displacement vector d from negative to positive. Its dipole moment is p = qd. Net charge is zero, but the fields of the separated charges do not cancel everywhere. Far from the pair, their leading inverse-square contributions cancel and the remaining dipole field falls as inverse distance cubed.
For separation 2a along the dipole axis and observation distance r from the centre with r ≫ a, the axial field magnitude is 2kp/r³ in the direction of p, while the equatorial field magnitude is kp/r³ opposite p. These are far-field approximations. At distances comparable to a, add the actual two Coulomb fields instead of using the compact dipole expressions.
In a uniform external field, the forces on the two charges cancel in translation but form a couple: τ = p × E. The torque tends to align p with E. Potential energy is U = −p·E up to an additive constant, so aligned orientation is stable and anti-aligned orientation unstable under small angular disturbances.
A nonuniform field can exert a net force on a dipole because the two charges encounter different field strengths or directions. The claim “a dipole has no net force” therefore requires uniformity of the external field. For a permanent fixed dipole in electrostatics, F = ∇(p·E) can express this effect under the usual small-dipole approximation.
τ = pE sinθ; U = −pE cosθ
Uniform external field, dipole moment from negative to positive. θ is the angle between p and E.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Directions are exact for the stated locations. The displayed field magnitudes are far-field approximations at center distance r much greater than the charge separation; arrow lengths are not magnitude-scaled.
ORIGINAL WORKED EXAMPLE
Rotate a dipole slowly
A dipole of moment 4 × 10⁻⁸ C m is in a uniform 2 × 10⁴ N C⁻¹ field. Find external work to rotate it slowly from aligned to perpendicular.
Initial energy is −pE and final energy is zero.
For quasistatic rotation with no change in kinetic energy, external work is ΔU = pE.
Compute 4 × 10⁻⁸ × 2 × 10⁴ = 8 × 10⁻⁴ J. The field does the opposite work.
Answer 8 × 10⁻⁴ J.
08 / UNDERSTAND & APPLY
8. Turn the field into a trajectory
A nonrelativistic particle of mass m and charge q in a prescribed uniform electric field has constant acceleration a = qE/m if other forces are neglected. The field direction determines acceleration for a positive charge and the opposite direction for a negative charge. With initial velocity perpendicular to E, the path is parabolic just as for a projectile in uniform gravity.
For parallel plates of length L with a uniform transverse field, a particle entering horizontally at speed u spends time L/u in the field if the horizontal force is zero. Its transverse deflection is qEL²/(2mu²), and transverse exit velocity is qEL/(mu). After leaving the field it travels along its exit direction if no other forces act. Fringing fields and collisions with plates are neglected in this idealisation.
Check that the particle remains between the plates and that speeds remain nonrelativistic. If both gravity and the electric force matter, add them as vectors before solving the motion. Electric equilibrium requires qE + mg = 0, so the necessary field direction depends on the sign of q. The trajectory is not generally a field line because inertia retains the initial velocity component.
ORIGINAL WORKED EXAMPLE
Deflection through a uniform field
A particle has q/m = +2 × 10⁶ C kg⁻¹ and enters horizontally at 10⁴ m s⁻¹ through a 0.10 m long region with upward E = 100 N C⁻¹. Find its exit deflection and transverse speed.
Upward acceleration is (q/m)E = 2 × 10⁸ m s⁻².
Transit time is L/u = 10⁻⁵ s.
Deflection is at²/2 = 0.010 m and transverse speed is at = 2 × 10³ m s⁻¹. These apply if the plates leave enough clearance.
Answer 1.0 cm upward; 2.0 × 10³ m s⁻¹ upward.
09 / UNDERSTAND & APPLY
9. Original practice and limiting-case checks
Each answer should include a direction or sign where appropriate. A correct magnitude with a reversed direction is not a complete electrostatics solution.
Superpose vectors, not unsigned field magnitudes.
Zero enclosed charge means zero net flux, not necessarily zero field.
Use Gauss’s law to extract E only when symmetry justifies taking E outside the integral.
Dipole far-field formulas need distance much larger than separation.
A prescribed uniform field gives constant acceleration only while its assumptions remain valid.
ORIGINAL PRACTICE · 1
The distance between two fixed point charges doubles. What happens to force magnitude?
Reveal answer & reasoning
It becomes one quarter.
Coulomb force varies as 1/r² when charges and medium are unchanged.
ORIGINAL PRACTICE · 2
Equal +Q charges are at x = ±a. What is the field at the origin?
Reveal answer & reasoning
Zero.
Their equal-magnitude fields point in opposite directions at the midpoint.
ORIGINAL PRACTICE · 3
Charges +Q at x = −a and −Q at x = +a produce what field at the origin?
Reveal answer & reasoning
2kQ/a² in the +x direction.
Both contributions point from the positive source toward the negative source at the midpoint.
ORIGINAL PRACTICE · 4
A closed surface encloses +3 nC and −1 nC. Find net flux.
Reveal answer & reasoning
(2 × 10⁻⁹ C)/ε₀.
Gauss’s law uses the algebraic enclosed charge. Outside charges do not alter total net flux.
ORIGINAL PRACTICE · 5
Can a closed surface have zero flux while E is nonzero on it?
Reveal answer & reasoning
Yes.
A closed box in a uniform electric field has equal entering and leaving flux.
ORIGINAL PRACTICE · 6
For a uniform infinite line charge, how does E change when radial distance triples?
Reveal answer & reasoning
It becomes one third.
Cylindrical symmetry gives E = λ/(2πε₀r), rather than an inverse-square law.
ORIGINAL PRACTICE · 7
A uniform solid nonconducting sphere has charge Q and radius R. What is E at its centre?
Reveal answer & reasoning
Zero.
Inside E = kQr/R³, which vanishes at r = 0 by symmetry.
ORIGINAL PRACTICE · 8
A dipole is perpendicular to a uniform field. What are net force and torque magnitude?
Reveal answer & reasoning
Net force zero; torque pE.
Equal and opposite forces form a couple. The maximum torque occurs at 90°.
ORIGINAL PRACTICE · 9
An electron initially at rest is placed in a rightward uniform field. Which way does it accelerate?
Reveal answer & reasoning
Leftward.
Its negative charge makes F = qE opposite the field.
ORIGINAL PRACTICE · 10
Two widely separated connected conducting spheres have radii R and 3R. In the isolated-sphere approximation, what is their charge ratio at equal potential?
Reveal answer & reasoning
1:3.
Equal kQ/R gives charge proportional to radius, not equal charge.
ORIGINAL PRACTICE · 11
A dipole’s far axial field is E at distance r. What is its far equatorial field magnitude at 2r?
Reveal answer & reasoning
E/16.
Axial E = 2kp/r³; equatorial at 2r is kp/(8r³), which is one sixteenth as large.
ORIGINAL PRACTICE · 12
What external work slowly rotates a dipole from perpendicular to anti-aligned?
Reveal answer & reasoning
+pE.
Energy rises from zero to +pE. Quasistatic external work equals that increase.
KEEP THE BIG IDEAS
Quick revision
Field is a local vector; flux is a signed surface integral.
Gauss’s law is general, but easy field extraction requires symmetry.
Electrostatic conductors are equipotentials with zero field in their material.
Dipoles and particle trajectories require explicit geometry, sign and approximation checks.
The teaching notes, worked explanations and original practice are written for this lesson. Separately labeled selected exam questions identify their exam, year, source and whether wording is paraphrased or an exact excerpt. Educational references do not certify current exam coverage.