# Electric Charges and Fields > Electric Charges and Fields for NEET-UG Physics — Coulomb's law, electric dipole, field lines, flux and Gauss's law from NCERT Class 12. - Canonical URL: https://prepelephant.com/topics/neet-ug/physics/electric-charges-and-fields - Exam / course: NEET-UG · Subject: Physics - Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team - First published: 2026-10-02 - Last updated: 2026-10-02 - How to cite: "Electric Charges and Fields", PrepElephant, https://prepelephant.com/topics/neet-ug/physics/electric-charges-and-fields ## Direct answer Charge comes in integral multiples of the electron charge — q = ne with e = 1.6 × 10^-19 C — is conserved, and pairwise attraction or repulsion follows Coulomb's law F = k q1 q2/r^2 with k = 1/(4πε0) = 9 × 10^9 N m^2 C^-2. The field E = F/q_0 maps the force per unit charge through space, the dipole is the chapter's standard extended object, and Gauss's law converts symmetric charge distributions into one-line field results. ## What you must remember - Quantisation: q = ±ne; conservation: charge is never created or destroyed, only transferred (charging by induction never touches the source). - Coulomb's law: F = (1/4πε0) q1 q2/r^2, ε0 = 8.85 × 10^-12 C^2 N^-1 m^-2; forces from multiple charges superpose pair by pair, vectorially. - Point-charge field E = kq/r^2, radially outward for positive charge; unit N/C, identical to V/m; field lines start on positive charges, end on negative ones, never cross and never loop in electrostatics. - Electric dipole: moment p = q × 2a pointing from -q to +q; axial field 2kp/r^3, equatorial field kp/r^3 — both falling as 1/r^3, faster than a point charge's 1/r^2. - Dipole in uniform field: net force zero, torque τ = pE sinθ aligning p with E. - Flux Φ = E·A cosθ (unit N m^2 C^-1); Gauss's law: total flux through any closed surface = q_enclosed/ε0, shape-independent. - Ready Gauss results: infinite line charge E = λ/(2πε0 r); infinite sheet E = σ/(2ε0) independent of distance; charged spherical shell: E = 0 everywhere inside, and outside E = kq/r^2 as though all charge sat at the centre. ## How Gauss's law earns its keep Ask for the field inside a charged spherical conducting shell. Draw a spherical Gaussian surface just inside the metal: it encloses no charge, so the total flux is zero, and by the perfect symmetry of the sphere the flux can only vanish if E = 0 at every point — no field inside, whatever the shell's charge. The same reasoning applied outside the shell gives E = kq/r^2, the entire shell masquerading as a point charge at its centre. Notice the entry ticket: symmetry. Gauss's law is always true, but it only yields numbers when the charge distribution lets you argue E is uniform and perpendicular over the chosen surface — spheres for spheres, cylinders for lines, pillboxes for sheets. That is why the three standard results above exhaust the examinable cases, and why a question asking for the field of a lumpy charged potato cannot be answered by Gauss. The dipole extends the story: at equal large distances, a dipole's field (falling as 1/r^3) beats the point charge's 1/r^2 into insignificance, which is why neutral matter's electrical effects are so short-ranged — the plus and minus almost cancel, and only the slight imbalance at close range survives. ## Where students slip Field lines are drawings, not trajectories — a charge released from rest moves along the force, which is along the field line only at the release point, and the line itself is not the path. Second, the dipole moment direction runs from negative to positive, opposite to the way intuition wants to point it; getting this backwards flips the torque's sign questions. Third, "the net force on a dipole in a uniform field is zero" coexists with a nonzero torque, and both statements appear in the options together — the pair is the point. Fourth, flux cares about enclosed charge only: moving the same charge anywhere inside the surface leaves the flux unchanged while completely changing the local field pattern. Finally, the sheet-field result E = σ/(2ε0) has no distance dependence — students who instinctively divide by r^2 again lose a free mark, and the shell's interior zero is the mirror trap. ## Frequently asked questions ### What is the value of the constant in Coulomb's law? k = 1/(4πε0) = 9 × 10^9 N m^2 C^-2, with ε0 = 8.85 × 10^-12 C^2 N^-1 m^-2 — the permittivity of free space. ### Why is the electric field inside a charged spherical shell zero? A Gaussian sphere inside the metal encloses no charge; by the symmetry of the shell the flux can vanish only if the field itself vanishes at every interior point. ### What torque acts on a dipole placed in a uniform electric field? τ = pE sinθ about the centre, tending to align the moment with the field; the net force is zero, so a uniform field turns but does not translate a dipole. ### How does the axial field of a dipole fall with distance? As 1/r^3 (E = 2kp/r^3), one power faster than a point charge's field — the two opposite charges nearly cancel at large distances. ### What does the number of electric field lines crossing a surface measure? The electric flux, proportional to E·A cosθ; over a closed surface the net count fixes the enclosed charge through Gauss's law.