PHYSICA

JEE Main chapters · Class 12 · Physics

Electrostatics: JEE Main Physics formulas and notes

Electrostatics (with capacitors) is one of the most important Class 12 chapters in JEE Mains: it usually gives 2 to 4 questions every year, almost always from capacitors (series, parallel, dielectrics, energy), Gauss’s law with spheres and shells, potential and potential energy of charges, and the dipole. The formulas are short; the marks come from choosing the correct symmetry and from sign care. About 14 hours is enough.

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Formulas with solved examples

Charge and Coulomb’s law

Properties of charge

How to apply. Convert between charge and the number of electrons with n = q ÷ e. Add charges with their signs; when two identical conductors touch, the total charge is shared equally.

Example. A body with 5 × 10¹² extra electrons has q = −5 × 10¹² × 1.6 × 10⁻¹⁹ = −8 × 10⁻⁷ C. A charge of 3.2 μC is 3.2 × 10⁻⁶ ÷ 1.6 × 10⁻¹⁹ = 2 × 10¹³ electrons.

Coulomb’s law

How to apply. Put the magnitudes of the charges in the formula for the size, and take the direction from the signs. In a medium divide by K; if the same pair is moved from air to a liquid and the force drops to 1/K, the liquid has constant K.

Example. 2 μC and 3 μC at 30 cm in air: F = 9 × 10⁹ × 2 × 10⁻⁶ × 3 × 10⁻⁶ ÷ 0.3² = 0.6 N. In water (K = 80) the force is 0.6 ÷ 80 = 7.5 × 10⁻³ N.

Superposition and equilibrium of charges on a line

How to apply. Equate the two forces on the third charge (kq₁q/x² = kq₂q/(d − x)²) to get √q₁ ÷ x = √q₂ ÷ (d − x). The position does not depend on the third charge or its sign. For all three to be free, also balance a force on an outer charge.

Example. +4 μC and +9 μC, 30 cm apart: x = 30 × 2 ÷ (2 + 3) = 12 cm from the 4 μC. For all three to be in equilibrium the middle charge is −(4 × 9) ÷ 5² = −1.44 μC.

Charged balls on strings and in a liquid

How to apply. Draw the three forces on a ball (weight, tension, Coulomb force) and use tanθ = F ÷ mg. For a small angle replace tanθ by x ÷ 2L. When the angle is unchanged on immersing, equate the tangents: F ÷ mg in air with (F ÷ K) ÷ (mg − upthrust) in the liquid, which gives K(mg − upthrust) = mg.

Example. q = 10⁻⁸ C on each ball of mass 0.18 g, L = 1 m: x³ = 2 × 9 × 10⁹ × 10⁻¹⁶ × 1 ÷ (1.8 × 10⁻⁴ × 10) = 10⁻³ m³, so x = 0.10 m. Balls of density 1600 kg/m³ keep the same angle in oil of density 800 kg/m³ when K = 1600 ÷ 800 = 2.

Charges at the corners of a square or polygon

How to apply. Resolve the forces of the neighbours along and perpendicular to the diagonal, using the symmetry, and balance the net outward push against the pull of the centre charge, which is a distance a/√2 away.

Example. Four charges of +4 μC at the corners of a square: the central charge needed is Q = −4 × (1 + 2.828) ÷ 4 = −3.83 μC.

Electric field

Electric field of a point charge

How to apply. Find the size of each field from kq/r², draw its direction, resolve into x and y parts and add. A point of zero field exists only where the fields point in opposite directions.

Example. The field 0.3 m from a 2 nC charge: E = 9 × 10⁹ × 2 × 10⁻⁹ ÷ 0.09 = 200 N/C. The force on a 5 μC charge placed there is 200 × 5 × 10⁻⁶ = 10⁻³ N.

Two equal charges: field on the bisector

How to apply. Add the two fields as vectors; the components along the line cancel by symmetry (like charges) or add (unlike charges). To find the maximum set dE/dy = 0.

Example. q = 2 μC at (±0.1 m, 0): at y = 0.1 m E = 2 × 9 × 10⁹ × 2 × 10⁻⁶ × 0.1 ÷ (0.02)^3/2 = 1.27 × 10⁶ N/C; the maximum is 4 × 9 × 10⁹ × 2 × 10⁻⁶ ÷ (3√3 × 0.01) = 1.39 × 10⁶ N/C, at y = 0.0707 m.

Electric field lines

How to apply. Use the picture to compare fields: closer lines mean a stronger field. Lines run from high potential to low potential.

Example. For +4q and −q, if 4n lines leave +4q then n of them end on −q, and 3n go to infinity.

Ring and arc

How to apply. Use symmetry: only the component along the axis survives, so E = ∫dE cosθ with cosθ = x/√(R² + x²). For an arc the perpendicular components cancel, and the length of the chord (2R sinα) replaces the circumference.

Example. Ring Q = 10 nC, R = 0.1 m, x = 0.1 m: E = 9 × 10⁹ × 10⁻⁸ × 0.1 ÷ (0.02)^3/2 = 3182 N/C. The maximum on the axis is 2 × 9 × 10⁹ × 10⁻⁸ ÷ (3√3 × 0.01) = 3464 N/C.

Disc, sheet and plates

How to apply. Add the sheet fields as vectors with directions: away from a positive sheet, towards a negative one. The field of an infinite sheet does not depend on the distance; a disc behaves like a sheet when x ≪ R.

Example. σ = 2 μC/m²: sheet E = 2 × 10⁻⁶ ÷ (2 × 8.85 × 10⁻¹²) = 1.13 × 10⁵ N/C. A disc with R = 0.3 m at x = 0.4 m: E = 1.13 × 10⁵ × (1 − 0.4/0.5) = 2.26 × 10⁴ N/C.

Line charge and rod

How to apply. Take a small element dq = λ dx, write its field and integrate with the angle as the variable. For the infinite line Gauss’s law with a cylinder gives the result faster.

Example. λ = 3 nC/m at r = 0.6 m: E = 2 × 9 × 10⁹ × 3 × 10⁻⁹ ÷ 0.6 = 90 N/C. A rod of 4 nC and length 0.2 m, point on its axis 0.1 m from the near end: E = 9 × 10⁹ × 4 × 10⁻⁹ ÷ (0.1 × 0.3) = 1200 N/C.

Charged particle in a uniform field

How to apply. Treat it like projectile motion: constant acceleration along E, constant speed across it. Neglect gravity for electrons and ions unless the problem includes the weight.

Example. An electron with v₀ = 2 × 10⁷ m/s enters plates 5 cm long with E = 2000 N/C: y = 1.6 × 10⁻¹⁹ × 2000 × 0.05² ÷ (2 × 9.1 × 10⁻³¹ × (2 × 10⁷)²) = 1.1 × 10⁻³ m. An oil drop of 9.6 × 10⁻¹⁵ kg in E = 6 × 10⁵ N/C balances with q = 9.6 × 10⁻¹⁵ × 10 ÷ (6 × 10⁵) = 1.6 × 10⁻¹⁹ C, one electron.

Small oscillations of a charge near a ring or a pair of charges

How to apply. Expand the field for a small displacement (x ≪ R, so (R² + x²)^3/2 ≈ R³), write F = −(constant)x and read ω² = constant ÷ m.

Example. Ring of 10 nC, R = 0.1 m, a particle of charge −1 nC and mass 1 mg: ω² = 9 × 10⁹ × 10⁻⁸ × 10⁻⁹ ÷ (10⁻³ × 10⁻⁶) = 90 s⁻², T = 2π ÷ √90 = 0.66 s.

Electric dipole

Field of a dipole

How to apply. Find the distance r from the centre of the dipole and whether the point is on the axis or on the equatorial line. For any other angle resolve p into a component along the line and one perpendicular to it.

Example. q = 2 μC, 2a = 2 mm so p = 4 × 10⁻⁹ C m. At r = 0.2 m: axial E = 2 × 9 × 10⁹ × 4 × 10⁻⁹ ÷ 0.008 = 9000 N/C and equatorial E = 4500 N/C.

Dipole in an electric field: torque and energy

How to apply. Take θ as the angle between p⃗ and E⃗ and write U = −pE cosθ. The dipole turns to line up with E⃗ and oscillates in SHM if disturbed a little.

Example. p = 4 × 10⁻⁹ C m in E = 5000 N/C at 30°: τ = 4 × 10⁻⁹ × 5000 × 0.5 = 10⁻⁵ N m. Turning it from 0° to 90° costs W = pE = 2 × 10⁻⁵ J.

Gauss’s law

Electric flux and Gauss’s law

How to apply. Choose a closed (Gaussian) surface through the point where E is wanted, with the same symmetry as the charge, so that E is constant on it and perpendicular or parallel to it. Then E × area = q(in) ÷ ε₀.

Example. A uniform field of 500 N/C at 60° to the normal through a plate of 0.2 m²: Φ = 500 × 0.2 × cos60° = 50 N m²/C. A closed surface enclosing +3 nC and −1 nC: Φ = 2 × 10⁻⁹ ÷ (8.85 × 10⁻¹²) = 226 N m²/C.

Flux through a cube, a disc and a surface

How to apply. Use symmetry: imagine the charge placed at the centre of a larger cube made of identical cubes (8 for a corner, 4 for an edge, 2 for a face), then divide q/ε₀ by that number.

Example. q = 5.31 nC at the centre of a cube: each face has 5.31 × 10⁻⁹ ÷ (6 × 8.85 × 10⁻¹²) = 100 N m²/C. With q at a corner, each far face has 5.31 × 10⁻⁹ ÷ (24 × 8.85 × 10⁻¹²) = 25 N m²/C.

Spheres and shells

How to apply. Draw a spherical Gaussian surface of radius r and find the charge inside it. For a solid sphere that charge is Q(r/R)³; for a shell it is zero if r < R.

Example. Q = 8 μC, R = 0.2 m (solid insulating): at r = 0.1 m E = 9 × 10⁹ × 8 × 10⁻⁶ × 0.1 ÷ 0.2³ = 9 × 10⁵ N/C; at r = R it is 1.8 × 10⁶ N/C; at r = 0.4 m it is 4.5 × 10⁵ N/C.

Cylinder and slab

How to apply. Use a coaxial cylinder (for line and cylinder) or a pillbox (for sheets and slabs) as the Gaussian surface. For the pillbox, flux comes out through both ends.

Example. A slab of ρ = 4 × 10⁻⁸ C/m³ and d = 0.2 m: outside E = 4 × 10⁻⁸ × 0.2 ÷ (2 × 8.85 × 10⁻¹²) = 452 N/C; at x = 0.05 m from the middle plane E = 4 × 10⁻⁸ × 0.05 ÷ (8.85 × 10⁻¹²) = 226 N/C.

Non-uniform charge density

How to apply. Integrate shell by shell: a thin shell of radius r and thickness dr carries 4πr²ρ dr. Then apply Gauss’s law with a sphere of radius r. Check that E is continuous at r = R.

Example. ρ = Ar with A = 2 × 10⁻⁶ C/m⁴ at r = 0.1 m: E = 2 × 10⁻⁶ × 0.01 ÷ (4 × 8.85 × 10⁻¹²) = 565 N/C.

Cavity in a charged sphere

How to apply. Replace the cavity by two superposed spheres of density +ρ and −ρ. Inside the cavity E⃗ = ρr⃗₁/3ε₀ − ρr⃗₂/3ε₀ = ρ(r⃗₁ − r⃗₂)/3ε₀ = ρd⃗/3ε₀, which does not depend on the position in the cavity.

Example. ρ = 5 × 10⁻⁶ C/m³ and d = 4 cm: E = 5 × 10⁻⁶ × 0.04 ÷ (3 × 8.85 × 10⁻¹²) = 7530 N/C, the same at every point of the cavity.

Potential and potential energy

Potential, field and the relation between them

How to apply. To find E from V(x, y, z) differentiate with respect to each coordinate and put in the signs; to find V from E integrate along any path (the work is path independent).

Example. V = 2x² + 3y volt: E(x) = −4x, E(y) = −3; at x = 1 m, E = (−4, −3) V/m, and |E| = √(16 + 9) = 5 V/m.

Potential of charge distributions

How to apply. Add the potentials of all the charges with their signs, which is simpler than adding fields. For a continuous charge use dV = k dq/r.

Example. A ring of 10 nC and R = 0.1 m at x = 0.1 m: V = 9 × 10⁹ × 10⁻⁸ ÷ √0.02 = 636 V. The axial point of a dipole of p = 4 × 10⁻⁹ C m, r = 0.2 m: V = 900 V. A line of 3 nC/m: V(0.3) − V(0.6) = 2 × 9 × 10⁹ × 3 × 10⁻⁹ × ln 2 = 37.4 V.

Potential of a sphere and shells

How to apply. For a point inside a set of shells, charges on shells of radius smaller than r count as kQ/r and those on larger shells count as kQ/R. Sum all terms.

Example. Q = 8 μC, R = 0.2 m: shell V = 9 × 10⁹ × 8 × 10⁻⁶ ÷ 0.2 = 3.6 × 10⁵ V. Insulating sphere: centre 5.4 × 10⁵ V; r = 0.1 m gives 4.95 × 10⁵ V. Shells of +4 nC (0.1 m) and −2 nC (0.2 m): at r = 0.15 m V = 9 × 10⁹ × (4 × 10⁻⁹ ÷ 0.15 − 2 × 10⁻⁹ ÷ 0.2) = 150 V.

Potential energy of a system of charges

How to apply. List every pair, find its distance and add the pair energies with their signs. For N identical charges the number of pairs is N(N − 1)/2.

Example. Four charges of 2 μC at the corners of a square of side 0.3 m: U = (4 + 1.414) × 9 × 10⁹ × (2 × 10⁻⁶)² ÷ 0.3 = 0.650 J.

Work-energy theorem for charges

How to apply. Conserve energy: K + U is constant, with U = qV. For two free bodies also conserve momentum. At the turning point the speed is zero (closest approach).

Example. A 5 MeV alpha particle (2e) heading for gold (79e): r(min) = 9 × 10⁹ × 79 × 2 × (1.6 × 10⁻¹⁹)² ÷ (5 × 1.6 × 10⁻¹³) = 4.55 × 10⁻¹⁴ m. Two 1 g particles with 2 μC each released at 0.2 m: each moves at v = √(9 × 10⁹ × 4 × 10⁻¹² ÷ (10⁻³ × 0.2)) = 13.4 m/s.

Equipotential surfaces and graphs of E and V

How to apply. Read E from the slope of a V–x graph (with the minus sign), and V from the area under an E–x graph. The zero of E on a graph is where V has an extreme value.

Example. Planes of 10 V and 20 V are 5 cm apart in a uniform field: E = 10 ÷ 0.05 = 200 V/m. If E = 200 V/m up to 0.3 m and 100 V/m from 0.3 m to 0.5 m, V falls by 60 + 20 = 80 V over 0.5 m.

Self energy of a sphere or shell

How to apply. Use u = ½ε₀E² integrated over the volume (shell element 4πr² dr) or build up the charge shell by shell with dW = V dq.

Example. A shell of 8 μC and R = 0.2 m: U = 9 × 10⁹ × (8 × 10⁻⁶)² ÷ (2 × 0.2) = 1.44 J. The same charge uniformly in a solid sphere: U = 3 × 9 × 10⁹ × (8 × 10⁻⁶)² ÷ (5 × 0.2) = 1.73 J.

Conductors

Conductors in electrostatic equilibrium

How to apply. Use these facts as the starting rules: the potential of a conductor is one number, and its charge is on the outside. For the pressure on a charged soap bubble or a conducting sphere use P = ½ε₀E².

Example. A sphere of radius 0.1 m and charge 8 nC: E at the surface = 9 × 10⁹ × 8 × 10⁻⁹ ÷ 0.01 = 7200 N/C, and the outward pressure ½ε₀E² = 0.5 × 8.85 × 10⁻¹² × 7200² = 2.3 × 10⁻⁴ Pa.

Induced charges and shells with charges inside

How to apply. Add the induced charges layer by layer using Gauss’s law inside the metal (E = 0 there). Then find the potential by adding contributions from each surface. Earthing means V = 0, not Q = 0.

Example. a = 0.1 m, b = 0.2 m, c = 0.3 m, q = 5 nC on the inner sphere and Q = 3 nC on the shell: outer surface 8 nC, V(shell) = 9 × 10⁹ × 8 × 10⁻⁹ ÷ 0.3 = 240 V, V(sphere) = 240 + 9 × 10⁹ × 5 × 10⁻⁹ × (10 − 5) = 465 V. If the shell is earthed: V(sphere) = 225 V.

Spheres joined by a wire; charge sharing

How to apply. Total charge is conserved. Equal potential means kq₁/R₁ = kq₂/R₂. Compute the energies before and after to get the heat lost.

Example. A 0.1 m sphere with 6 μC joined to an uncharged 0.2 m sphere: V = 9 × 10⁹ × 6 × 10⁻⁶ ÷ 0.3 = 1.8 × 10⁵ V; charges 2 μC and 4 μC. Energy changes from 1.62 J to 0.18 + 0.36 = 0.54 J, so 1.08 J is lost.

Earthing and induced charge on an earthed sphere

How to apply. Write the potential at a convenient point (usually the centre) as a sum of all contributions and set it equal to the given V (0 for earth). The charge on a surface far from the others is found from this single equation.

Example. An earthed sphere of R = 0.1 m and a point charge of +2 μC at 0.5 m from its centre: induced charge = −2 × 0.1 ÷ 0.5 = −0.4 μC.

Capacitance and combinations

Capacitance and units

How to apply. Find the potential difference that a charge Q produces between the conductors (use Gauss’s law); C = Q/V is then constant. For a single conductor the second plate is at infinity.

Example. A sphere of radius 9 cm in air: C = 0.09 ÷ (9 × 10⁹) = 10⁻¹¹ F = 10 pF. At 1000 V it holds Q = 10 nC.

Parallel plate capacitor

How to apply. Take the area of overlap and the separation in metres. A change of d, A or K changes C in the same proportion. Plate charges follow Q = CV.

Example. A = 100 cm² = 10⁻² m² and d = 1 mm: C = 8.85 × 10⁻¹² × 10⁻² ÷ 10⁻³ = 88.5 pF. At 100 V: Q = 8.85 nC and E = 100 ÷ 10⁻³ = 10⁵ V/m.

Spherical and cylindrical capacitors

How to apply. Find V between the surfaces from V = kq(1/a − 1/b) for spheres, or from the line-charge field for cylinders, and use C = Q/V. Check which conductor is earthed.

Example. a = 0.1 m, b = 0.2 m: C = 0.1 × 0.2 ÷ (9 × 10⁹ × 0.1) = 22.2 pF. If the inner sphere is earthed instead: C = 4π × 8.85 × 10⁻¹² × 0.04 ÷ 0.1 = 44.5 pF. Cylinder a = 1 mm, b = 10 mm: C per metre = 5.56 × 10⁻¹¹ ÷ ln 10 = 24.1 pF/m.

Series and parallel combinations

How to apply. Join the points of equal potential (a wire joins them), then simplify step by step. In series every capacitor has the same charge; in parallel every one has the same voltage. Find the total, then work backwards.

Example. 2 μF and 3 μF in series across 100 V: C = 6 ÷ 5 = 1.2 μF, Q = 120 μC on each, and V = 60 V on the 2 μF and 40 V on the 3 μF.

Parallel plates: charge on each face

How to apply. Find the total charge, put half on each outer face, then work plate by plate from one end, because E must be zero inside every metal plate.

Example. Plates of +6 μC and +2 μC: outer faces 4 μC each; plate 1 has 4 μC on its left face and 2 μC on its right face; plate 2 has −2 μC on its left face and 4 μC on its right face (sum 2 μC). The gap field is 2 × 10⁻⁶ ÷ (ε₀A).

Energy and dielectrics

Energy stored and energy density

How to apply. Pick the form of U that contains the quantity that stays constant (V for a battery, Q for an isolated capacitor). The loss of energy on charging is always ½CV².

Example. 4 μF at 200 V: U = ½ × 4 × 10⁻⁶ × 200² = 0.08 J, the battery gave 0.16 J and 0.08 J was lost as heat. A field of 10⁵ V/m in air stores u = ½ × 8.85 × 10⁻¹² × 10¹⁰ = 0.0443 J/m³.

Dielectric between the plates

How to apply. First decide what is held constant: Q (battery off) or V (battery on). Then change C by the factor K and find Q, V, E and U from Q = CV.

Example. An air capacitor charged to 300 V and disconnected is filled with oil, K = 3: V = 100 V and E falls to 1/3. With Q = 3 μC the induced charge is Q′ = 3 × (1 − 1/3) = 2 μC.

Battery connected or disconnected

How to apply. Write which of Q and V stays fixed, then express U, E and the work in terms of the changing C. Energy balance: W(battery) = ΔU + W(mechanical).

Example. C₀ = 10 μF at 100 V: U₀ = 0.05 J. Fill with K = 4 with the battery on: Q = 4 mC, U = 0.2 J, the battery gives 3 × 10⁻³ × 100 = 0.3 J, so 0.15 J is mechanical work. Battery off: U = 0.05 ÷ 4 = 0.0125 J.

Dielectric or conducting slab between the plates

How to apply. Treat the layers (air, slab, air) as capacitors in series, or equivalently take the effective gap d − t + t/K. A thin metal sheet of thickness t reduces the gap by t.

Example. A = 0.01 m², d = 5 mm, t = 3 mm of K = 3: d − t + t/K = 2 + 1 = 3 mm, so C = 8.85 × 10⁻¹² × 0.01 ÷ 0.003 = 29.5 pF. With a metal slab: 2 mm gives 44.3 pF.

Layers and side-by-side dielectrics

How to apply. Decide if the dielectrics are stacked along the field (add dᵢ/Kᵢ) or side by side (add Kᵢ Aᵢ/d). Mixed cases are series and parallel combinations of the pieces.

Example. A = 0.01 m²; layers 2 mm (K = 2) and 3 mm (K = 6): Σd/K = 1 + 0.5 = 1.5 mm, C = 8.85 × 10⁻¹² × 0.01 ÷ (1.5 × 10⁻³) = 59 pF. Two halves with K = 2 and K = 4 side by side, d = 1 mm: C = 8.85 × 10⁻¹² × (2 × 0.005 + 4 × 0.005) ÷ 10⁻³ = 265 pF.

Force between the plates

How to apply. Use half of the field between the plates when finding the force on one plate. Then work = force × displacement for small changes, or the energy change for large ones.

Example. A = 0.01 m², d = 1 mm, V = 1000 V: E = 10⁶ V/m, F = ½ × 8.85 × 10⁻¹² × 0.01 × (10⁶)² = 0.0443 N.

Joining, charging and circuits

Joining charged capacitors and the heat loss

How to apply. Conserve the charge on the joined plates, put the two capacitors at the same final V, and find the energy before and after. Take care of the sign for opposite connection.

Example. 2 μF at 100 V joined to 3 μF at 50 V (same polarity): V = (200 + 150) ÷ 5 = 70 V; loss = 2 × 10⁻⁶ × 3 × 10⁻⁶ × 50² ÷ (2 × 5 × 10⁻⁶) = 1.5 × 10⁻³ J. With opposite polarity V = (200 − 150) ÷ 5 = 10 V.

Charging and discharging through a resistor

How to apply. At t = 0 an uncharged capacitor behaves as a wire, and after a long time every capacitor branch carries no current. For other times use the exponentials with τ = RC.

Example. R = 1 kΩ, C = 100 μF, V = 10 V: τ = 0.1 s, final charge 1 mC. At t = 0.1 s: q = 1 × (1 − e⁻¹) = 0.632 mC and I = (10 ÷ 1000) × e⁻¹ = 3.68 mA.

Steady-state circuits with capacitors

How to apply. Redraw the circuit with the capacitors removed, find the currents with Ohm’s law, then put the voltages across the capacitors from the potentials at their ends. Q = CV for each capacitor.

Example. 10 V battery with 3 Ω and 2 Ω in series; a 5 μF capacitor across the 2 Ω resistor: I = 10 ÷ 5 = 2 A, V(C) = 4 V, Q = 5 × 4 = 20 μC.

Most asked question types

How do I solve “Coulomb force and equilibrium of charges” questions in JEE Main?

Two or three point charges lie on a line or at the corners of a figure. Find the force on one of them, the position of equilibrium or the third charge needed. 1. Draw the force of each charge on the one in question. 2. Add as vectors; use symmetry to cancel components. 3. For equilibrium equate the two forces and use x = d√q₁/(√q₁ + √q₂). Example: +4 μC and +9 μC, 30 cm apart: the zero-force point is 12 cm from the 4 μC.

How do I solve “Charged balls on strings, in air and in a liquid” questions in JEE Main?

Two charged balls hang apart. Find the angle, the charge or the dielectric constant of the liquid in which the angle stays the same. 1. Balance mg, tension and the Coulomb force: tanθ = F/mg. 2. In the liquid divide F by K and reduce the weight by the upthrust. 3. For the same angle K = ρ/(ρ − σ). Example: Balls of density 1600 kg/m³ in oil of 800 kg/m³: K = 2.

How do I solve “Electric field of point charges and the neutral point” questions in JEE Main?

Charges are placed at points on a line, on a square or on a triangle. Find the field at a point, where the field is zero, or where it is greatest. 1. Find each field from kq/r² with its direction. 2. Resolve into components and add. 3. For a zero use the opposite directions; for a maximum differentiate. Example: Two charges of 2 μC at (±0.1 m, 0): on the bisector at y = 0.1 m E = 1.27 × 10⁶ N/C.

How do I solve “Field of a ring, a disc, a rod or an arc” questions in JEE Main?

A charged ring, disc, rod or semicircle is given. Find the field on its axis or at its centre. 1. Use the symmetry to keep one component. 2. Take dq and integrate, or use the standard result. 3. Check the limits: far away it is kq/x², near the centre of a ring zero. Example: Ring of 10 nC, R = 0.1 m, at x = 0.1 m: E = 3182 N/C.

How do I solve “Charged particle in a uniform field” questions in JEE Main?

An electron, proton or ion enters or is released in a uniform field between plates. Find its acceleration, speed, deflection or time. 1. a = qE/m, direction from the sign of the charge. 2. Along the field use v² = u² + 2as; across it the speed is constant. 3. Use the energy relation qEd = ΔK when only the speed is asked. Example: An electron entering at 2 × 10⁷ m/s in 2000 N/C over 5 cm is deflected 1.1 mm.

How do I solve “Electric dipole: field, torque and energy” questions in JEE Main?

A dipole of given moment is placed in a field or at a point near another charge. Find the field at a point, the torque, the energy or the work to turn it. 1. Axial 2kp/r³, equatorial kp/r³ (opposite directions). 2. τ = pE sinθ, U = −pE cosθ. 3. Work = pE(cosθ₁ − cosθ₂); oscillation T = 2π√(I/pE). Example: p = 4 × 10⁻⁹ C m in 5000 N/C at 30°: τ = 10⁻⁵ N m.

How do I solve “Electric flux and Gauss’s law” questions in JEE Main?

A charge is inside or near a cube, a sphere or another closed surface. Find the flux through the surface or through a part of it. 1. Φ = q(in)/ε₀ for a closed surface. 2. Use symmetry: cube faces, 8 cubes at a corner, 4 at an edge. 3. Flux through a plane in a uniform field is EA cosθ. Example: q at a corner of a cube: q/24ε₀ through each far face.

How do I solve “Field of spheres, shells, cylinders and sheets” questions in JEE Main?

A uniformly charged sphere, shell, cylinder, wire or plane is given. Find the field at several distances and compare them. 1. Choose the Gaussian surface of the same shape. 2. Find q inside the surface: for a solid sphere Q(r/R)³. 3. Write the field and compare the ratios. Example: Solid sphere Q = 8 μC, R = 0.2 m: E(r = 0.1) = 9 × 10⁵ N/C, E(R) = 1.8 × 10⁶ N/C.

How do I solve “Non-uniform charge density and cavity” questions in JEE Main?

The density depends on r (ρ = Ar or ρ₀(1 − r/R)) or a sphere has a spherical hole. Find the field or the position where it is maximum. 1. Integrate 4πr²ρ dr to get q(r). 2. E = kq(r)/r²; differentiate for the maximum. 3. For a cavity: E = ρd/3ε₀, uniform and along d. Example: ρ = Ar: E = Ar²/4ε₀ inside; with A = 2 × 10⁻⁶, r = 0.1 m: 565 N/C.

How do I solve “Potential of charges, rings and shells” questions in JEE Main?

Charges, a ring or a charged conducting sphere are given. Find the potential at a point or the potential difference between two points. 1. V = kq/r for each charge; add with signs. 2. Inside a shell V = kQ/R; on the ring axis kQ/√(R² + x²). 3. For a solid insulating sphere V(centre) = 1.5 kQ/R. Example: Shells of +4 nC (0.1 m) and −2 nC (0.2 m): at r = 0.15 m, V = 150 V.

How do I solve “Potential energy of a system and work done” questions in JEE Main?

Several charges are assembled, moved or released. Find the energy of the arrangement, the work needed, or the speed gained. 1. U = Σ kqᵢqⱼ/rᵢⱼ over all pairs. 2. Work by the agent = ΔU = qΔV. 3. Conserve energy (and momentum for free particles). Example: Four 2 μC charges on a square of 0.3 m side: U = 0.650 J.

How do I solve “Equipotentials, E from V and graphs” questions in JEE Main?

V(x, y, z) or a V–x or E–x graph is given. Find E, the direction of the field, or V at some point. 1. E = −dV/dx for each coordinate. 2. Read slope for E and area for ΔV. 3. E is perpendicular to the equipotential, towards lower V. Example: V = 2x² + 3y: at x = 1, E = (−4, −3), |E| = 5 V/m.

How do I solve “Conducting spheres, shells and charge sharing” questions in JEE Main?

Spheres or shells are connected by wires, earthed, or have charges inside cavities. Find charges, potentials or the heat lost. 1. Use conservation of charge and equal potential. 2. Induced charges: −q inside, +q outside the shell. 3. Earthing: set V = 0 to find the charge. Example: 6 μC on a 0.1 m sphere joined to an uncharged 0.2 m sphere: 2 μC and 4 μC; 1.08 J lost.

How do I solve “Capacitance of geometry and combinations” questions in JEE Main?

A capacitor of a given shape or a network of capacitors is given. Find the capacitance between two points, the charge or the voltage on one of them. 1. Parallel plate ε₀A/d, sphere 4πε₀R, spherical 4πε₀ab/(b − a). 2. Simplify series and parallel groups step by step. 3. Q is the same in series and V is the same in parallel. Example: 2 μF and 3 μF in series across 100 V: 60 V and 40 V.

How do I solve “Dielectric slab, conducting slab and layers” questions in JEE Main?

A slab or several layers fill the gap of a parallel plate capacitor. Find C, the field, the charge or the energy, with the battery on or off. 1. C = ε₀A/(d − t + t/K); layers: Σdᵢ/Kᵢ. 2. Say what stays constant: V (battery on) or Q (battery off). 3. Find the energy and the work done on the slab. Example: A = 0.01 m², d = 5 mm, slab 3 mm of K = 3: C = 29.5 pF.

How do I solve “Energy, force and heat loss in capacitors” questions in JEE Main?

A capacitor is charged, its plates are pulled apart, or two capacitors are joined. Find the energy stored, the force or the heat produced. 1. U = ½CV² or Q²/2C, as the case requires. 2. Force between plates Q²/2ε₀A. 3. Joined: conserve charge and find the common V. Example: 2 μF at 100 V joined to 3 μF at 50 V: V = 70 V, loss 1.5 mJ.

How do I solve “Charging, discharging and steady-state circuits” questions in JEE Main?

A capacitor is in a circuit with resistors. Find the charge or current at a time, or the charge once a steady state has been reached. 1. Steady state: no current through the capacitor. 2. Voltage across it = potential difference of its ends. 3. Time dependence: q = Q(1 − e^−t/RC) or Qe^−t/RC. Example: R = 1 kΩ, C = 100 μF, 10 V: at t = RC the charge is 0.632 mC.

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Simulations for Electrostatics

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