Electrostatics: Charge, Fields, Gauss's Law, and Capacitance

Situation: You are asked for the total charge on an object made of extra or missing electrons.

Use q = ne. Why: total charge comes from the number of elementary charges. Say: q equals n e. q = total charge; n = number of extra or missing electrons/protons; e = elementary charge = 1.60 × 10^-19 C. If electrons were added, the total charge is negative. If electrons were removed, the total charge is positive.

Situation: You need the charge conservation idea for rubbing, touching, or transfer problems.

Use q_total,before = q_total,after. Why: charge is conserved. It can move, but it is not created or destroyed in ordinary physics problems. Say: total charge before equals total charge after.

Situation: A charged object is brought near a conductor and charges shift around without touching.

Use induction reasoning. Why: charges in a conductor move and redistribute when an external charge is nearby. Remember: conductors let charge move, insulators do not let charge move freely, and grounding lets excess charge leave or enter.

Situation: Two point charges are a distance r apart and you need the electric force between them.

Use F = k|q1q2|/r^2. Why: Coulomb's law gives the magnitude of the force between two point charges. Say: F equals k times the absolute value of q one q two over r squared. F = electric force magnitude; k = 8.99 × 10^9 N·m^2/C^2; q1 and q2 = charges; r = distance. Same signs repel, opposite signs attract.

Situation: There are more than two point charges and you need the net force on one of them.

Use Coulomb's law on each pair, then vector addition. Why: electric forces obey superposition. Steps: find each force with F = k|q1q2|/r^2, resolve into components if needed, add components, then rebuild magnitude and direction.

Situation: A charge placed at a point feels a force, and you want the electric field at that point.

Use E = F/q. Why: electric field is force per unit charge. Say: E equals F over q. E = electric field; F = force on the test charge; q = test charge. Field direction is the direction a positive test charge would move.

Situation: A point charge creates a field, and you need the field strength at distance r.

Use E = k|q|/r^2. Why: a point charge produces an inverse-square electric field. Say: E equals k times the absolute value of q over r squared. E = electric field magnitude; k = Coulomb's constant; q = source charge; r = distance. Direction: away from positive, toward negative.

Situation: You know the electric field and a charge placed there, and you need the force on the charge.

Use F = qE. Why: force on a charge in a field equals charge times field. Say: F equals q E. If q is negative, the force points opposite the field direction.

Situation: Charge is spread out continuously, not concentrated at one point.

Use dE = k dq/r^2, then integrate. Why: break the object into tiny pieces of charge and add their field contributions. Say: d E equals k d q over r squared. Full path: choose dq, express r using geometry, keep needed components, then integrate.

Situation: Charge is spread along a line and you need charge per length.

Use λ = Q/L. Why: linear charge density tells how charge is distributed along a length. Say: lambda equals Q over L. λ = linear charge density; Q = total charge; L = length. Full solve path: 1) find λ = Q/L, 2) choose tiny length dx or dl, 3) convert to dq = λdx or λdl, 4) plug into the main formula like dE = k dq/r^2 or dV = k dq/r, 5) write r from geometry, 6) resolve components if needed, 7) integrate over the full object.

Situation: Charge is spread over a surface and you need charge per area.

Use σ = Q/A. Why: surface charge density tells how charge is distributed over an area. Say: sigma equals Q over A. Full solve path: 1) find σ = Q/A, 2) choose tiny patch dA, 3) convert to dq = σdA, 4) plug into the main formula like dE = k dq/r^2 or dV = k dq/r, 5) write r from geometry, 6) resolve components if needed, 7) integrate over the whole surface.

Situation: Charge is spread through a volume and you need charge per volume.

Use ρ = Q/V. Why: volume charge density tells how charge is distributed through a volume. Say: rho equals Q over V. Full solve path: 1) find ρ = Q/V, 2) choose tiny volume dV, 3) convert to dq = ρdV, 4) plug into the main formula like dE = k dq/r^2 or dV_potential = k dq/r, 5) write r from geometry, 6) use symmetry if possible, 7) integrate over the full volume.

Situation: You are interpreting or drawing electric field lines.

Use the rules: lines start on positive charges, end on negative charges, tangent to the line gives field direction, and closer lines mean stronger field.

Situation: You have equal and opposite charges separated by distance d and need dipole moment.

Use p = qd. Why: dipole moment measures separation of equal and opposite charge. Say: p equals q d. p = dipole moment; q = magnitude of one charge; d = separation distance. Direction goes from negative to positive.

Situation: A uniform electric field passes through a flat surface and you need electric flux.

Use Φ_E = EA cosθ. Why: electric flux measures how much electric field passes through a surface. Say: phi E equals E A cosine theta. Φ_E = electric flux; E = electric field magnitude; A = area; θ = angle between the field and the area vector. θ = 0° gives maximum flux, θ = 90° gives zero flux.

Situation: The electric field is not uniform or the surface is curved.

Use Φ_E = ∮ E · dA. Why: use the surface-integral definition of flux for non-uniform fields or closed surfaces. Say: phi E equals closed surface integral of E dot d A.

Situation: You want the total flux through a closed surface due to enclosed charge.

Use ∮ E · dA = q_enc/ε0. Why: Gauss's law connects total flux through a closed surface to the charge inside it. Say: closed integral of E dot d A equals q enclosed over epsilon naught. q_enc = enclosed charge; ε0 = 8.85 × 10^-12 C^2/(N·m^2). Only enclosed charge matters for net flux.

Situation: You have a highly symmetric charge distribution and want electric field fast.

Use Gauss's law plus symmetry. Why: if the field has constant magnitude over the Gaussian surface, Gauss's law is much easier than integrating Coulomb's law. Full solve path: 1) identify spherical, cylindrical, or planar symmetry, 2) choose a matching Gaussian surface, 3) decide where E is constant and where E · dA = 0, 4) simplify ∮ E · dA, 5) find q_enc, 6) set ∮ E · dA = q_enc/ε0, 7) solve for E.

Situation: You need the field outside a point charge or spherical charge distribution.

Use E(4πr^2) = q_enc/ε0, then E = (1/4πε0) q_enc/r^2. Why: spherical symmetry makes E constant over a sphere. Say: E times four pi r squared equals q enclosed over epsilon naught.

Situation: You need the field near an infinite line of charge.

Use E(2πrL) = λL/ε0, so E = λ/(2π ε0 r). Why: cylindrical symmetry around a line charge. Say: E equals lambda over two pi epsilon naught r.

Situation: You need the field near an infinite sheet of charge.

Use 2EA = σA/ε0, so E = σ/(2ε0). Why: field from an infinite sheet is uniform and perpendicular on both sides. Say: E equals sigma over two epsilon naught.

Situation: You need the field inside a uniformly charged solid sphere at radius r.

Use q_enc = ρ(4/3 π r^3) and E(4πr^2) = q_enc/ε0, giving E = ρr/(3ε0). Why: only enclosed charge within radius r contributes to Gauss's law. Say: E equals rho r over three epsilon naught.

Situation: Charge is uniformly spread through a sphere of total charge Q and radius R, and you need enclosed charge at radius r less than R.

Use q_enc = Q(r^3/R^3). Why: enclosed charge is proportional to enclosed volume. Say: q enclosed equals Q times r cubed over R cubed.

Situation: You are asked about electric field inside a conductor at electrostatic equilibrium.

Use E_inside = 0. Why: charges have rearranged until there is no net field inside the conductor.

Situation: You are asked where excess charge sits on a conductor in electrostatic equilibrium.

Use the concept rule: excess charge is on the surface. Why: if excess charge stayed inside, it would keep moving until equilibrium was reached.

Situation: You need the electric field just outside a conductor from its surface charge density.

Use E = σ/ε0. Why: right outside a conductor, the field is perpendicular to the surface and depends on the surface charge density. Say: E equals sigma over epsilon naught.

Situation: Two point charges are separated by distance r and you need electric potential energy.

Use U = k q1q2/r. Why: this gives the electric potential energy of a two-charge system. Say: U equals k q one q two over r. Opposite signs give negative U; same signs give positive U.

Situation: You need the work required to assemble several point charges from far away.

Add the pairwise potential energies. Why: total electric potential energy of a system is the sum over interacting pairs, k q_i q_j / r_ij.

Situation: You know potential energy and charge and need electric potential.

Use V = U/q. Why: electric potential is potential energy per unit charge. Say: V equals U over q.

Situation: A charge moves between two points and you need potential difference.

Use ΔV = ΔU/q. Why: potential difference is change in potential energy per unit charge. Say: delta V equals delta U over q.

Situation: You know work done by the electric field and need the potential difference.

Use ΔV = -W_field/q. Why: work done by the field lowers electric potential energy. Say: delta V equals negative W over q.

Situation: A charge moves through a voltage and you need the gained or lost energy.

Use ΔU = qΔV. Why: energy change equals charge times potential difference. Say: delta U equals q delta V.

Situation: A point charge creates a potential and you need the electric potential at distance r.

Use V = kq/r. Why: a point charge creates scalar electric potential. Say: V equals k q over r. Potential keeps the sign of q.

Situation: There are several point charges and you need the total electric potential at a point.

Use V = Σ k q_i/r_i. Why: electric potential is a scalar, so add algebraically. Say: V equals the sum of k q i over r i.

Situation: You have a continuous charge distribution and need electric potential.

Use V = ∫ k dq/r. Why: break the distribution into tiny charge pieces and add the scalar potentials. Full solve path: 1) decide line/surface/volume, 2) write λ = Q/L or σ = Q/A or ρ = Q/V, 3) convert to dq = λdx or λdl or σdA or ρdV, 4) substitute into V = ∫ k dq/r, 5) write r in terms of the integration variable, 6) set limits, 7) integrate.

Situation: You are given an electric dipole and asked for its dipole moment.

Use p = qd. Why: dipole moment is charge times separation. Say: p equals q d.

Situation: You need the electric potential due to a dipole far away from it.

Use V = k p cosθ / r^2. Why: far from a dipole, this gives the dipole's potential. Say: V equals k p cosine theta over r squared.

Situation: You know how potential changes with position and need the x-component of electric field.

Use E_x = -dV/dx. Why: electric field points in the direction of decreasing potential. Say: E x equals negative d V over d x.

Situation: You know potential as a full function of position and need the vector electric field.

Use E = -∇V. Why: electric field is the negative gradient of potential. Say: E equals negative grad V.

Situation: You know the electric field and need potential difference between two points.

Use ΔV = -∫ E · dl. Why: potential difference comes from integrating the electric field along a path. Say: delta V equals negative integral of E dot d l.

Situation: You are checking whether charges moving along a surface do electric work.

On an equipotential surface, use ΔV = 0, so W = 0 for motion along it. Why: no change in potential means no work by the electric field for motion along the surface.

Situation: You need the energy from an electron accelerated through a potential difference.

Use ΔU = qΔV. Why: this is the main electrostatics energy equation behind accelerated-charge applications.

Situation: A capacitor stores charge Q at voltage V, and you need capacitance.

Use C = Q/V. Why: capacitance tells how much charge is stored per volt. Say: C equals Q over V. C = capacitance; Q = stored charge magnitude on one plate; V = potential difference.

Situation: You have a parallel-plate capacitor with plate area A and plate spacing d in vacuum or air.

Use C = ε0A/d. Why: parallel-plate capacitance increases with area and decreases with plate separation. Say: C equals epsilon naught A over d.

Situation: You have a spherical capacitor with inner radius a and outer radius b.

Use C = 4π ε0 ab/(b - a). Why: this is the vacuum capacitance of concentric spheres.

Situation: You have a cylindrical capacitor with inner radius a, outer radius b, and length L.

Use C = 2π ε0 L / ln(b/a). Why: this is the vacuum capacitance of coaxial cylinders.

Situation: Several capacitors are connected in series and you need one equivalent capacitor.

Use 1/C_eq = 1/C1 + 1/C2 + 1/C3 + ... Why: series capacitors all carry the same charge, and the total voltage splits. Say: one over C equivalent equals one over C one plus one over C two plus ...

Situation: Several capacitors are connected in parallel and you need one equivalent capacitor.

Use C_eq = C1 + C2 + C3 + ... Why: parallel capacitors all share the same voltage, and their stored charges add. Say: C equivalent equals C one plus C two plus C three ...

Situation: You know the capacitor's charge and voltage and need stored energy.

Use U = 1/2 QV. Why: one form of capacitor energy. Say: U equals one half Q V.

Situation: You know charge and capacitance and need capacitor energy.

Use U = Q^2/(2C). Why: same energy, rewritten in terms of Q and C. Say: U equals Q squared over two C.

Situation: You know capacitance and voltage and need capacitor energy.

Use U = 1/2 C V^2. Why: same energy, rewritten in terms of C and V. Say: U equals one half C V squared.

Situation: You need energy per volume stored in the electric field between capacitor plates.

Use u_E = 1/2 ε0 E^2. Why: electric-field energy density in vacuum. Say: u sub E equals one half epsilon naught E squared.

Situation: A dielectric is inserted between capacitor plates and you need the new capacitance.

Use C = κC0. Why: the dielectric increases capacitance by the dielectric constant factor. Say: C equals kappa C naught. κ = dielectric constant; C0 = original capacitance without dielectric.

Situation: You know the original electric field in a vacuum capacitor and need the field inside after inserting a dielectric.

Use E = E0/κ. Why: the dielectric reduces the net electric field inside. Say: E equals E naught over kappa.

Situation: An isolated capacitor gets a dielectric inserted and you need the new stored energy.

Use U = U0/κ. Why: for an isolated capacitor, charge stays fixed, capacitance rises, and energy drops. Say: U equals U naught over kappa.

Situation: You need the induced field inside a dielectric.

Use E_ind = E0 - E = E0(1 - 1/κ). Why: the dielectric creates an opposing field, reducing the net internal field. Say: E induced equals E naught minus E.

Situation: Charge Q passes through a wire in time t, and you need current.

Use I = ΔQ/Δt. Why: current is the rate of charge flow. Say: I equals delta Q over delta t.

Situation: You need the current density in a wire carrying current uniformly over area A.

Use J = I/A. Why: current density is current per cross-sectional area. Say: J equals I over A.

Situation: You need current density from moving charges in a conductor.

Use J = nqv_d. Why: current density depends on number density, charge, and drift speed. Say: J equals n q v sub d.

Situation: You need current from carrier density and drift speed in a wire.

Use I = nqv_d A. Why: multiply current density by area. Say: I equals n q v d A.

Situation: You need drift velocity in a wire.

Use v_d = I/(nqA). Why: rearranged from I = nqv_d A. Say: v d equals I over n q A.

Situation: You know resistivity, length, and cross-sectional area and need a wire's resistance.

Use R = ρL/A. Why: resistance depends on material and geometry. Say: R equals rho L over A.

Situation: You need resistivity from measured resistance, length, and area.

Use ρ = RA/L. Why: rearranged resistance formula. Say: rho equals R A over L.

Situation: You need the microscopic form of Ohm's law linking field and current density.

Use J = σE or E = ρJ. Why: these connect material response to electric field. Say: J equals sigma E; E equals rho J. Conductivity σ = 1/ρ.

Situation: You know voltage and resistance and need current through an ohmic resistor.

Use V = IR or I = V/R. Why: Ohm's law connects voltage, current, and resistance for ohmic devices. Say: V equals I R.

Situation: You know voltage and current and need the resistance.

Use R = V/I. Why: rearranged Ohm's law.

Situation: You know current and voltage and need electric power.

Use P = IV. Why: power is the rate electrical energy is delivered. Say: P equals I V.

Situation: You know current and resistance and need power dissipated by a resistor.

Use P = I^2R. Why: power in a resistor expressed using Ohm's law. Say: P equals I squared R.

Situation: You know voltage and resistance and need resistor power.

Use P = V^2/R. Why: another Ohm's-law form of power. Say: P equals V squared over R.

Situation: You know power and time and need electrical energy used.

Use E = Pt. Why: energy equals power times time. Say: E equals P t.

Situation: A battery with internal resistance r is delivering current I, and you need terminal voltage.

Use V_term = ε - Ir. Why: internal resistance lowers the output voltage when current leaves the positive terminal. Say: V terminal equals epsilon minus I r.

Situation: A battery is being charged and current enters its positive terminal.

Use V_term = ε + Ir. Why: in charging mode, terminal voltage exceeds emf.

Situation: Several resistors are in series and you want one equivalent resistor.

Use R_eq = R1 + R2 + R3 + ... Why: same current flows through series resistors, and voltage drops add.

Situation: Several resistors are in parallel and you want one equivalent resistor.

Use 1/R_eq = 1/R1 + 1/R2 + 1/R3 + ... Why: same voltage across parallel resistors, currents add.

Situation: You are at a junction and need a current equation.

Use Σ I_in = Σ I_out. Why: charge is conserved at a junction. Say: sum of currents in equals sum of currents out.

Situation: You go around a complete loop and need the voltage equation.

Use Σ ΔV = 0. Why: energy is conserved around a closed loop. Say: sum of delta V around a loop equals zero.

Situation: You are reasoning about ideal meters.

Use the concept rules: ideal ammeter has R ≈ 0, ideal voltmeter has R ≈ ∞. Why: an ammeter should not change current much, and a voltmeter should not draw current.

Situation: A capacitor is charging through a resistor from a battery and you need charge as a function of time.

Use q(t) = Cε(1 - e^(-t/RC)). Why: this describes charging in an RC circuit. Say: q of t equals C epsilon times one minus e to the negative t over R C.

Situation: A capacitor is charging through a resistor and you need current versus time.

Use I(t) = (ε/R)e^(-t/RC). Why: charging current starts maximum and decays exponentially. Say: I of t equals epsilon over R times e to the negative t over R C.

Situation: A charged capacitor is discharging through a resistor and you need charge versus time.

Use q(t) = Q0 e^(-t/RC). Why: discharging charge decays exponentially. Say: q of t equals Q naught e to the negative t over R C.

Situation: A charged capacitor is discharging through a resistor and you need current versus time.

Use I(t) = -(Q0/RC)e^(-t/RC). Why: the negative sign shows discharge direction opposite the charging sign convention.

Situation: You need the characteristic timescale for an RC circuit.

Use τ = RC. Why: the RC time constant sets how fast charging and discharging happen. Say: tau equals R C.

Situation: A moving charge enters a magnetic field and you need magnetic force magnitude.

Use F = |q|vB sinθ. Why: magnetic force depends on charge, speed, field strength, and angle. Say: F equals absolute value of q times v B sine theta.

Situation: A moving charge in a magnetic field needs the full vector force, not just magnitude.

Use F = q v × B. Why: magnetic force uses a cross product. Say: F equals q v cross B. Direction: right-hand rule for positive charge, reverse for negative charge.

Situation: A charged particle moves perpendicular to a magnetic field and follows a circle; you need the radius.

Use r = mv/(|q|B). Why: magnetic force provides centripetal force. Say: r equals m v over absolute value of q B.

Situation: A charged particle moves perpendicular to B and you need its cyclotron angular frequency.

Use ω = |q|B/m. Why: circular magnetic motion has this angular frequency. Say: omega equals absolute value of q B over m.

Situation: A charged particle moves perpendicular to B and you need period or frequency.

Use T = 2πm/(|q|B) and f = |q|B/(2πm). Why: these come from circular magnetic motion.

Situation: A straight wire of length L carries current I in magnetic field B and you need the magnetic force.

Use F = ILB sinθ. Why: a current-carrying wire in a field experiences magnetic force. Say: F equals I L B sine theta.

Situation: You need the full vector force on a current segment in a magnetic field.

Use F = I L × B. Why: vector form of magnetic force on a straight conductor.

Situation: A current loop in a magnetic field experiences torque and you need its magnetic dipole moment.

Use μ = IA, or for N turns μ = NIA. Why: magnetic dipole moment of a loop equals current times area. Say: mu equals I A.

Situation: A current loop in a magnetic field is trying to rotate and you need torque.

Use τ = μB sinθ, or τ = NIAB sinθ for N turns. Why: magnetic torque aligns the dipole moment with the field. Say: tau equals mu B sine theta.

Situation: You need magnetic potential energy of a dipole in a field.

Use U = -μ · B or U = -μB cosθ. Why: this tells which orientation is energetically favorable.

Situation: Charges in a current-carrying strip are deflected sideways by a magnetic field and you need Hall voltage or field relation.

Use qE_H = qv_dB so E_H = v_dB, then V_H = E_H w. Why: electric and magnetic forces balance at steady Hall deflection. Say: E H equals v d B.

Situation: A current element creates a magnetic field and you need the tiny field contribution.

Use dB = (μ0/4π)(I dl sinθ/r^2). Why: Biot-Savart law gives the magnetic field from a small current segment. Say: d B equals mu naught over four pi times I d l sine theta over r squared.

Situation: You need the total magnetic field from a shaped wire carrying current.

Use B = ∫ dB using Biot-Savart law. Why: add all tiny field contributions from the current distribution. Full solve path: 1) start with dB = (μ0/4π)(I dl sinθ/r^2), 2) choose tiny wire element dl, 3) write r to the field point, 4) write sinθ from geometry, 5) decide which components cancel by symmetry, 6) keep only the surviving components, 7) integrate over the whole wire or loop.

Situation: You need the magnetic field at distance r from a long straight wire carrying current I.

Use B = μ0I/(2πr). Why: long straight wire result from Biot-Savart or Ampère's law. Say: B equals mu naught I over two pi r.

Situation: Two long parallel wires carry currents and you need force per length between them.

Use F/L = μ0 I1 I2/(2πr). Why: one wire creates a field and the other feels force from it. Same-direction currents attract; opposite-direction currents repel.

Situation: You need the magnetic field at the center of a circular loop of radius R carrying current I.

Use B = μ0I/(2R), or B = μ0NI/(2R) for N loops. Why: circular loop center-field formula.

Situation: You need the field on the axis of a circular current loop at distance x from the center.

Use B = μ0 I R^2 / [2(R^2 + x^2)^(3/2)], multiplied by N for N loops. Why: axis field of a loop.

Situation: A current configuration has strong symmetry and you need the magnetic field.

Use ∮ B · dl = μ0 I_enc. Why: Ampère's law is the magnetic analogue of Gauss's law for highly symmetric cases. Say: closed integral of B dot d l equals mu naught I enclosed.