Coulomb's Law
Statement
The electrostatic force between two point charges is directly proportional to the product of their magnitudes, and inversely proportional to the square of the distance between them, acting along the line joining them.
Key Properties
Unlike gravity, Coulomb's force can be attractive (opposite charges) or repulsive (like charges). It obeys the principle of superposition — net force on any charge from several others is the vector sum of individual pairwise forces.
Solved Example
Calculate the electrostatic force between two charges of $2\ \mu\text{C}$ and $3\ \mu\text{C}$ placed 0.3 m apart.
$F = k\dfrac{q_1q_2}{r^2} = (9\times10^9)\dfrac{(2\times10^{-6})(3\times10^{-6})}{(0.3)^2} = (9\times10^9)\dfrac{6\times10^{-12}}{0.09}$
$= (9\times10^9)(6.67\times10^{-11}) \approx 0.6\ \text{N}$
Q. If the distance between two point charges is doubled, how does the force between them change?
Ans: Since F ∝ 1/r², doubling the distance reduces the force to 1/4 of its original value.
Q. Why is Coulomb's force considered a "central force," and what conservation-related consequence does this have?
Ans: A central force always acts along the line joining the two interacting particles (here, the two point charges) — this is exactly what Coulomb's law describes. Since torque about the source charge is always zero for a purely central force (force and position vector are parallel/antiparallel), angular momentum about that point is conserved for a charge moving solely under this force, similar to gravitational orbital motion.
Coulomb's Law in a Medium
When two charges are placed in a medium (rather than vacuum), the force between them is reduced by a factor equal to the medium's relative permittivity (dielectric constant, $K$): $F_{medium} = \dfrac{F_{vacuum}}{K}$. This is why electrostatic forces are noticeably weaker when charges are immersed in an insulating liquid or placed within a solid dielectric, compared to the same charges and separation in vacuum or air.
Solved Example
Two charges experience a force of 10 N in vacuum. Calculate the force between them if they are instead placed in a medium of dielectric constant 5, at the same separation.
$F_{medium} = \dfrac{F_{vacuum}}{K} = \dfrac{10}{5} = 2\ \text{N}$
Q. Why does placing two charges in a dielectric medium (rather than vacuum) reduce the force between them?
Ans: The dielectric medium's molecules become polarised in response to the charges' electric field, creating induced dipoles that partially oppose (screen) the original field between the two charges — this partial screening effectively weakens the net electrostatic interaction between the charges, reducing the force compared to what it would be in vacuum, where no such polarisation/screening effect exists.
Electric Field
Definition
The electric field at a point is the force experienced per unit positive test charge placed at that point: $\vec{E} = \vec{F}/q_0$. It is a vector field that exists at every point in space around a charge, regardless of whether a test charge is actually present there.
Electric Field Due to a Point Charge
$$E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2} = k\dfrac{q}{r^2}$$
Solved Example
Calculate the electric field at a distance of 0.5 m from a point charge of $4\ \mu\text{C}$.
$E = k\dfrac{q}{r^2} = (9\times10^9)\dfrac{4\times10^{-6}}{(0.5)^2} = (9\times10^9)\dfrac{4\times10^{-6}}{0.25} = (9\times10^9)(1.6\times10^{-5})$
$= 1.44\times10^5\ \text{N/C}$
Q. Why is the electric field considered to exist at a point in space even before any test charge is actually placed there?
Ans: The electric field is treated as a genuine physical property of the space surrounding a source charge, existing independently of whether a test charge is present to detect it — the test charge is merely a conceptual tool used to define and measure the field's strength and direction at any given point; the field itself is understood to be created by the source charge and to permeate all of space around it.
Q. A test charge of 2 μC experiences a force of 0.4 N at a certain point. Calculate the electric field at that point.
Ans: E = F/q₀ = 0.4/(2×10⁻⁶) = 2×10⁵ N/C.
Electric Field Due to Multiple Charges — Superposition
The net electric field at a point due to several charges is the vector sum of the individual fields each charge would independently produce at that point.
Solved Example
Two equal charges of $+2\ \mu\text{C}$ are placed 0.4 m apart. Calculate the electric field at the midpoint between them.
Each charge produces a field of magnitude $E = k\dfrac{q}{r^2} = (9\times10^9)\dfrac{2\times10^{-6}}{(0.2)^2} = (9\times10^9)\dfrac{2\times10^{-6}}{0.04} = 4.5\times10^5\ \text{N/C}$ at the midpoint, but pointing in opposite directions (away from each positive charge, towards the other).
Since both charges are equal and the point is exactly midway, these two fields are equal in magnitude and opposite in direction, giving a net field of zero at the midpoint.
Q. Why does the electric field at the midpoint between two equal, like (same-sign) charges become exactly zero?
Ans: By symmetry, each charge produces a field of equal magnitude at the midpoint (since both charges are equal and equidistant from that point) — but since both charges have the same sign, their individual fields at the midpoint point in exactly opposite directions (each pointing away from its own source charge, towards the other charge). These two equal-magnitude, opposite-direction field vectors cancel completely, giving a net field of zero at that specific point.
Electric Field Lines & Electric Flux
Electric Field Lines
Imaginary lines used to visualise an electric field — drawn such that the tangent at any point gives the field's direction there, and the density of lines (how closely packed they are) represents the field's relative strength. Field lines always originate from positive charges and terminate on negative charges, never cross each other, and are always perpendicular to a conductor's surface.
Electric Flux
Electric flux ($\Phi_E$) is a measure of the total number of field lines passing through a given surface: $\Phi_E = \vec{E}\cdot\vec{A} = EA\cos\theta$, where $\theta$ is the angle between the field and the surface's normal (perpendicular) direction.
Solved Example
An electric field of 200 N/C passes through a flat surface of area 0.5 m², oriented so the field makes an angle of 60° with the surface's normal. Calculate the electric flux through the surface.
$\Phi_E = EA\cos\theta = (200)(0.5)\cos60° = (100)(0.5) = 50\ \text{N m}^2/\text{C}$
Q. Why can electric field lines never cross each other?
Ans: The direction of the electric field at any point is uniquely defined (a single, definite direction, given by the tangent to the field line there). If two field lines were to cross, that intersection point would have to represent two different field directions simultaneously — which is physically impossible, since the electric field at any single point can only point in one specific direction at a time.
Q. Why is the electric flux through a surface zero when the field is exactly parallel to the surface (perpendicular to its normal)?
Ans: Flux depends on Ecosθ, where θ is measured from the surface's normal direction. If the field is parallel to the surface itself, it is perpendicular to the surface's normal, meaning θ = 90° and cosθ = 0 — so no field lines actually pass through the surface in this orientation, giving zero flux, regardless of the field's actual magnitude.
Gauss's Law
Statement
The total electric flux through any closed surface (a "Gaussian surface") equals the total charge enclosed within that surface, divided by $\varepsilon_0$ — regardless of the surface's exact shape or size, and regardless of any charges located outside it.
Gauss's Law
$$\oint \vec{E}\cdot d\vec{A} = \dfrac{Q_{enclosed}}{\varepsilon_0}$$
Why Gauss's Law Is So Powerful
For situations with high symmetry (spherical, cylindrical, or planar charge distributions), Gauss's law provides a vastly simpler way to calculate electric fields than direct application of Coulomb's law summed over every charge element — by choosing a Gaussian surface that matches the symmetry, the field can often be pulled out of the flux integral entirely, reducing the problem to simple algebra.
Solved Example
Using Gauss's law, derive the electric field at a distance $r$ from an infinitely long, uniformly charged straight wire with linear charge density $\lambda$.
Choose a cylindrical Gaussian surface of radius $r$ and length $L$, coaxial with the wire. By symmetry, the field is radially outward and has the same magnitude everywhere on the curved surface, with zero flux through the flat end-caps (field is parallel to them).
$\oint \vec{E}\cdot d\vec{A} = E(2\pi rL) = \dfrac{\lambda L}{\varepsilon_0}$
$E = \dfrac{\lambda}{2\pi\varepsilon_0 r}$
Q. Why does the electric flux through a closed Gaussian surface depend only on the charge enclosed, and not on any charges located outside the surface?
Ans: Any field lines originating from a charge outside the closed surface must both enter and exit the surface (since the charge itself is not enclosed) — the flux contribution from such a charge entering the surface is exactly cancelled by an equal and opposite flux contribution as the same field lines exit, resulting in zero net flux contribution from any external charge, regardless of its magnitude or position outside the surface.
Q. A Gaussian sphere encloses a net charge of 4 μC. Calculate the total electric flux through this sphere. (ε₀ = 8.854×10⁻¹² C²/N m²)
Ans: Φ_E = Q/ε₀ = (4×10⁻⁶)/(8.854×10⁻¹²) ≈ 4.52×10⁵ N m²/C.
Electric Field Inside a Charged Conductor
Using Gauss's law, the electric field inside the material of any charged conductor in electrostatic equilibrium is always zero — any excess charge resides entirely on the conductor's outer surface, since a Gaussian surface drawn just inside the conductor's boundary must enclose zero net charge (as free charges within a conductor rearrange themselves until the internal field vanishes and equilibrium is reached).
Solved Example
Explain, using Gauss's law, why the electric field just inside the surface of a charged conducting sphere is exactly zero, even though the sphere carries a non-zero net charge.
In electrostatic equilibrium, no current flows within a conductor, meaning the field within the conductor's material must be exactly zero (otherwise, free charges would continue to experience a force and keep moving). All of the conductor's excess charge therefore resides on its outer surface. Drawing a Gaussian surface just inside the conductor's boundary (within its material) encloses zero net charge (since all charge is on the outer surface, outside this inner Gaussian surface) — by Gauss's law, this means the flux (and hence the field) through and within this inner surface must be zero.
Q. Why does all the excess charge on a conductor reside on its outer surface, rather than being distributed throughout its volume?
Ans: Since free charges within a conductor are mobile and repel each other (for like charges), they naturally redistribute themselves to minimise their mutual repulsion, moving as far apart from each other as the conductor's geometry allows — this pushes all excess charge to the outermost surface, the location that maximises the average distance between individual charges, while the interior of the conductor remains at zero net charge and zero electric field in electrostatic equilibrium.
Electric Potential & Potential Energy
Electric Potential
Electric potential (V) at a point is the work done in bringing a unit positive test charge from infinity to that point, without any acceleration: $V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}$ (for a point charge). Unlike electric field (a vector), potential is a scalar quantity.
Relation Between Field & Potential
$E = -\dfrac{dV}{dr}$ — the electric field points in the direction of decreasing potential, and its magnitude equals the negative rate of change of potential with distance.
Electric Potential Energy
The potential energy of a charge $q$ at a point with potential $V$: $U = qV$. For two point charges: $U = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r}$.
Solved Example
Calculate the electric potential at a distance of 0.2 m from a point charge of $5\ \mu\text{C}$.
$V = k\dfrac{q}{r} = (9\times10^9)\dfrac{5\times10^{-6}}{0.2} = (9\times10^9)(2.5\times10^{-5}) = 2.25\times10^5\ \text{V}$
Q. Why is no work done in moving a charge along an equipotential surface?
Ans: An equipotential surface, by definition, has the same electric potential at every point on it. Since work done by an external agent in moving a charge equals qΔV (change in potential energy), and ΔV = 0 for any two points on the same equipotential surface, the work done is exactly zero — this is also consistent with the fact that the electric field (and hence electric force) is always perpendicular to an equipotential surface, so no component of force acts along any displacement within that surface.
Q. Calculate the potential energy of a system of two point charges, +3 μC and −2 μC, separated by 0.5 m.
Ans: U = kq₁q₂/r = (9×10⁹)(3×10⁻⁶)(−2×10⁻⁶)/0.5 = (9×10⁹)(−6×10⁻¹²)/0.5 = −0.108 J.
Equipotential Surfaces
An equipotential surface is a surface on which every point has the same electric potential. For a point charge, equipotential surfaces are concentric spheres centred on the charge. A key property: the electric field is always perpendicular to an equipotential surface at every point, since any component of field along the surface would imply a potential difference within that surface, contradicting its defining property.
Solved Example
Two points, A and B, lie on the same equipotential surface around a point charge. Explain why no work is required to move a test charge from A to B.
Since A and B are both on the same equipotential surface, they are, by definition, at the same electric potential ($V_A = V_B$). Work done in moving a charge $q$ between two points is $W = q(V_B-V_A)$; since $V_A = V_B$, this gives $W = 0$ — no work is needed, regardless of the specific path taken between A and B along that surface.
Q. Why must the electric field always be perpendicular to an equipotential surface at every point?
Ans: If the field had any component lying along (tangent to) the equipotential surface, that component would do work on a charge moved along the surface, which would necessarily change the charge's potential energy — but since every point on the surface shares the same potential, no such change (and hence no such work) can occur along the surface. This forces the field to have no component tangent to the surface at all, meaning it must be entirely perpendicular to it everywhere.
Electric Dipole
Definition
An electric dipole consists of two equal and opposite point charges separated by a small distance. Its strength is described by the dipole moment, $\vec{p} = q(2\vec{a})$, a vector pointing from the negative to the positive charge, where $2a$ is the separation.
Torque on a Dipole in a Uniform Field
A dipole placed in a uniform external electric field experiences a torque tending to align it with the field: $\vec{\tau} = \vec{p} \times \vec{E}$, with magnitude $\tau = pE\sin\theta$. The dipole experiences zero net force in a uniform field (since the two equal, opposite charges experience equal and opposite forces), but a non-zero torque unless already aligned with the field.
Solved Example
A dipole of moment $4\times10^{-6}\ \text{C m}$ is placed in a uniform field of $2\times10^5\ \text{N/C}$ at an angle of 30° to the field. Calculate the torque experienced.
$\tau = pE\sin\theta = (4\times10^{-6})(2\times10^5)\sin30° = (0.8)(0.5) = 0.4\ \text{N m}$
Q. Why does an electric dipole experience zero net force in a uniform electric field, despite each individual charge experiencing a force?
Ans: In a uniform field, both the positive and negative charges of the dipole experience forces of exactly equal magnitude (since the field strength is the same at both charge locations), but in opposite directions (since the charges themselves are opposite in sign) — these two equal and opposite forces exactly cancel when considering the dipole as a whole system, resulting in zero net force, even though each individual charge does experience its own non-zero force.
Q. At what orientation angle (relative to the field) does a dipole experience maximum torque, and at what angle does it experience zero torque?
Ans: Maximum torque occurs at θ = 90° (dipole moment perpendicular to the field, since sin90° = 1, giving τ = pE). Zero torque occurs at θ = 0° or 180° (dipole moment parallel or antiparallel to the field, since sin0° = sin180° = 0) — these represent the dipole's stable and unstable equilibrium orientations respectively.
Capacitors & Capacitance
Definition
A capacitor is a device for storing electric charge (and hence energy) in the electric field between two conductors. Capacitance (C) is the ratio of charge stored to the potential difference across the capacitor: $C = Q/V$, measured in farads (F).
Solved Example
Calculate the capacitance of a parallel plate capacitor with plate area $0.02\ \text{m}^2$ and separation 1 mm. ($\varepsilon_0 = 8.854\times10^{-12}\ \text{F/m}$)
$C = \dfrac{\varepsilon_0 A}{d} = \dfrac{(8.854\times10^{-12})(0.02)}{1\times10^{-3}} = \dfrac{1.771\times10^{-13}}{1\times10^{-3}} \approx 1.77\times10^{-10}\ \text{F} = 177\ \text{pF}$
Q. Why does the capacitance of a parallel plate capacitor decrease as the separation between the plates increases?
Ans: Since C = ε₀A/d, capacitance is inversely proportional to plate separation d — increasing d weakens the electric field between the plates for a given charge, meaning a larger potential difference (V) results from the same stored charge (Q), and since C = Q/V, a larger V for the same Q gives a smaller capacitance.
Q. A capacitor stores 4×10⁻⁴ C of charge at a potential difference of 200 V. Calculate its capacitance.
Ans: C = Q/V = (4×10⁻⁴)/200 = 2×10⁻⁶ F = 2 μF.
Combination of Capacitors
Series Combination
Capacitors in series carry the same charge, but the total voltage divides among them. Effective capacitance is smaller than any individual capacitance: $\dfrac{1}{C_{eq}} = \dfrac{1}{C_1}+\dfrac{1}{C_2}+\cdots$
Parallel Combination
Capacitors in parallel have the same voltage across each, but the total charge divides among them. Effective capacitance is larger than any individual capacitance: $C_{eq} = C_1+C_2+\cdots$
Solved Example
Two capacitors of 4 μF and 6 μF are connected in series. Calculate the equivalent capacitance.
$\dfrac{1}{C_{eq}} = \dfrac{1}{4}+\dfrac{1}{6} = \dfrac{3}{12}+\dfrac{2}{12} = \dfrac{5}{12}$
$C_{eq} = \dfrac{12}{5} = 2.4\ \mu\text{F}$
Q. Why is the equivalent capacitance of a series combination always smaller than the smallest individual capacitance in that combination?
Ans: Adding capacitors in series is mathematically analogous to increasing the effective distance between the outermost plates while the charge remains the same throughout — this always results in a smaller net capacitance, since 1/C_eq is the sum of reciprocals (each individually positive), making 1/C_eq always larger than any single 1/C term, and hence C_eq always smaller than the smallest individual capacitance.
Q. Two capacitors of 3 μF and 5 μF are connected in parallel. Calculate the equivalent capacitance.
Ans: C_eq = C₁ + C₂ = 3 + 5 = 8 μF.
Solved Example
Two capacitors of 2 μF and 8 μF are connected in series to a 100 V battery. Calculate the charge on each capacitor and the voltage across each.
$\dfrac{1}{C_{eq}} = \dfrac{1}{2}+\dfrac{1}{8} = \dfrac{4}{8}+\dfrac{1}{8} = \dfrac{5}{8} \Rightarrow C_{eq} = \dfrac{8}{5} = 1.6\ \mu\text{F}$
Since capacitors in series carry the same charge: $Q = C_{eq}V = (1.6\times10^{-6})(100) = 1.6\times10^{-4}\ \text{C}$ on both capacitors.
$V_1 = Q/C_1 = (1.6\times10^{-4})/(2\times10^{-6}) = 80\ \text{V}$; $V_2 = Q/C_2 = (1.6\times10^{-4})/(8\times10^{-6}) = 20\ \text{V}$ (sum = 100 V, confirming the total voltage is correctly split).
Q. Why do capacitors in series always carry exactly the same charge, regardless of their individual capacitance values?
Ans: In a series connection, the plates of adjacent capacitors are connected only to each other (not to the battery directly) — any charge that accumulates on one plate must come from an equal and opposite induced charge on the facing plate of the neighbouring capacitor, since these connected plates form an isolated conductor with zero net charge. This charge conservation on the connecting plates forces every capacitor in the series chain to acquire exactly the same magnitude of charge, regardless of their individual capacitance values.
Dielectrics
Definition
A dielectric is an insulating material that, when inserted between a capacitor's plates, increases its capacitance by a factor called the dielectric constant ($K$) — the dielectric's molecules polarise in response to the field, creating an internal field that partially opposes the original, reducing the net field (and hence voltage for the same charge), thereby increasing capacitance.
Solved Example
A parallel plate capacitor has a capacitance of 5 μF in vacuum. Calculate its new capacitance when a dielectric of dielectric constant 4 completely fills the gap between the plates.
$C_{new} = KC_{old} = (4)(5) = 20\ \mu\text{F}$
Q. Why does inserting a dielectric between a capacitor's plates increase its capacitance?
Ans: The dielectric's molecules become polarised in response to the capacitor's electric field, creating an induced internal field that opposes (partially cancels) the original field between the plates. This reduces the net field, and hence the voltage needed to maintain a given amount of stored charge — since capacitance C = Q/V, a smaller required V for the same Q results in a larger capacitance.
Q. A capacitor with air between its plates has capacitance C₀. If a dielectric of constant K = 3 is inserted, and the capacitor is connected to a battery maintaining constant voltage, how does the stored charge change?
Ans: Since Q = CV, and C increases by a factor of K = 3 (to 3C₀) while V remains constant (fixed by the battery), the stored charge Q also increases by a factor of 3.
Energy Stored in a Capacitor
Energy Formula
The energy stored in a charged capacitor equals the work done in charging it, which can be expressed in several equivalent forms:
$$U = \tfrac{1}{2}QV = \tfrac{1}{2}CV^2 = \dfrac{Q^2}{2C}$$
Solved Example
A 10 μF capacitor is charged to a potential difference of 100 V. Calculate the energy stored.
$U = \tfrac{1}{2}CV^2 = \tfrac{1}{2}(10\times10^{-6})(100)^2 = \tfrac{1}{2}(10\times10^{-6})(10{,}000) = 0.05\ \text{J}$
Q. Why is the energy stored in a capacitor given by ½CV², rather than simply QV (which would be the total work if charge were transferred all at once at constant voltage V)?
Ans: As a capacitor charges, the voltage across it increases progressively from 0 up to its final value V, since V = Q/C depends on the charge already accumulated at each instant — the work done to add each successive small amount of charge increases as charging proceeds, rather than staying constant at the final voltage V throughout. Integrating this gradually increasing work over the entire charging process (rather than simply multiplying the total charge by the final voltage) gives exactly half of QV, hence the factor of ½ in the energy formula.
Q. If the voltage across a capacitor is doubled, by what factor does the stored energy increase?
Ans: Since U ∝ V², doubling the voltage increases stored energy by a factor of 2² = 4.
Q. A capacitor stores 0.02 J of energy at a charge of 4×10⁻⁴ C. Calculate its capacitance.
Ans: U = Q²/2C → 0.02 = (4×10⁻⁴)²/2C → 0.02 = (1.6×10⁻⁷)/2C → 0.02 = 8×10⁻⁸/C → C = 8×10⁻⁸/0.02 = 4×10⁻⁶ F = 4 μF.
⚡ Formula Cheat-Sheet
Coulomb's lawF = kq₁q₂/r²
Electric field (point charge)E = kq/r²
Gauss's lawΦ = Q_enc/ε₀
Electric potentialV = kq/r
Dipole torqueτ = pE sinθ
Parallel plate capacitanceC = ε₀A/d
Series capacitors1/C_eq = Σ(1/Cᵢ)
Parallel capacitorsC_eq = ΣCᵢ
Capacitance with dielectricC = Kε₀A/d
Energy in capacitorU = ½CV² = Q²/2C