Electrostatics¶

Exam-ready answer

Assume stationary point charges in a homogeneous, linear and isotropic medium of permittivity \(\epsilon\) in farads per metre (F/m). Let \(\vec R_{12}=\vec r_2-\vec r_1\), \(R_{12}=|\vec R_{12}|\), and let \(\hat a_{12}\) point from source charge \(Q_1\) toward \(Q_2\). Coulomb's law states that the force on \(Q_2\) due to \(Q_1\) is

Like charges therefore repel along \(+\hat a_{12}\) and unlike charges attract along \(-\hat a_{12}\). The inverse-square result applies when charge dimensions are negligible compared with separation; for a distributed charge, each differential contribution must be integrated.

Electric field intensity at a point is the force per unit positive test charge as that test charge tends to zero, so it does not disturb the source distribution:

Putting a positive test charge \(q_t\) at distance \(R\) from a source \(Q\) gives

where \(\hat a_R\) is radially outward from \(Q\); the algebraic sign of \(Q\) reverses the direction. By linear superposition,

or the sum becomes \(\int dQ\,\hat a_R/(4\pi\epsilon R^2)\) for line, surface or volume charge.

Check: in free space, \(Q=1\,\mu\text{C}\) at \(R=0.10\,\text{m}\) produces \(E\simeq(8.99\times10^9)(10^{-6})/(0.10)^2=8.99\times10^5\,\text{V/m}\) radially outward. Coulomb's direct formula is unsuitable near material boundaries or for time-varying fields; there one solves the field equations with boundary conditions. Applications include force calculation, electrostatic sensors and the starting field model for capacitance.

Practice target: 8 minutes; define the source-to-field orientation before writing the vector law and finish with units and superposition.

Exam-ready answer

Electric flux \(\Psi\) measures the total electric displacement crossing an oriented surface. If \(d\vec S=\hat n\,dS\) points outward from a closed surface, then

The sign is positive for flux leaving in the \(+\hat n\) direction and negative for entering flux. Electric flux density or electric displacement \(\vec D\) is flux per unit area normal to the flux; \([\vec D]=\text{C/m}^2\). For a point free charge \(Q\) in a uniform medium, spherical symmetry gives \(\vec D=Q\hat a_r/(4\pi r^2)\).

In matter, the constitutive relation separates free-charge field from material polarization:

where \(\vec P\) is polarization density in C/m\(^2\). For a linear, isotropic dielectric, \(\vec P=\epsilon_0\chi_e\vec E\), hence

Thus \(\vec D\) and \(\vec E\) are parallel only in an isotropic medium. In free space \(\epsilon_r=1\); in an anisotropic dielectric, \(\epsilon\) is a tensor and the vectors need not be parallel.

At an interface whose unit normal \(\hat n\) points from medium 1 to medium 2, an infinitesimal pillbox and loop give

where \(\rho_s\) is free surface charge density in C/m\(^2\). With \(\rho_s=0\), normal \(D\) is continuous while normal \(E\) changes inversely with permittivity.

Check: if \(E=2\,\text{kV/m}\) in a dielectric with \(\epsilon_r=4\), then \(D=\epsilon_0\epsilon_rE=70.8\,\text{nC/m}^2\). In a uniform capacitor, \(C=Q/V\), \(D=Q/A\), and stored energy density is \(w_e=\tfrac12\vec E\cdot\vec D\) J/m\(^3\). These relations support capacitor and insulation design, but nonlinear or hysteretic dielectrics require their measured \(D\)-\(E\) law.

Practice target: 8–9 minutes; show the outward surface normal, both constitutive forms, units, and the two interface conditions.

Exam-ready answer

Let \(S\) be a closed surface bounding volume \(V\), with \(d\vec S=\hat n\,dS\) directed outward. Gauss's law states that the net outward electric displacement flux equals the enclosed free charge:

where \(\vec D\) is in C/m\(^2\), \(\rho_v\) is free volume-charge density in C/m\(^3\), and \(Q_{\text{enc}}\) is in coulombs. Charges outside \(S\) can produce local flux, but their entering and leaving contributions cancel in the net integral.

The divergence theorem converts the flux of any differentiable vector field through a closed surface into a volume integral:

Combining this with Gauss's law gives

Because the equality holds for every arbitrary infinitesimal volume, the integrands must be equal:

This is Maxwell's first equation in differential or point form. It says that positive free charge is a source of \(\vec D\), negative free charge is a sink, and a charge-free point has zero net local divergence. For a homogeneous linear dielectric, \(\vec D=\epsilon\vec E\), so \(\nabla\cdot\vec E=\rho_v/\epsilon\); if \(\epsilon\) varies spatially, one must retain \(\nabla\cdot(\epsilon\vec E)=\rho_v\) rather than pulling \(\epsilon\) outside the divergence.

Symmetry check: enclose a positive point charge \(Q\) by a sphere of radius \(r\). The field is radial and has constant magnitude on the sphere, so

The outward direction follows from \(Q>0\); a negative \(Q\) reverses it. The \(1/r^2\) decrease exactly offsets the \(4\pi r^2\) growth of spherical area, leaving total flux equal to \(Q\).

Across a dielectric boundary, the same law applied to a vanishing pillbox whose normal points from region 1 to 2 gives \(\hat n\cdot(\vec D_2-\vec D_1)=\rho_s\). This is the surface-charge counterpart of the volume equation and is essential in capacitor and dielectric-interface problems.

Gauss's law is universally valid in electrostatics, but it is an efficient field-solving tool only for spherical, cylindrical or planar symmetry, where the normal component is constant or zero on chosen surface pieces. For arbitrary geometry it still checks flux and charge conservation, while Poisson/Laplace methods or numerical field solvers determine \(\vec E\). Applications include coaxial capacitance, charged conductors, dielectric boundaries and verification of finite-element solutions.

Practice target: 16–18 minutes; reserve one-third for the integral-to-differential derivation and draw the symmetry-matched Gaussian surfaces.

Exam-ready answer

Assume an infinite homogeneous medium of permittivity \(\epsilon\) and use outward-oriented Gaussian surfaces. Gauss's law is

1. Point charge \(Q\). Choose a concentric sphere of radius \(r\). Spherical symmetry makes \(\vec D=D_r\hat a_r\) constant and normal to the surface:

2. Infinite line charge \(\lambda\) C/m. Choose a coaxial cylinder of radius \(\rho\) and length \(L\). Flux through the end caps is zero because \(\vec D=D_\rho\hat a_\rho\) is parallel to them. Therefore

3. Uniformly charged solid sphere. Let its radius be \(a\) and volume density be \(\rho_v\) C/m\(^3\). For \(r<a\), the enclosed charge is \(\rho_v(4\pi r^3/3)\); for \(r\ge a\), the total charge is \(Q=\rho_v4\pi a^3/3\). Hence

For a conducting sphere or thin spherical shell in electrostatic equilibrium, \(E=0\) inside and the outside expression is unchanged. Positive \(Q\), \(\lambda\) or \(\rho_v\) gives outward fields; negative charge reverses each unit vector.

Check: the solid-sphere formulas agree at \(r=a\), both giving \(E(a)=\rho_v a/(3\epsilon)\), so there is no discontinuity when no surface-charge sheet is present. The three decays differ because enclosed charge and Gaussian area scale differently: \(1/r^2\) for a localized charge, \(1/\rho\) for a line, and linear growth inside a uniform sphere. These ideal results model conductors and coaxial systems; finite lines or nonspherical charge distributions lack the assumed symmetry and require Coulomb integration or numerical solution.

Practice target: 9 minutes; draw all three Gaussian surfaces and show the enclosed-charge step for the solid sphere.

Exam-ready answer

In electrostatics, \(\nabla\times\vec E=0\), so the field is conservative and work is path independent. The potential difference from point \(A\) to point \(B\) is the negative work done by the electric field per unit positive test charge:

The minus sign means potential decreases when \(d\vec l\) points along \(\vec E\). Equivalently, the external work per unit charge in a quasistatic displacement is \(+\Delta V\). For two neighboring points separated by \(d\vec l\),

But the total differential of a scalar field is \(dV=\nabla V\cdot d\vec l\). Since this equality holds for every displacement direction,

Thus \(\vec E\) is normal to an equipotential surface and points toward the greatest decrease of \(V\); no work is done along an equipotential. With reference \(V(\infty)=0\), integration of the radial point-charge field gives

Check: differentiating this result yields \(E_r=-dV/dr=Q/(4\pi\epsilon r^2)\), outward for \(Q>0\), confirming both sign and units. Potential is a scalar and therefore usually easier to superpose than vector fields: \(V=\sum_iQ_i/(4\pi\epsilon R_i)\), followed by \(\vec E=-\nabla V\).

The relation is used in boundary-value, capacitor and insulation calculations. If two conductors carry \(\pm Q\) at potential difference \(V\), \(C=Q/V\) in farads and stored energy is \(W=\tfrac12CV^2=\tfrac12QV\) joules. Absolute potential depends on the chosen reference, whereas potential difference and \(\vec E\) are physical. The electrostatic scalar-potential description alone is insufficient for a general time-varying magnetic vector potential, where \(\vec E=-\nabla V-\partial\vec A/\partial t\).

Practice target: 8 minutes; state whose work fixes the sign, derive the gradient relation, and verify it with the point-charge potential.

Exam-ready answer

For an electrostatic field, Maxwell's first equation and the potential relation are

where \(\rho_v\) is free charge density in C/m\(^3\), \(V\) is potential in volts and \(\vec E\) is in V/m. In a linear medium \(\vec D=\epsilon\vec E\). Substitution, with the gradient orientation chosen so \(\vec E\) points from high to low potential, gives the general material equation

If \(\epsilon\) is constant, it can be taken outside the divergence:

In a charge-free region, \(\rho_v=0\), so

In Cartesian coordinates, \(\nabla^2V=\partial^2V/\partial x^2+\partial^2V/\partial y^2+\partial^2V/\partial z^2\). Poisson's equation describes regions containing charge; Laplace's equation determines source-free fields from boundary values. A unique solution follows when conductor potentials are specified (Dirichlet data), normal field or charge is specified (Neumann data), or compatible mixed data are supplied. At a dielectric interface, \(V\) and tangential \(E\) are continuous, while \(\hat n\cdot(\vec D_2-\vec D_1)=\rho_s\) for a normal directed from medium 1 to 2.

Worked check: parallel plates. Let \(V(0)=0\) and \(V(d)=V_0\) in a charge-free uniform dielectric. One-dimensional Laplace equation gives \(d^2V/dx^2=0\), hence \(V=Ax+B\). Applying the boundaries yields

For plate area \(A\) with negligible fringing, \(C=\epsilon A/d\) F, \(W=\tfrac12CV_0^2\) J and \(w_e=\tfrac12\epsilon E^2\) J/m\(^3\). Applications include capacitors, coaxial cables, shielding and semiconductor depletion regions. The simple constant-\(\epsilon\) Poisson form is not valid across abrupt or nonlinear material changes; use \(\nabla\cdot(\epsilon\nabla V)=-\rho_v\) piecewise with the interface conditions, often numerically for irregular geometry.

Practice target: 9 minutes; derive both equations before using the plate boundary conditions to check the sign of the field.