Electrostatics formulas

Force on $Q_2$ due to $Q_1$; ${\hat{a}}_{12}$ is the unit vector from $Q_1$ toward $Q_2$ and $R=|{\mathbf{r}}_2-{\mathbf{r}}_1|$. A positive product $Q_1{Q}_2$ means repulsion. Valid for point charges (or bodies small compared with $R$). In a homogeneous dielectric filling all space replace ${\epsilon }_0$ by $\epsilon$: the force falls by the factor ${\epsilon }_r$.

Field at distance $R$ from a point charge $Q$: away from a positive charge, toward a negative one. A charge $q$ placed in the field feels $\mathbf{F}=q\mathbf{E}$. Units: V/m, the same as N/C.

$\rho$ is the perpendicular distance from the line and ${\hat{a}}_{\rho }$ points perpendicularly away from it (for ${\rho }_L>0$). For a line not on the $z$-axis, use the perpendicular vector from the line to the field point.

Ring of radius $a$ in the plane $z=0$, centred on the origin, total charge $Q=2\pi a{\rho }_L$. Zero at the centre; $|\mathbf{E}|$ is maximum at $z=\pm a/\sqrt{2}$, where it equals $Q/(6\sqrt{3}\pi {\epsilon }_0{a}^2)$; for $|z|\gg a$ it looks like a point charge.

Disc of radius $a$ in the plane $z=0$. Below the disc ($z<0$) the field has the same magnitude with $|z|$ in place of $z$ and points along $-{\hat{a}}_z$. As $a\to \infty$ it becomes the infinite sheet.

${\hat{a}}_n$ is the unit normal pointing away from the sheet, on each side. The magnitude does not depend on the distance from the sheet.

Radius $a$, total charge $Q$ (for the solid sphere $Q=\frac{4}{3}\pi a^3{\rho }_v$). The shell is a thin surface carrying a uniform ${\rho }_S$; an isolated charged conducting sphere, solid or hollow, gives the same field because its charge spreads uniformly over its surface. A shell whose charge is not uniform can have a field inside. Outside, any spherically symmetric charge acts as a point charge at the centre. Inside the solid sphere $E$ grows linearly with $r$ and peaks at the surface.

Units C/m². For a point charge $\mathbf{D}=\frac{Q}{4\pi r^2}{\hat{a}}_r$, whatever the homogeneous medium around it.

$S$ is any closed surface, $d\mathbf{S}$ points outward and $Q_{\text{enc}}$ is the total free charge inside (the last integral is written for a volume charge; enclosed point, line and surface charges count too). Always true, but it yields $\mathbf{D}$ only when symmetry makes $\mathbf{D}$ normal to the surface with constant magnitude (or tangential) on every part of $S$.

Follows from the integral form and the divergence theorem. To get the charge density from a given field, differentiate; do not integrate.

The factors $\rho$, $r^2$ and $\sin \theta$ go INSIDE the derivative. A purely radial field with $D_r\propto 1/r^2$ has zero divergence everywhere except $r=0$.

Work done per unit charge by an external agent in moving a positive test charge from $B$ to $A$. Integrate along any convenient path; the lower limit is the reference point $B$.

Work done by an external agent moving $Q$ slowly (no change in kinetic energy) from $A$ to $B$; the field itself does $-W_{\text{ext}}$. Positive when a positive charge is moved toward higher potential; zero along an equipotential and around any closed path.

Reference at infinity. $V$ is a scalar, so add the contributions with their signs and no components; this is often easier than integrating $\mathbf{E}$ directly, and $-\nabla V$ then gives the field.

The minus sign makes $\mathbf{E}$ point toward decreasing $V$. $|\nabla V|$ is the largest rate of change of $V$ with distance at that point.

${\rho }_A$ and ${\rho }_B$ are the perpendicular distances of $A$ and $B$ from the line. The reference cannot be at infinity (the logarithm diverges); only differences are meaningful.

Radius $a$, total charge $Q$, reference at infinity; the shell result also holds for an isolated charged conducting sphere. Inside the shell $V$ is constant and equal to its surface value, although $\mathbf{E}=0$ there. At the centre of the solid sphere $V=3Q/(8\pi {\epsilon }_{0}a)$, 1.5 times its surface value. Stored energy: $Q^2/(8\pi {\epsilon }_{0}a)$ for the shell or conducting sphere, $3Q^2/(20\pi {\epsilon }_{0}a)$ for the uniformly charged solid sphere.

Charges $+Q$ and $-Q$ a distance $d$ apart, centred on the origin along the $z$-axis; the dipole moment $\mathbf{p}=Qd{\hat{a}}_z$ points from $-Q$ to $+Q$, and $\theta$ is measured from $+z$. Valid for $r\gg d$. $V$ falls as $1/r^2$ and $E$ as $1/r^3$; $V=0$ on the whole plane $\theta =90^{\circ }$. In a uniform external field ${\mathbf{E}}_0$ a dipole feels no net force, but a torque $\mathbf{T}=\mathbf{p}\times {\mathbf{E}}_0$, and has potential energy $U=-\mathbf{p}\cdot {\mathbf{E}}_0$, lowest when $\mathbf{p}$ is aligned with ${\mathbf{E}}_0$.

In the sum, $V_k$ is the potential at $Q_k$ due to all the OTHER charges; for two point charges $W_E=\frac{Q_1{Q}_2}{4\pi {\epsilon }_{0}R}$. The first integral runs over the charge, the second over all space where the field exists. Applied to point charges, the integral forms would include each charge's infinite self-energy, which the sum leaves out. $w_E$ is the energy density, to be integrated over the whole region where the field exists.

From $\nabla \cdot (\epsilon \mathbf{E})={\rho }_v$ and $\mathbf{E}=-\nabla V$ in a homogeneous medium. Cartesian: ${\nabla }^{2}V=\frac{{\partial }^{2}V}{\partial x^2}+\frac{{\partial }^{2}V}{\partial y^2}+\frac{{\partial }^{2}V}{\partial z^2}$. When $V$ depends only on the radius: $\frac{1}{\rho }\frac{d}{d\rho }\left(\rho \frac{dV}{d\rho }\right)$ (cylindrical) and $\frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{dV}{dr}\right)$ (spherical). When $V$ depends on one coordinate only, Laplace's equation integrates directly: $V=Ax+B$ (planar), $V=A\ln \rho +B$ (cylindrical), $V=A/r+B$ (spherical), with $A$ and $B$ fixed by the two boundary potentials. So between parallel plates with one homogeneous, charge-free dielectric in the gap (fringing neglected), $V$ is linear in distance.

${\chi }_e$ is the electric susceptibility (zero in vacuum). The last two relations hold for linear isotropic media; $\epsilon ={\epsilon }_0{\epsilon }_r$ is the permittivity.

${\hat{a}}_n$ is the outward normal of the dielectric's surface. The total bound charge of a dielectric body is zero. Gauss's law in terms of total charge: $\nabla \cdot ({\epsilon }_0\mathbf{E})={\rho }_v+{\rho }_{pv}$.

${\hat{a}}_n$ is the unit normal pointing from region 2 into region 1, and ${\rho }_S$ is the FREE surface charge on the interface. Between two perfect dielectrics ${\rho }_S=0$, so $D_n$ is continuous and ${\epsilon }_1{E}_{n1}={\epsilon }_2{E}_{n2}$. Tangential $\mathbf{E}$ is always continuous; tangential $\mathbf{D}$ is not (unless ${\epsilon }_1={\epsilon }_2$): $D_{t1}/{\epsilon }_1=D_{t2}/{\epsilon }_2$.

Angles measured from the NORMAL to the interface; valid when ${\rho }_S=0$. The field line lies farther from the normal in the medium with the larger permittivity.

Equal to $\frac{1}{2}\int \epsilon E^{2}dv$ over the region between the conductors. Capacitances in parallel add; in series their reciprocals add.

Plate area $A$, gap $d$, one homogeneous dielectric; uniform field, fringing neglected.

Layers parallel to the plates act as capacitors in series. $D$ is the same in every layer, so the field is larger in the layer of lower permittivity. For any number of layers, $C=A/{\sum }_k(d_k/{\epsilon }_k)$. Special cases in an air gap $d$: a dielectric slab of thickness $t$ and relative permittivity ${\epsilon }_r$, parallel to the plates, gives $C={\epsilon }_{0}A/(d-t+t/{\epsilon }_r)$; an uncharged metal slab of thickness $t$ gives $C={\epsilon }_{0}A/(d-t)$. Neither depends on where the slab sits in the gap.

Each dielectric fills the whole gap $d$ over part of the plate area ($A_1+A_2=A$): capacitors in parallel. $E=V/d$ is the same in both; $D$ differs.

Inner radius $a$, outer radius $b$, length $L$. The field is largest at the surface of the inner conductor, $\rho =a$.

Concentric spheres of radii $a<b$; letting $b\to \infty$ gives the isolated sphere. With two layers (${\epsilon }_1$ for $a<r<c$, ${\epsilon }_2$ for $c<r<b$): $C=4\pi /\left[\frac{1}{{\epsilon }_1}\left(\frac{1}{a}-\frac{1}{c}\right)+\frac{1}{{\epsilon }_2}\left(\frac{1}{c}-\frac{1}{b}\right)\right]$.

In F/m. Wire radius $a$, centre-to-centre spacing $d$, height $h$ of the wire axis above a grounded plane; the approximations need $d\gg a$ and $h\gg a$. The wire-plane result follows from the image method: the plane lies midway between the wire and its image, at half the voltage, so the capacitance doubles.

${\rho }_S={\epsilon }_0{E}_z$ at $z=0^{+}$. Its magnitude is largest, $Q/(2\pi h^2)$, directly below the charge, and it integrates to $-Q$ over the whole plane. The force on $Q$ is the Coulomb attraction of its image, a distance $2h$ away.