Electric Field from Continuous Charge Distributions - Quick Notes

Continuous Charge Distributions: Quick Reference

• Key idea: Electric field from a continuous distribution obtained by superposition of many infinitesimal charges dq.

  • At P, dE = $\displaystyle d\mathbf{E} = \frac{k\,dq}{r^{2}}\ \hat{\mathbf{r}}$ where $k = \frac{1}{4\pi\varepsilon_{0}}$ and $\hat{\mathbf{r}}$ points from dq to P.
  • Total field: $\displaystyle \mathbf{E} = \int d\mathbf{E} = \int \frac{k\,dq}{r^{2}}\ \hat{\mathbf{r}}.$

• Differential charge elements and densities

  • Linear: $dq = \lambda\,dl\,.$
  • Surface: $dq = \sigma\,dA\,.$
  • Volume: $dq = \rho\,dV\,.$

• General procedure (continuous distributions)

  • Divide the charge into small elements dq.
  • Compute the field dE due to one element at P.
  • Sum (integrate) contributions from all elements: $\mathbf{E} = \int d\mathbf{E}.

• Symmetry and direction of E

  • For symmetric objects, E points along the symmetry axis or perpendicular to symmetry planes.
  • Use components: sum along directions that survive symmetry; cancel others.

• Common densities

  • Linear density: $\lambda = \dfrac{dq}{dl}$
  • Surface density: $\sigma = \dfrac{dq}{dA}$
  • Volume density: $\rho = \dfrac{dq}{dV}$

• Key geometries and axis-field formulas (along symmetry axis)

  • Long straight wire (line charge density $\lambda$):
  • Magnitude: $\displaystyle E = \frac{\lambda}{2\pi\varepsilon_{0}\,r}$
  • Direction: radial outward from wire (if $\lambda>0$).
  • Ring of radius a
    with total charge $Q$, field on axis at distance $x$:
  • $\displaystyle E(x) = \frac{k Q x}{(x^{2} + a^{2})^{3/2}}$
  • Charged disk (radius $R$), uniform surface density $\sigma$, axis distance $z$:
  • $\displaystyle E(z) = \frac{\sigma}{2\varepsilon_{0}}\left(1 - \frac{z}{\sqrt{z^{2} + R^{2}}}\right)$
  • Infinite plane with surface charge density $\sigma$:
  • Magnitude: $\displaystyle E = \frac{\sigma}{2\varepsilon_{0}}$
  • Direction: perpendicular to the plane, away from plane if $\sigma>0$.

• Spherical and cylindrical cases

  • Solid sphere of radius $R$ with uniform density $\rho$:
  • Outside (r > R): $\displaystyle E(r) = \frac{1}{4\pi\varepsilon_{0}}\frac{Q}{r^{2}}, \quad Q = \tfrac{4}{3}\pi R^{3}\rho.$
  • Inside (r < R): $\displaystyle E(r) = \frac{\rho\,r}{3\varepsilon<em>{0}} = \frac{1}{4\pi\varepsilon</em>{0}}\frac{Q r}{R^{3}}.$
  • Hollow spherical shell with inner radius $R<em>{1}$ and outer radius $R</em>{2}$, density $\rho = \rho_{0}\frac{6}{r}$ (spherical symmetry):
  • Total charge: $\displaystyle Q<em>{\text{shell}}= \int</em>{R<em>{1}}^{R</em>{2}} \rho \ dV = 12\pi\rho<em>{0}\,(R</em>{2}^{2}-R_{1}^{2}).$
  • For r > R2: $\displaystyle E(r) = \frac{1}{4\pi\varepsilon<em>{0}}\frac{Q</em>{\text{shell}}}{r^{2}}.$
  • For r < R1: $\displaystyle E(r) = 0.$

• Nonuniform and composite shapes (examples)

  • Cylindrical shell with density $\rho = \rho_{0}\,r$ for $a<r<b$ and height $h$:
  • Total charge: $\displaystyle Q = Q<em>{shell} = \int</em>{a}^{b} \rho\,dV = 2\pi h \rho<em>{0} \int</em>{a}^{b} r^{2}\,dr = \frac{2\pi h\rho_{0}}{3}\,(b^{3}-a^{3}).$
  • If h = 2b, then $Q = \frac{4\pi b\rho_{0}}{3}\,(b^{3}-a^{3}).$

• Point charge (reference)

  • For a point charge $Q$ at the origin, field at position $\mathbf{r}$:
  • $\displaystyle \mathbf{E}(\mathbf{r}) = \frac{1}{4\pi\varepsilon_{0}}\frac{Q}{|\mathbf{r}|^{3}}\,\mathbf{r} = \frac{kQ}{r^{2}}\hat{\mathbf{r}}.$

• Quick example reference

  • If a single charge of $Q=100\,\text{C}$ is at the origin and you evaluate at $P=(3,4)\,\text{m}$, the distance is $r = \sqrt{3^{2}+4^{2}} = 5\,\text{m}$ and the field magnitude is $E = \dfrac{k|Q|}{r^{2}} = \dfrac{(8.988\times10^{9})\times100}{25} = 3.5952\times10^{10}\ \text{N/C}$ toward/away from the origin depending on the sign of Q.

• Notes on applying to exam problems

  • Identify symmetry to pick the direction of E.
  • Choose appropriate dq, express dq in terms of the natural density for the geometry, then integrate over the region of interest.
  • Remember to include vector directions when adding contributions (use components if necessary).