1.1 Electric field

The electric field \(\vb E(\vb x,t)\) maps a point \(\vb x\) and a time \(t\) to a vector \(\vb E\).

Electric charge

Electric charge plays two roles:

1. it is the propensity for producing an electric field;

2. it determines the force, \(\vb F\) in response to an applied electric field. From (1.1), \(q = |\vb F|/|\vb E|\).

Charges, at rest or in motion, are the sources of all electromagnetic fields.

1.2 Coulomb’s law

Coulomb’s law gives the electric force \(\vb F_E\) that static point charges (i.e. in vacuum, at rest) exert on other charges.

where

• \(Q\) is the charge of the particle experiencing \(\vb F_E\);

• \(\vb r_i\) is the separation vector from charge \(q_i\) to \(Q\), with length \(r_i = |\vb r_i|\) and unit vector \(\displaystyle \uv r_i = \frac{\vb r_i}{|\vb r_i|}\);

• \(\varepsilon_0\) is the permittivity of free space, \(\varepsilon_0 = 8.854\times 10^{-12}\,\mathrm{F/m}\).

Combining (1.1) and (1.2) gives the electric field produced by \(N\) charged particles: \[\keyeq{\displaystyle \vb E(\vb r) = \frac{\vb F_E}{Q} = \frac{1}{4\pi\varepsilon_0}\sum_{i=1}^{N}\frac{q_i}{r_i^2}\,\uv r_i}\]

Example: two equal charges

Two charges \(q\) sit on the \(x\)-axis at \(x=\pm L\). Find the field at point \(P\), a height \(D\) above their midpoint.

By superposition, \(\vb E(P) = \vb E_1 + \vb E_2 = (E_x^1 + E_x^2)\,\uv\imath + (E_y^1 + E_y^2)\,\uv\jmath\). By symmetry \(E_x^2 = -E_x^1\) and \(E_y^2 = E_y^1\), so \[\vb E(P) = 2E_y^1\,\uv\jmath .\] With \(\sin\theta = D/\sqrt{D^2+L^2}\) and \(r_1^2 = D^2+L^2\), \[E_y^1 = \frac{q}{4\pi\varepsilon_0}\,\frac{\sin\theta}{r_1^2} = \frac{q}{4\pi\varepsilon_0}\,\frac{D}{(D^2+L^2)^{3/2}} ,\] and therefore \[\keyeq{\displaystyle \vb E(P) = \frac{q}{2\pi\varepsilon_0}\,\frac{D}{(D^2+L^2)^{3/2}}\;\uv\jmath}\]

1.3 Voltage

The electric potential difference, a.k.a. “voltage”, is the (negative of the) work done per unit charge: \[\begin{equation} \tag{1.3} \keyeq{\displaystyle \Delta V = -\frac{W_E}{q} = -E\,d\cos\theta} \end{equation}\]

1.4 Capacitors

Parallel-plate capacitor

For a parallel-plate capacitor with plate area \(A\) and separation \(d\), the field between the plates is uniform: \[E = \frac{Q}{\varepsilon_0 A}.\] Using (1.3) (magnitude, with \(\theta=0\)), \(V = Ed\), so \[C_{\parallel} = \frac{Q}{V} = \frac{Q}{Ed} = \frac{Q}{\dfrac{Q}{\varepsilon_0 A}\,d} \quad\Longrightarrow\quad \keyeq{\displaystyle C_{\parallel} = \frac{\varepsilon_0 A}{d}}\] Note that \(Q\) cancels: capacitance depends only on geometry (a general result).

Capacitors in series

\[C_1 = \frac{Q}{V_1}, \qquad C_2 = \frac{Q}{V_2} \qquad\Longrightarrow\qquad C = \frac{Q}{V} = \frac{Q}{\dfrac{Q}{C_1}+\dfrac{Q}{C_2}} = \frac{1}{\dfrac{1}{C_1}+\dfrac{1}{C_2}}\] \[\keyeq{\displaystyle \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2}}\]

Capacitors in parallel

Example

Find the effective capacitance of the circuit below.

\(C_2\) and \(C_3\) are in parallel, so they combine to \(C_4 = C_2 + C_3 = 6\,\mathrm{\mu F}\). \(C_4\) is then in series with \(C_1\): \[\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_4} \;\Longrightarrow\; C = \frac{C_1 C_4}{C_1 + C_4} = \frac{12\cdot 6}{12+6}\,\mathrm{\mu F} \qquad\Longrightarrow\qquad \keyeq{C = 4\,\mathrm{\mu F}}\]

Problems

Problems are taken from the sources listed under Sources and Attribution. Numbers refer to the end-of-chapter problems; figures have been redrawn.

OpenStax University Physics Vol. 2, Chapter 5: Electric Charges and Fields

OpenStax University Physics Vol. 2, Chapter 8: Capacitance