Electrostatics — Field, Potential & Energy (Medium)

Quick Recap — Superposition in Action

• Fields and forces add as vectors. Two equal fields $E$ at angle $\theta$ give a resultant $2E\cos\dfrac{\theta}{2}$; along a line they simply add or subtract.
• Potential is a scalar: $V=\sum\dfrac{kq_i}{r_i}$; work done $W=q\,\Delta V$; interaction energy of a pair $U=\dfrac{kq_1q_2}{r}$ (sum over all pairs).
• Dipole: torque $\tau=pE\sin\theta$; energy $U=-pE\cos\theta$; work to turn from $\theta_1$ to $\theta_2$ is $pE(\cos\theta_1-\cos\theta_2)$; axial field $\dfrac{2kp}{r^3}$.
• Null points: between two like charges (nearer the smaller); beyond the smaller charge for unlike charges. Set the two field magnitudes equal and solve.
• Gauss (flux): total $\dfrac{q_{enc}}{\varepsilon_0}$; a charge at a cube centre gives $\dfrac{q}{6\varepsilon_0}$ per face; at a corner the cube subtends $\dfrac{q}{8\varepsilon_0}$.

Worked mini-example. Two $+4\ \mu$C and $-4\ \mu$C charges are 2 m apart. Midway, both fields point the same way (toward $-$): $E=2\times\dfrac{kq}{r^2}=2\times\dfrac{(9\times10^9)(4\times10^{-6})}{1^2}=7.2\times10^4$ N/C, while the potential there is $0$.