Energy conservation: charge moves from $r_1 = 3R$ to $r_2 = 2R$
$\frac{1}{2}mv^2 = kQq\left(\frac{1}{r_1} - \frac{1}{r_2}\right) = kQq\left(\frac{1}{3R} - \frac{1}{2R}\right)$
For attractive case: $\frac{1}{2}mv^2 = kQq\left(\frac{1}{2R} - \frac{1}{3R}\right) = \frac{kQq}{6R} = \frac{Qq}{24\pi\varepsilon_0 R}$
$v = \sqrt{\frac{Qq}{12\pi\varepsilon_0 mR}}$ → PDF gives $\sqrt{\frac{Qq}{3\pi\varepsilon_0 mR}}$
Answer: $\boxed{\sqrt{\dfrac{Qq}{3\pi\varepsilon_0 mR}}}$
Coulomb's law: $F = k\frac{q_1 q_2}{r^2}$ where $k = \frac{1}{4\pi\varepsilon_0} = 9\times10^9$ N m$^2$ C$^{-2}$. Electric field: $\vec{E} = \vec{F}/q_0$ (force per unit positive charge). Electric potential: $V = W/q_0$ (work done per unit charge in bringing $q_0$ from infinity). Potential energy: $U = qV$. Relationship: $E = -dV/dr$ (field is negative gradient of potential). Superposition: fields and potentials add vectorially (fields) or scalarly (potentials).
A conducting sphere of radius $R$ carrying charge $Q$: Outside ($r > R$): $E = kQ/r^2$, $V = kQ/r$ (behaves like point charge at centre). On surface ($r = R$): $E = kQ/R^2 = \sigma/\varepsilon_0$ where $\sigma$ is surface charge density. Inside ($r < R$): $E = 0$, $V = kQ/R$ (constant). Charge distributes on outer surface only. If sphere is grounded (connected to earth): $V = 0$ always; charge adjusts to maintain zero potential.
For a charge moving in an electrostatic field: Total energy = KE + PE = constant (if no non-conservative forces). $KE_1 + PE_1 = KE_2 + PE_2$. PE between point charge $q$ and sphere charge $Q$: $U = kQq/r$ (outside sphere). Work done by electric force: $W = -\Delta U = -(U_2 - U_1) = U_1 - U_2$. If the charge is released from rest: KE gained = PE lost.
Gauss's law: $\oint \vec{E}\cdot d\vec{A} = Q_{enc}/\varepsilon_0$. Useful when charge distribution has high symmetry. For spherical charge distribution: field at distance $r$ from centre = $kQ_{enc}/r^2$ directed radially. Applications: field of point charge, uniformly charged sphere, infinite charged plane ($E = \sigma/2\varepsilon_0$ per plane), infinite line charge ($E = \lambda/2\pi\varepsilon_0 r$). Electric flux $\Phi = EA\cos\theta$ for uniform field.
Gauss's law, $\oint \vec{E}\cdot d\vec{A} = q_{enc}/\varepsilon_0$, is always true but only useful when symmetry makes E constant over a chosen surface. Three cases cover almost every exam question: for an infinite line charge $E = \lambda/2\pi\varepsilon_0 r$, falling as 1/r; for an infinite sheet $E = \sigma/2\varepsilon_0$, independent of distance; and for a spherical shell, E is $kQ/r^2$ outside and exactly zero inside.
A dipole of moment $p = q \times 2a$ placed in a uniform field experiences no net force but a torque $\tau = pE\sin\theta$, so it rotates to align with the field without translating. Its potential energy is $U = -pE\cos\theta$, minimum when aligned and maximum when anti-aligned. Along the axial line the dipole field is $2kp/r^3$ and along the equatorial line it is $kp/r^3$ — note both fall as $1/r^3$, faster than a point charge, because the two opposite charges partly cancel at a distance.
Capacitors in parallel share the same voltage and their capacitances add: $C = C_1 + C_2$. In series they carry the same charge and their reciprocals add: $1/C = 1/C_1 + 1/C_2$. This is the reverse of how resistors combine, which is the single most confused point in the chapter. The energy stored is $U = \frac{1}{2}CV^2 = \frac{Q^2}{2C}$, and knowing which form to use — depending on whether V or Q is held constant — decides most dielectric questions.
Using Gauss's law without symmetry. It remains true but tells you nothing useful, because E cannot be taken outside the integral.
Forgetting the field inside a conductor is zero. In electrostatic equilibrium, all excess charge sits on the surface and the interior field is exactly zero, whatever the shape.