Original: sphere 1 and 2 each have charge $q$, separated by distance $r$.
$$F = \frac{kq^2}{r^2}$$Third uncharged sphere touched to sphere 1: charges split equally.
Sphere 1 → $q/2$, third sphere → $q/2$, sphere 2 → still $q$.
Third sphere (charge $q/2$) placed midway, at distance $r/2$ from each.
$$F_1 = \frac{k(q/2)(q/2)}{(r/2)^2} = \frac{kq^2/4}{r^2/4} = \frac{kq^2}{r^2} = F \text{ (repulsion toward sphere 2)}$$$$F_2 = \frac{k(q/2)(q)}{(r/2)^2} = \frac{kq^2/2}{r^2/4} = \frac{2kq^2}{r^2} = 2F \text{ (repulsion toward sphere 1)}$$Net force on middle sphere $= 2F - F = F$ ... but wait, the problem says between the two original spheres. Checking NEET answer key: $3F/8$.
Reconsidering: if third sphere is not placed midway but rather placed between them at midpoint of original $r$ separation, same calculation gives net $F$ on middle sphere. PDF answer = $3F/8$ suggests different reading.
$$F = \frac{1}{4\pi\epsilon_0}\frac{q_1 q_2}{r^2} = \frac{kq_1 q_2}{r^2}$$ $k = 9\times 10^9$ N m$^2$ C$^{-2}$. Force along line joining charges. Like charges repel, unlike attract. Vector form: $\vec{F}_{12} = k\frac{q_1 q_2}{r^2}\hat{r}_{12}$. Superposition principle: net force = vector sum of individual forces.
When two identical conducting spheres are touched: charges redistribute equally (by symmetry). If charges are $q_1$ and $q_2$: after touching each gets $(q_1+q_2)/2$. For uncharged sphere ($q_2 = 0$) touching sphere with charge $q$: both get $q/2$. After separation: charges are conserved individually.
$\vec{E} = \vec{F}/q_0$ (force per unit positive test charge). For point charge: $E = kq/r^2$. Superposition: $\vec{E}_{net} = \sum \vec{E}_i$. Field lines: start at positive, end at negative. Never cross. Density of lines proportional to field strength. Uniform field between parallel plates: $E = \sigma/\epsilon_0 = V/d$.
$V = W/q_0 = kq/r$ for point charge. Potential energy $U = kq_1q_2/r$. $E = -dV/dr$ (field = negative gradient of potential). Equipotential surfaces perpendicular to field lines. Work done moving charge between equipotentials = $q\Delta V$. Potential at surface of conductor = constant (equipotential surface).
$$\oint \vec{E}\cdot d\vec{A} = \frac{Q_{enc}}{\epsilon_0}$$ Net electric flux through closed surface = enclosed charge/\epsilon_0. Applications: field due to infinite line charge $E = \lambda/(2\pi\epsilon_0 r)$. Infinite sheet: $E = \sigma/(2\epsilon_0)$. Uniformly charged sphere: outside $E = kq/r^2$, inside $E = kqr/R^3$.
Capacitance $C = Q/V$. Parallel plate: $C = \epsilon_0 A/d$. With dielectric: $C = K\epsilon_0 A/d$ ($K$ = dielectric constant). Series: $1/C_{eff} = \sum 1/C_i$. Parallel: $C_{eff} = \sum C_i$. Energy stored: $U = Q^2/(2C) = CV^2/2 = QV/2$. Dielectric increases capacitance and reduces field (polarisation opposes external field).
Dipole moment $\vec{p} = q\vec{d}$ (from $-q$ to $+q$). Torque in uniform field: $\tau = pE\sin\theta = \vec{p}\times\vec{E}$. PE: $U = -pE\cos\theta = -\vec{p}\cdot\vec{E}$. Field on axis (end-on): $E = 2kp/r^3$. Field on equator (broad-side): $E = kp/r^3$. Electric field of dipole falls as $1/r^3$ (faster than monopole $1/r^2$).
In electrostatic equilibrium: field inside conductor = 0. Charges reside on surface. Field just outside surface = $\sigma/\epsilon_0$ (normal to surface). Entire conductor at same potential. Cavity inside conductor: field = 0 inside cavity (Faraday cage). Electrostatic shielding: sensitive instruments kept in grounded metal enclosures. Sharing of charge: when conductors connected, charges distribute to equalise potential.