ELECTROSTATICS CYU 27

Question 27 Solution

Goal: Find the radius $r$ that minimizes the total energy (Electrostatic + Surface Tension).

Step 1: Define Energies

Let $N$ be the number of drops and $r$ be the radius.
Total Volume $V_{tot}$ is constant $\implies N \cdot \frac{4}{3}\pi r^3 = V_{tot} \implies N \propto r^{-3}$.

• Surface Energy ($U_S$): $$U_S = N \cdot (4\pi r^2 \sigma) = \left(\frac{V_{tot}}{\frac{4}{3}\pi r^3}\right) 4\pi r^2 \sigma = \frac{3 \sigma V_{tot}}{r}$$
• Electrostatic Energy ($U_E$): Self-energy of one drop is $\frac{3}{5} \frac{k q_{drop}^2}{r}$.
  Charge per drop $q_{drop} = \rho \cdot \frac{4}{3}\pi r^3$. $$U_{E, one} = \frac{3}{5 \cdot 4\pi\epsilon_0 r} \left( \rho \frac{4}{3}\pi r^3 \right)^2 = \frac{4\pi \rho^2 r^5}{15\epsilon_0}$$ Total $U_E = N \cdot U_{E, one} = \left(\frac{V_{tot}}{\frac{4}{3}\pi r^3}\right) \frac{4\pi \rho^2 r^5}{15\epsilon_0} = \frac{\rho^2 V_{tot} r^2}{5\epsilon_0}$

Total Energy $U(r) = \frac{A}{r} + B r^2$, where $A = 3\sigma V_{tot}$ and $B = \frac{\rho^2 V_{tot}}{5\epsilon_0}$.

Step 2: Minimize Energy

Differentiate $U(r)$ with respect to $r$ and set to zero:

$$\frac{dU}{dr} = -\frac{A}{r^2} + 2Br = 0$$ $$\frac{3\sigma V_{tot}}{r^2} = 2 \left( \frac{\rho^2 V_{tot}}{5\epsilon_0} \right) r$$ $$\frac{3\sigma}{r^2} = \frac{2\rho^2 r}{5\epsilon_0}$$ $$r^3 = \frac{15\sigma\epsilon_0}{2\rho^2}$$