## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

$r = 5.21\times 10^{-7}~m$
The force exerted upward on the droplet by the electric field must be equal in magnitude to the weight of the droplet. We can find the mass of the droplet: $mg = E~q$ $mg = \frac{\Delta V~q}{d}$ $m = \frac{\Delta V~q}{g~d}$ $m = \frac{(25~V)(15)(1.6\times 10^{-19}~C)}{(9.8~m/s^2)~(12\times 10^{-3}~m)}$ $m = 5.1\times 10^{-16}~kg$ We can find the radius of the droplet: $\rho = \frac{m}{V}$ $V = \frac{m}{\rho}$ $\frac{4}{3}~\pi~r^3 = \frac{m}{\rho}$ $r^3 = \frac{3m}{4\pi~\rho}$ $r = \sqrt[3] {\frac{3m}{4\pi~\rho}}$ $r = \sqrt[3] {\frac{(3)(5.1\times 10^{-16}~kg)}{(4\pi)(860~kg/m^3)}}$ $r = 5.21\times 10^{-7}~m$