Problem: Free charges do not remain stationary when close together. To illustrate this, calculate the magnitude of the instantaneous acceleration, in meters per second squared, of two isolated protons separated by 2.5 nm.

FREE Expert Solution

Newton's second law:

$\boxed{\mathbf{\Sigma }\mathbf{F}\mathbf{=}\mathbf{m}\mathbf{a}}$

Coulomb's law:

$\boxed{\mathbf{F}\mathbf{=}\frac{\mathbf{k}{\mathbf{q}}_{\mathbf{1}}{\mathbf{q}}_{\mathbf{2}}}{{\mathbf{r}}^{\mathbf{2}}}}$

q₁ = q₂ = +e = 1.6 × 10^-19 C

k = 8.99 × 10⁹ N•m²/C² 

r = 2.5nm(10^-9m/1nm) = 2.5 × 10^-9 m

m_p = 1.67 × 10^-27 kg