 Solution

First hold the surface electric charge fixed and increase the bulk salt concentration. Since ${\kappa }^2\propto c$, the Debye–Hückel screening length decreases as $c^{-1/2}$. For surface gap $h=R-2a\ge 0$,Thus both the range and the strength of repulsion decrease. In the repulsive phase diagram, the sample moves downwards at fixed volume fraction and can melt from a crystal into a fluid. Because screening length is already a diagram axis, changing salt moves the state through that diagram rather than independently changing its fixed-charge coexistence curves. If Van der Waals forces become dominant, however, adding salt may instead destabilize the dispersion and cause aggregation. These are distinct physical regimes.

Second increase $\sigma$ at fixed salt concentration and particle size. The electric charge is $Q=4\pi a^2\sigma$, so the pair repulsion increases approximately as ${\sigma }^2$ within the Debye–Hückel approximation, while the reservoir screening length remains unchanged. Freezing then becomes possible at lower volume fraction or shorter screening length. At fixed ${\lambda }_D$, the coexistence curves shift towards lower $\eta$; a previously fluid sample may crystallize.

These statements assume fixed, uniform charge and a salt reservoir dominating the screening. Strong charging requires nonlinear screening and possibly charge regulation. At very low added salt or appreciable particle concentration, particle-released counterions also contribute to ionic strength, so the screening length is no longer independently fixed by the added salt alone.