As $\mathrm{q}$ is constant, so $E \propto \frac{1}{R^{2}}$
Radius is halved. Therefore, electric field will becomes $4$ times or $4 \mathrm{\,E}.$
Further, $\quad \mathrm{V}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{q}}{\mathrm{R}}.$
As $\mathrm{q}$ is constant, so $\mathrm{V} \propto \frac{1}{\mathrm{R}}.$
Radius is halved, so potential will becomes two time or $2 \mathrm{\,V}.$

$(i)$ The potential difference
$(ii)$ The capacitance
$(iii)$ The charge on the plates


