Answer :
[tex]\boxed{\sf C=\dfrac{Q}{V}}[/tex]
But
[tex]\boxed{\sf \Delta V_{R_2\to R_1}={\displaystyle{\int}^{R_1}_{R_2}}dV=-{\displaystyle{\int}^{R_1}_{R_2}}\dfrac{kQ}{r^2}dR}[/tex]
- Hence higher the radius lower the voltage
- Lower the voltage higher the capacitance .