(N/A) Given:
Radius of the inner sphere,$r_1 = 12 \;cm = 0.12 \;m$
Radius of the outer sphere,$r_2 = 13 \;cm = 0.13 \;m$
Charge on the inner sphere,$q = 2.5 \;\mu C = 2.5 \times 10^{-6} \;C$
Dielectric constant,$K = 32$
$(a)$ The capacitance $C$ of a spherical capacitor is given by:
$C = \frac{4 \pi \epsilon_0 K r_1 r_2}{r_2 - r_1}$
Using $\frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \;N \;m^2 \;C^{-2}$:
$C = \frac{32 \times 0.12 \times 0.13}{9 \times 10^9 \times (0.13 - 0.12)} = \frac{0.4992}{9 \times 10^7} \approx 5.55 \times 10^{-9} \;F$
$(b)$ The potential $V$ of the inner sphere is:
$V = \frac{q}{C} = \frac{2.5 \times 10^{-6}}{5.55 \times 10^{-9}} \approx 450 \;V = 4.5 \times 10^2 \;V$
$(c)$ Capacitance of an isolated sphere of radius $r = 0.12 \;m$ is:
$C' = 4 \pi \epsilon_0 r = \frac{0.12}{9 \times 10^9} \approx 1.33 \times 10^{-11} \;F$
The capacitance of the spherical capacitor is much larger because the earthed outer sphere reduces the potential of the inner sphere for the same charge,thereby increasing the capacitance $(C = q/V)$.