$\therefore q=\frac{4}{3} \pi \rho R^{3}$
$V_{C}+V_{S}=\frac{3}{2}\left(\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{R}\right)-\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{R}$
$=\frac{q}{8 \pi \varepsilon_{0} R}$
Substituting the value of $q$ we have
$V_{C}-V_{S}=\frac{\rho R^{3}}{6 \varepsilon_{0}}$



$(A)$ the electrostatic field is zero
$(B)$ the electrostatic potential is constant
$(C)$ the electrostatic field is constant in magnitude
$(D)$ the electrostatic field has same direction
$\left\{ Use : \frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} Nm ^{2} C ^{-2}\right\}$