Electric Potential And Potential Energy

Electric potential due to a point charge is $V=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r}$, with $\frac{1}{4\pi {\epsilon }_0}=9\times 10^9\mathrm{N}{\mathrm{m}}^2/{\mathrm{C}}^2$, a formula stated in NCERT Class 12, Chapter 2 (Electrostatic Potential and Capacitance). Here $q=5\times 10^{-9}$ C and $r=0.5$ m. Substituting, $V=9\times 10^9\times \frac{5\times 10^{-9}}{0.5}$. The product $9\times 10^9\times 5\times 10^{-9}=45$, and dividing by $0.5$ gives $V=90$ V. The value 45 V is wrong because it forgets to divide by $r=0.5$, effectively using $r=1$ m. The value 180 V is wrong because it divides by $0.25$, mistakenly squaring the distance as if computing a field. The value 9 V is wrong because it drops the factor of 5 from the charge. As potential is a scalar, no direction enters the result. A sign and magnitude check confirms a positive charge gives positive potential, and nano-coulomb charges at half a metre yielding tens of volts is entirely reasonable.