(I) What is the electric potential 15.0 cm from a \({\bf{3}}{\bf{.00}}\;{\bf{\mu C}}\) point charge?

The electric potential at any point in space relies on the charge and the distance of the point from the charge.

The electric potential due to a point charge is given by,

\(V = k\frac{Q}{r} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{Q}{r}\) … (i)

Here, k is electrostatic force constant whose value is \(9.0 \times {10^9}\;{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}\), \({\varepsilon _0}\)is the absolute electrical permittivity of the free space, Q is the charge and r is the distance.

The point charge is, \(Q = 3.00\;\mu {\rm{C}}\)

The distance of point from the point charge is, \(r = 15.0\;{\rm{cm}}\)

The electric potential at a distance of r is given by,

\(V = k\frac{Q}{r}\)

Substitute the values in the above expression.

\(\begin{aligned}V &= \left( {9.0 \times {{10}^9}\;{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right) \times \frac{{3.{\rm{00}}\;\mu {\rm{C}} \times \frac{{{\rm{1}}{{\rm{0}}^{ - 6}}\;{\rm{C}}}}{{1\;\mu {\rm{C}}}}}}{{15.0\;{\rm{cm}} \times \frac{{{{10}^{ - 2}}\;{\rm{m}}}}{{1\;{\rm{cm}}}}}}\\ &= 9.0 \times {10^9} \times 0.20 \times {10^{ - 4}}\\ &= 1.80 \times {10^5}\;{\rm{V}}\end{aligned}\)

Thus, the value of electric potential is \(1.80 \times {10^5}\;{\rm{V}}\).