(I) A point charge Q creates an electric potential of +165 V at a distance of 15 cm. What is Q?

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 electric potential at any point is, \(V = + 165\;{\rm{V}}\)

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

From equation (i), the point charge Q is given as:

\(Q = \frac{{Vr}}{k}\)

Substitute the values in the above expression.

\(\begin{aligned}Q &= \frac{{\left( {165\;{\rm{V}}} \right) \times \left( {15.0\;{\rm{cm}} \times \frac{{{{10}^{ - 2}}\;{\rm{m}}}}{{1\;{\rm{cm}}}}} \right)}}{{\left( {9.0 \times {{10}^9}\;{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)}}\\ &= 2.8 \times {10^{ - 9}}\;{\rm{C}}\end{aligned}\)

Thus, the point charge is\(2.8 \times {10^{ - 9}}\;{\rm{C}}\).