91Ó°ÊÓ

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\(2.8 \times {10^{ - 9}}\;{\rm{C}}\).