(II) Two identical \({\bf{ + 9}}{\bf{.5}}\;{\bf{\mu C}}\) point charges are initially 5.3 cm from each other. If they are released at the same instant from rest, how fast will each be moving when they are very far away from each other? Assume they have identical masses of 1.0 mg.

The electrical potential energy may be described as the energy owned by the electric charges due to conservative Coulomb force and also due to the configuration of the point charges within a defined system.

The magnitude of both point charges is \(Q = 9.5\;{\rm{\mu C}}\).

The separation between the charges is \(r = 5.3\;{\rm{cm}}\).

The mass of both charges is \(m = 1.0\;{\rm{mg}}\).

The potential energy of two identical charges can be calculated as:

\(PE = \frac{{k{Q^2}}}{r}\)

Here, k is the Coulomb’s constant.

Substitute the values in the above equation.

\(\begin{aligned}PE &= \frac{{\left( {9 \times {{10}^9}\;{{{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}} \mathord{\left/{\vphantom {{{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}} {{{\rm{C}}^{\rm{2}}}}}} \right.} {{{\rm{C}}^{\rm{2}}}}}} \right){{\left( {\left( {{\rm{9}}{\rm{.5}}\;{\rm{\mu C}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 6}}}}\;{\rm{C}}}}{{{\rm{1}}\;{\rm{\mu C}}}}} \right)} \right)}^2}}}{{\left( {{\rm{5}}{\rm{.3}}\;{\rm{cm}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}\;{\rm{m}}}}{{{\rm{1}}\;{\rm{cm}}}}} \right)}}\\ &= 15.32\;{\rm{J}}\end{aligned}\)

Now apply the conservation of energy principle to calculate the velocity of the both point charges.

\(\begin{aligned}PE &= 2KE\\PE &= 2\left( {\frac{1}{2}m{v^2}} \right)\\v &= \sqrt {\frac{{PE}}{m}} \end{aligned}\)

Now apply the conservation of energy principle to calculate the velocity of the both point charges.

\(\begin{aligned}PE &= 2KE\\PE &= 2\left( {\frac{1}{2}m{v^2}} \right)\\v &= \sqrt {\frac{{PE}}{m}} \end{aligned}\)

Substitute the values in the above equation.

\(\begin{aligned}v &= \sqrt {\frac{{\left( {15.32\;{\rm{J}}} \right)}}{{\left( {1\;{\rm{mg}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 3}}}}\;{\rm{g}}}}{{{\rm{1}}\;{\rm{mg}}}}} \right)\left( {\frac{{{\rm{1}}\;{\rm{kg}}}}{{{\rm{1}}{{\rm{0}}^{\rm{3}}}\;{\rm{g}}}}} \right)}}} \\v &= 3.91 \times {10^3}\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}\end{aligned}\)

Thus, the velocity of point charges after they are released is \(3.91 \times {10^3}\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}\).