(II) A \({\bf{ + 35}}\;{\bf{\mu C}}\) point charge is placed 46 cm from an identical \({\bf{ + 35}}\;{\bf{\mu C}}\) charge. How much work would be required to move a \({\bf{ + 0}}{\bf{.50}}\;{\bf{\mu C}}\) test charge from a point midway between them to a point 12 cm closer to either of the charges?

The work required to move a point charge from one point to another point through an electric field is equal to the product of the charge and change in the electric potential between two points.

The magnitude of the first charge is \({q_1} = 35\;{\rm{\mu C}}\).

The magnitude of the second charge is \({q_2} = 35\;{\rm{\mu C}}\).

The magnitude of the test charge is \(q = 0.5\;{\rm{\mu C}}\).

Since the distance between the two charges is 46 cm, the midway distance between two charges will be:

\(\begin{aligned}{r_1} &= {r_2} = \frac{{46\;{\rm{cm}}}}{2}\\{r_1} &= {r_2} = 23\;{\rm{cm}}\end{aligned}\)

The electric potential at midway between two charges can be calculated as:

\(\begin{aligned}{V_1} &= \frac{{k{q_1}}}{{{r_1}}} + \frac{{k{q_2}}}{{{r_2}}}\\ &= k\left( {\frac{{{q_1}}}{{{r_1}}} + \frac{{{q_2}}}{{{r_2}}}} \right)\end{aligned}\)

Substitute the values in the above expression.

\(\begin{aligned}{V_1} &= \left( {9 \times {{10}^9}\;{{{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}} \mathord{\left/{\vphantom {{{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}} {{{\rm{C}}^{\rm{2}}}}}} \right.} {{{\rm{C}}^{\rm{2}}}}}} \right)\left( {\frac{{\left( {{\rm{35}}\;{\rm{\mu C}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 6}}}}\;{\rm{C}}}}{{{\rm{1}}\;{\rm{\mu C}}}}} \right)}}{{\left( {23\;{\rm{cm}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}\;{\rm{m}}}}{{{\rm{1}}\;{\rm{cm}}}}} \right)}} + \frac{{\left( {{\rm{35}}\;{\rm{\mu C}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 6}}}}\;{\rm{C}}}}{{{\rm{1}}\;{\rm{\mu C}}}}} \right)}}{{\left( {23\;{\rm{cm}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}\;{\rm{m}}}}{{{\rm{1}}\;{\rm{cm}}}}} \right)}}} \right)\\ &= 2.739 \times {10^6}\;{\rm{V}}\end{aligned}\)

Assume that the charge moved 12 cm closer to first charge, and then the distance of the first charge from the second charge will be:

\(\begin{aligned}{r_1} &= \frac{{{\rm{46}}\;{\rm{cm}}}}{{\rm{2}}} - 12\;{\rm{cm}}\\ &= 23\;{\rm{cm}} - 12\;{\rm{cm}}\\ &= 11\;{\rm{cm}}\end{aligned}\)

The distance of the second charge from the first charge will be:

\(\begin{aligned}{r_2} &= \frac{{{\rm{46}}\;{\rm{cm}}}}{{\rm{2}}} + 12\;{\rm{cm}}\\ &= 23\;{\rm{cm}} + 12\;{\rm{cm}}\\ &= 35\;{\rm{cm}}\end{aligned}\)

The electric potential at point 12 cm from first charge can be calculated as:

\(\begin{aligned}{V_2} &= k\left( {\frac{{{q_1}}}{{{r_1}}} + \frac{{{q_2}}}{{{r_2}}}} \right)\\ &= \left( {9 \times {{10}^9}\;{{{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}} \mathord{\left/{\vphantom {{{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}} {{{\rm{C}}^{\rm{2}}}}}} \right.} {{{\rm{C}}^{\rm{2}}}}}} \right)\left( {\frac{{\left( {{\rm{35}}\;{\rm{\mu C}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 6}}}}\;{\rm{C}}}}{{{\rm{1}}\;{\rm{\mu C}}}}} \right)}}{{\left( {11\;{\rm{cm}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}\;{\rm{m}}}}{{{\rm{1}}\;{\rm{cm}}}}} \right)}} + \frac{{\left( {{\rm{35}}\;{\rm{\mu C}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 6}}}}\;{\rm{C}}}}{{{\rm{1}}\;{\rm{\mu C}}}}} \right)}}{{\left( {35\;{\rm{cm}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}\;{\rm{m}}}}{{{\rm{1}}\;{\rm{cm}}}}} \right)}}} \right)\\ &= 3.763 \times {10^6}\;{\rm{V}}\end{aligned}\)

The difference in the electric potential can be calculated as:

\(\begin{aligned}\Delta V &= {V_2} - {V_1}\\ &= \left( {3.763 \times {{10}^6}\;{\rm{V}}} \right) - \left( {2.739 \times {{10}^6}\;{\rm{V}}} \right)\\& = 1.024 \times {10^6}\;{\rm{V}}\end{aligned}\)

\(\begin{aligned}W &= q\Delta V\\ &= \left( {{\rm{0}}{\rm{.5}}\;{\rm{\mu C}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 6}}}}\;{\rm{C}}}}{{{\rm{1}}\;{\rm{\mu C}}}}} \right)\left( {1.024 \times {{10}^6}\;{\rm{V}}} \right)\\ &= 0.512\;{\rm{J}}\end{aligned}\)

Thus, the work required to move a test charge is \(0.512\;{\rm{J}}\).