Question

(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?

Short answer

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

Step-by-step solution

Step 1:Understanding of work done

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.

Given Information

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 \(q = 0.5\;{\rm{\mu C}}\).

Evaluation of the electric potential for the initial condition

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.

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

Evaluation of the electric potential for the final condition

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:

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

Evaluation of the difference in the electric potential

The difference in the electric potential can be calculated as:

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

Evaluation of the work required to move a test charge

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

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