Question: (III) Use Coulomb’s law to determine the magnitude and direction of the electric field at points A and B in Fig. 16–60 due to the two positive charges \(\left( {Q = 4.7\;{\rm{\mu C}}} \right)\) shown. Are your results consistent with Fig. 16–32b?

FIGURE 16–60 Problem 35.

The value of the Coulomb force can be obtained by evaluating the value of the charges, the Coulomb’s constant, and the separation between the charges.

Its value is altered inversely to the value of the separation between the charges.

Given data:

The magnitude of the charge is \(Q = 4.7\;{\rm{\mu C}}\).

The net electric field at point A is shown below:

The angle made by the electric charge with the horizontal can be calculated as:

\(\begin{aligned}{l}\theta &= {\tan ^{ - 1}}\left( {\frac{{5\;{\rm{cm}}}}{{10\;{\rm{cm}}}}} \right)\\\theta &= 26.6^\circ \end{aligned}\)

The shortest distance of the electric charge from point A can be calculated as:

\(\begin{aligned}{l}d &= \sqrt {{{\left( {\left( {{\rm{5}}\;{\rm{cm}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}\;{\rm{m}}}}{{{\rm{1}}\;{\rm{cm}}}}} \right)} \right)}^2} + {{\left( {\left( {10\;{\rm{cm}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}\;{\rm{m}}}}{{{\rm{1}}\;{\rm{cm}}}}} \right)} \right)}^2}} \\d &= 0.1118\;{\rm{m}}\end{aligned}\)

The electric field at point A can be calculated as:

The net electric field at point B is shown below:

Point B is not symmetrical. Therefore, both vertical and horizontal components show the effect on point B.

The angle made by the charge on the left side can be calculated as:

\(\begin{aligned}{l}{\theta _{{\rm{left}}}} &= {\tan ^{ - 1}}\left( {\frac{{5\;{\rm{cm}}}}{{5\;{\rm{cm}}}}} \right)\\{\theta _{{\rm{left}}}} &= 45^\circ \end{aligned}\)

The angle made by the charge on the right side can be calculated as:

\(\begin{aligned}{l}{\theta _{{\rm{right}}}} &= {\tan ^{ - 1}}\left( {\frac{{5\;{\rm{cm}}}}{{15\;{\rm{cm}}}}} \right)\\{\theta _{{\rm{right}}}} &= 18.4^\circ \end{aligned}\)

The distance of the charge from point B on the left side can be calculated as:

\(\begin{aligned}{l}{d_{{\rm{left}}}} &= \sqrt {{{\left( {\left( {{\rm{5}}\;{\rm{cm}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}\;{\rm{m}}}}{{{\rm{1}}\;{\rm{cm}}}}} \right)} \right)}^2} + {{\left( {\left( {5\;{\rm{cm}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}\;{\rm{m}}}}{{{\rm{1}}\;{\rm{cm}}}}} \right)} \right)}^2}} \\{d_{{\rm{left}}}} &= 0.0707\;{\rm{m}}\end{aligned}\)

The distance of the charge from point B on the right side can be calculated as:

\(\begin{aligned}{l}{d_{{\rm{right}}}} &= \sqrt {{{\left( {\left( {{\rm{5}}\;{\rm{cm}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}\;{\rm{m}}}}{{{\rm{1}}\;{\rm{cm}}}}} \right)} \right)}^2} + {{\left( {\left( {15\;{\rm{cm}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}\;{\rm{m}}}}{{{\rm{1}}\;{\rm{cm}}}}} \right)} \right)}^2}} \\{d_{{\rm{right}}}} &= 0.1582\;{\rm{m}}\end{aligned}\)

The x-component of the electric field at point B can be calculated as:

The y-component of the electric field at point B can be calculated as:

The magnitude of the electric field at point B can be calculated as:

The direction of the electric field at point B can be calculated as:

The results are consistent with Fig. 16–32b. The field at point B should be to the right and vertical in the diagram, meeting the calculations. The field at point A is directed straight up, meeting the calculations. Ultimately, the field lines are closer together at point B than at point A, showing that the field is stronger, and meet the calculations.