Question

Calculate the magnitude and direction of the Coulomb force on each of the three charges shown in the figure below.
6.00 µC charge:
magnitude


Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. N
direction


1.50 µC charge:
magnitude N
direction

-2.00 µC charge:
magnitude N
direction



(6)------(1.5)-------(-2)
<---3cm--><----2cm-->

Answer 1

Concepts and reason

The concepts required to solve the given question is electrostatic force.

Initially, write an expression for the electrostatic forces between two charges and calculate the net force on the $6.00{\rm{ }}\mu {\rm{C}}$ charge. Later, calculate the net force on the $15{\rm{ }}\mu {\rm{C}}$ charge. Finally, calculate the force of attraction between $- 2.00{\rm{ }}\mu {\rm{C}}$ and $15.0{\rm{ }}\mu {\rm{C}}$ charges.

Fundamentals

The expression for the electrostatic force is as follows:

$F = \frac{{qq'}}{{4\pi {\varepsilon _0}{r^2}}}$

Here, q and q’ are the magnitude of the charges, r is the distance between the charges, and ${\varepsilon _0}$ is the permittivity of free space.

Assume ${F_1}$ be the force acting on the $6.00{\rm{ }}\mu {\rm{C}}$ charge along the negative x axis and ${F_2}$ be the force along the positive x axis between the $6.00{\rm{ }}\mu {\rm{C}}$ and $- 2.00{\rm{ }}\mu {\rm{C}}$.

Thus, the net force on the $6.00{\rm{ }}\mu {\rm{C}}$ charge is as follows:

${F_{{\rm{net}}}} = - {F_1} + {F_2}$

Substitute $\frac{{qq'}}{{4\pi {\varepsilon _0}{r^2}}}$ for ${F_1}$ and ${F_2}$ in the equation ${F_{{\rm{net}}}} = - {F_1} + {F_2}$.

${F_{{\rm{net}}}} = - \frac{{{q_1}{q_1}'}}{{4\pi {\varepsilon _0}{r_1}^2}} + \frac{{{q_2}{q_2}'}}{{4\pi {\varepsilon _0}{r_2}^2}}$

Substitute $9 \times {10^9}{\rm{ N}} \cdot {{\rm{m}}^2} \cdot {{\rm{C}}^{ - 2}}$ for $\frac{1}{{4\pi {\varepsilon _0}}}$, $6.00{\rm{ }}\mu {\rm{C}}$ for ${q_1}$ and ${q_2}$, $1.50{\rm{ }}\mu {\rm{C}}$ for ${q_1}'$ and $2.00{\rm{ }}\mu {\rm{C}}$ for ${q_2}$, 3.00 cm for ${r_1}$, and 5.00 cm for ${r_2}$ in the equation ${F_{{\rm{net}}}} = - \frac{{{q_1}{q_1}'}}{{4\pi {\varepsilon _0}{r_1}^2}} + \frac{{{q_2}{q_2}'}}{{4\pi {\varepsilon _0}{r_2}^2}}$

$\begin{array}{c}\\{F_{{\rm{net}}}} = - \frac{{{q_1}{q_1}'}}{{4\pi {\varepsilon _0}{r_1}^2}} + \frac{{{q_2}{q_2}'}}{{4\pi {\varepsilon _0}{r_2}^2}}\\\\ = 9 \times {10^9}{\rm{ N}} \cdot {{\rm{m}}^2} \cdot {{\rm{C}}^{ - 2}}\left( { - \frac{{\left( {6.00{\rm{ }}\mu {\rm{C}}} \right)1.50{\rm{ }}\mu {\rm{C}}}}{{{{\left( {3.00{\rm{ cm}}} \right)}^2}}} + \frac{{\left( {6.00{\rm{ }}\mu {\rm{C}}} \right)\left( {2.00{\rm{ }}\mu {\rm{C}}} \right)}}{{{{\left( {5.00{\rm{ cm}}} \right)}^2}}}} \right)\\\\ = 9 \times {10^9}{\rm{ N}} \cdot {{\rm{m}}^2} \cdot {{\rm{C}}^{ - 2}}\left( { - \frac{{\left( {6.00{\rm{ }}\mu {\rm{C}}} \right)\left( {1.50{\rm{ }}\mu {\rm{C}}} \right)}}{{{{\left( {3.00{\rm{ cm}}} \right)}^2}{{\left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1.00{\rm{ cm}}}}} \right)}^2}}} + \frac{{\left( {6.00{\rm{ }}\mu {\rm{C}}} \right)\left( {2.00{\rm{ }}\mu {\rm{C}}} \right)}}{{{{\left( {5.00{\rm{ cm}}} \right)}^2}{{\left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1.00{\rm{ cm}}}}} \right)}^2}}}} \right){\left( {\frac{{{{10}^{ - 6}}{\rm{ C}}}}{{1.00{\rm{ }}\mu {\rm{C}}}}} \right)^2}\\\\ = - 46.8{\rm{ N}}\\\end{array}$

The net force on the $1.50{\rm{ }}\mu {\rm{C}}$ charge is as follows:

${F_{{\rm{net}}}} = {F_1} + {F_2}$

Substitute $\frac{{qq'}}{{4\pi {\varepsilon _0}{r^2}}}$ for ${F_1}$ and ${F_2}$ in the equation ${F_{{\rm{net}}}} = {F_1} + {F_2}$.

${F_{{\rm{net}}}} = - \frac{{{q_1}{q_1}'}}{{4\pi {\varepsilon _0}{r_1}^2}} + \frac{{{q_2}{q_2}'}}{{4\pi {\varepsilon _0}{r_2}^2}}$

Substitute $9 \times {10^9}{\rm{ N}} \cdot {{\rm{m}}^2} \cdot {{\rm{C}}^{ - 2}}$ for $\frac{1}{{4\pi {\varepsilon _0}}}$, $6.00{\rm{ }}\mu {\rm{C}}$ for ${q_1}$, $1.50{\rm{ }}\mu {\rm{C}}$ for ${q_1}'$ and ${q_2}'$, 3.00 cm for ${r_1}$, $2.00{\rm{ }}\mu {\rm{C}}$ for ${q_2}$ , and 2.00 cm for ${r_2}$ in the equation ${F_{{\rm{net}}}} = - \frac{{{q_1}{q_1}'}}{{4\pi {\varepsilon _0}{r_1}^2}} + \frac{{{q_2}{q_2}'}}{{4\pi {\varepsilon _0}{r_2}^2}}$.

$\begin{array}{c}\\{F_{{\rm{net}}}} = \frac{{{q_1}{q_1}'}}{{4\pi {\varepsilon _0}{r_1}^2}} + \frac{{{q_2}{q_2}'}}{{4\pi {\varepsilon _0}{r_2}^2}}\\\\ = \left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2} \cdot {{\rm{C}}^{ - 2}}} \right)\left( {\frac{{\left( {6.00{\rm{ }}\mu {\rm{C}}} \right)1.50{\rm{ }}\mu {\rm{C}}}}{{{{\left( {3.00{\rm{ cm}}} \right)}^2}}} + \frac{{\left( {2.00{\rm{ }}\mu {\rm{C}}} \right)1.50{\rm{ }}\mu {\rm{C}}}}{{{{\left( {2.00{\rm{ cm}}} \right)}^2}}}} \right)\\\\ = \left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2} \cdot {{\rm{C}}^{ - 2}}} \right)\left( {\frac{{\left( {6.00{\rm{ }}\mu {\rm{C}}} \right)\left( {1.50{\rm{ }}\mu {\rm{C}}} \right)}}{{{{\left( {3.00{\rm{ cm}}} \right)}^2}{{\left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1.00{\rm{ cm}}}}} \right)}^2}}} + \frac{{\left( {2.00{\rm{ }}\mu {\rm{C}}} \right)\left( {1.50{\rm{ }}\mu {\rm{C}}} \right)}}{{{{\left( {2.00{\rm{ cm}}} \right)}^2}{{\left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1.00{\rm{ cm}}}}} \right)}^2}}}} \right){\left( {\frac{{{{10}^{ - 6}}{\rm{ C}}}}{{1.00{\rm{ }}\mu {\rm{C}}}}} \right)^2}\\\\ = 157.5{\rm{ N}}\\\end{array}$

Assume ${F_1}$ be the force of attraction acting on the $- 2.00{\rm{ }}\mu {\rm{C}}$ charge due to $6.00{\rm{ }}\mu {\rm{C}}$along the negative x axis and ${F_2}$ be the force of attraction between $- 2.00{\rm{ }}\mu {\rm{C}}$ and $1.50{\rm{ }}\mu {\rm{C}}$charges along the negative x axis.

Thus, the net force is as follows:

${F_{{\rm{net}}}} = - {F_1} - {F_2}$

Substitute $\frac{{qq'}}{{4\pi {\varepsilon _0}{r^2}}}$ for ${F_1}$ and ${F_2}$ in the equation ${F_{{\rm{net}}}} = {F_1} + {F_2}$.

${F_{{\rm{net}}}} = - \left( {\frac{{{q_1}{q_1}'}}{{4\pi {\varepsilon _0}{r_1}^2}} + \frac{{{q_2}{q_2}'}}{{4\pi {\varepsilon _0}{r_2}^2}}} \right)$

Substitute $9 \times {10^9}{\rm{ N}} \cdot {{\rm{m}}^2} \cdot {{\rm{C}}^{ - 2}}$ for $\frac{1}{{4\pi {\varepsilon _0}}}$, $6.00{\rm{ }}\mu {\rm{C}}$ for ${q_1}$, ${\rm{2}}{\rm{.00 }}\mu {\rm{C}}$ for ${q_1}'$ and ${q_2}'$, 5.00 cm for ${r_1}$, ${\rm{1}}{\rm{.50 }}\mu {\rm{C}}$ for ${q_2}$ , and 2.00 cm for ${r_2}$ in the equation ${F_{{\rm{net}}}} = - \left( {\frac{{{q_1}{q_1}'}}{{4\pi {\varepsilon _0}{r_1}^2}} + \frac{{{q_2}{q_2}'}}{{4\pi {\varepsilon _0}{r_2}^2}}} \right)$.

$\begin{array}{c}\\{F_{{\rm{net}}}} = - \left( {\frac{{{q_1}{q_1}'}}{{4\pi {\varepsilon _0}{r_1}^2}} + \frac{{{q_2}{q_2}'}}{{4\pi {\varepsilon _0}{r_2}^2}}} \right)\\\\ = - \left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2} \cdot {{\rm{C}}^{ - 2}}} \right)\left( {\frac{{\left( {6.00{\rm{ }}\mu {\rm{C}}} \right)\left( {2.00{\rm{ }}\mu {\rm{C}}} \right)}}{{{{\left( {5.00{\rm{ cm}}} \right)}^2}}} + \frac{{\left( {2.00{\rm{ }}\mu {\rm{C}}} \right)\left( {1.50{\rm{ }}\mu {\rm{C}}} \right)}}{{{{\left( {2.00{\rm{ cm}}} \right)}^2}}}} \right)\\\\ = - \left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2} \cdot {{\rm{C}}^{ - 2}}} \right)\left( {\frac{{\left( {6.00{\rm{ }}\mu {\rm{C}}} \right)\left( {2.00{\rm{ }}\mu {\rm{C}}} \right)}}{{{{\left( {5.00{\rm{ cm}}} \right)}^2}{{\left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1.00{\rm{ cm}}}}} \right)}^2}}} + \frac{{\left( {2.00{\rm{ }}\mu {\rm{C}}} \right)\left( {1.50{\rm{ }}\mu {\rm{C}}} \right)}}{{{{\left( {2.00{\rm{ cm}}} \right)}^2}{{\left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1.00{\rm{ cm}}}}} \right)}^2}}}} \right){\left( {\frac{{{{10}^{ - 6}}{\rm{ C}}}}{{1.00{\rm{ }}\mu {\rm{C}}}}} \right)^2}\\\\ = - 110.7{\rm{ N}}\\\end{array}$

Ans:

The magnitude of the force is equal to 46.8 N and is directed in the negative x direction.

The magnitude of the force is equal to 157.5 N and is directed in the positive x direction.

The magnitude of the force is equal to 110.7 N and is directed in the negative x direction.