Question

Three charges are arranged as shown in the figure below. Find the magnitude and direction of the electrostatic force on the charge q = 4.86 nC at the origin. (Let r12 = 0.285 m.)

Answer 1

Concepts and reason

The concepts required to solve the given question is Coulomb’s law.

Initially, calculate the magnitude of the force in the horizontal and the vertical direction. Later, calculate the magnitude of net force by taking the resultant of two forces in the x and y direction. Finally, calculate the direction of force.

Fundamentals

The expression for the electrostatic force in the horizontal direction is as follows:

${F_{\rm{x}}} = \frac{{{q_1}{q_2}}}{{4\pi {\varepsilon _0}{r^2}}}$

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.

The expression for the electrostatic force in the vertical direction is as follows:

${F_{\rm{y}}} = \frac{{{q_1}{q_2}}}{{4\pi {\varepsilon _0}{r^2}}}$

The magnitude of the net force is as follows:

$F = \sqrt {F_{\rm{x}}^2 + F_{\rm{y}}^2}$

The expression for the calculation of the direction is as follows:

$\theta = 180^\circ + {\tan ^{ - 1}}\left( {\frac{{{F_{\rm{y}}}}}{{{F_{\rm{x}}}}}} \right)$

Substitute $9 \times {10^9}{\rm{ N}} \cdot {{\rm{m}}^2} \cdot {{\rm{C}}^{ - 2}}$ for $\frac{1}{{4\pi {\varepsilon _0}}}$, $4.86{\rm{ nC}}$ for ${q_1}$, ${\rm{6}}{\rm{.00 nC}}$for ${q_2}$, and 0.285 m for r in the equation ${F_{\rm{x}}} = \frac{{{q_1}{q_2}}}{{4\pi {\varepsilon _0}{r^2}}}$ to calculate the horizontal component of the force.

$\begin{array}{c}\\{F_{\rm{x}}} = \frac{{\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2} \cdot {{\rm{C}}^{ - 2}}} \right)\left( {4.86{\rm{ nC}}} \right)\left( {{\rm{6}}{\rm{.00 nC}}} \right)}}{{{{\left( {0.285{\rm{ m}}} \right)}^2}}}\left( { - \hat i} \right)\\\\ = \frac{{\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2} \cdot {{\rm{C}}^{ - 2}}} \right)\left( {4.86{\rm{ nC}}} \right)\left( {\frac{{{{10}^{ - 9}}{\rm{ C}}}}{{1.00{\rm{ nC}}}}} \right)\left( {{\rm{6}}{\rm{.00 nC}}} \right)\left( {\frac{{{{10}^{ - 9}}{\rm{ C}}}}{{1.00{\rm{ nC}}}}} \right)}}{{{{\left( {0.285{\rm{ m}}} \right)}^2}}}\left( { - \hat i} \right)\\\\ = - 3.23 \times {10^{ - 6}}\left( {\hat i} \right){\rm{ N}}\\\end{array}$

Substitute $9 \times {10^9}{\rm{ N}} \cdot {{\rm{m}}^2} \cdot {{\rm{C}}^{ - 2}}$ for $\frac{1}{{4\pi {\varepsilon _0}}}$, $4.86{\rm{ nC}}$ for ${q_1}$, ${\rm{3}}{\rm{.00 nC}}$for ${q_2}$, and 0.100 m for r in the equation ${F_{\rm{y}}} = \frac{{{q_1}{q_2}}}{{4\pi {\varepsilon _0}{r^2}}}$ to calculate the vertical component of the force.

$\begin{array}{c}\\{F_{\rm{y}}} = \frac{{\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2} \cdot {{\rm{C}}^{ - 2}}} \right)\left( {4.86{\rm{ nC}}} \right)\left( {{\rm{6}}{\rm{.00 nC}}} \right)}}{{{{\left( {0.285{\rm{ m}}} \right)}^2}}}\left( { - \hat i} \right)\\\\ = \frac{{\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2} \cdot {{\rm{C}}^{ - 2}}} \right)\left( {4.86{\rm{ nC}}} \right)\left( {\frac{{{{10}^{ - 9}}{\rm{ C}}}}{{1.00{\rm{ nC}}}}} \right)\left( {{\rm{3}}{\rm{.00 nC}}} \right)\left( {\frac{{{{10}^{ - 9}}{\rm{ C}}}}{{1.00{\rm{ nC}}}}} \right)}}{{{{\left( {0.100{\rm{ m}}} \right)}^2}}}\left( { - \hat i} \right)\\\\ = - 1.31 \times {10^{ - 5}}\left( {\hat j} \right){\rm{ N}}\\\end{array}$

Thus, the magnitude of the net force is obtained by substituting $- 3.23 \times {10^{ - 6}}{\rm{ N}}$ for ${F_{\rm{x}}}$ and $- 1.31 \times {10^{ - 5}}{\rm{ N}}$ for ${F_{\rm{y}}}$ in the equation $F = \sqrt {F_{\rm{x}}^2 + F_{\rm{y}}^2}$.

$\begin{array}{c}\\F = \sqrt {{{\left( { - 3.23 \times {{10}^{ - 6}}{\rm{ N}}} \right)}^2} + {{\left( { - 1.31 \times {{10}^{ - 5}}{\rm{ N}}} \right)}^2}} \\\\ = 1.35 \times {10^{ - 5}}{\rm{ N}}\\\end{array}$

Substitute$- 3.23 \times {10^{ - 6}}{\rm{ N}}$ for ${F_{\rm{x}}}$ and $- 1.31 \times {10^{ - 5}}{\rm{ N}}$ for ${F_{\rm{y}}}$ in the equation $\theta = 180^\circ + {\tan ^{ - 1}}\left( {\frac{{{F_{\rm{y}}}}}{{{F_{\rm{x}}}}}} \right)$.

$\begin{array}{c}\\\theta = 180^\circ + {\tan ^{ - 1}}\left( {\frac{{ - 1.31 \times {{10}^{ - 5}}{\rm{ N}}}}{{ - 3.23 \times {{10}^{ - 6}}{\rm{ N}}}}} \right)\\\\ = 180^\circ + {\tan ^{ - 1}}\left( {4.055} \right)\\\\ = 180^\circ + 76.1^\circ \\\\ = 256.1^\circ \\\end{array}$

Ans:

The magnitude of the force is equal to $1.35 \times {10^{ - 5}}{\rm{ N}}$ and the direction of the force is equal to $256.1^\circ$.