Question

A charge of 6.04 mC is placed at each corner of a square 0.180 m on a side. Determine the magnitude of the force on each charge.

Answer 1

Concepts and reason

The concept of electric force is required to solve the problem.

First, determine the length of the diagonal of the square. Then determine the force on one of the charge due to each remaining charge. Then determine the direction of each force and find out the magnitude of net force on that charge.

Fundamentals

Electrostatic force is the force between two charges separated by a distance r. The force on charge ${q_2}$ due to the charge ${q_1}$ is given as,

$F = \frac{{k{q_1}{q_2}}}{{{r_{12}}^2}}$

Here, k is the coulomb’s constant and ${r_{12}}$ is the distance between two charges.

The figure given below shows the arrangement of charges on the corners of square of side a.

The force on charge 4 due to charge 1 is,

${F_{14}} = \frac{{k{q_1}{q_4}}}{{{r_{14}}^2}}$

Substitute q for ${q_1}$ and ${q_4}$ , and a for ${r_{14}}$ in the above equation.

${F_{14}} = \frac{{k{q^2}}}{{{a^2}}}$

The force on charge 4 due to charge 3 is,

${F_{34}} = \frac{{k{q_3}{q_4}}}{{{r_{14}}^2}}$

Substitute q for ${q_3}$ and ${q_4}$ , and a for ${r_{14}}$ in the above equation.

${F_{34}} = \frac{{k{q^2}}}{{{a^2}}}$

Determine the distance between the charge 2 and charge 4 using Pythagoras theorem.

$\begin{array}{c}\\{r_{24}} = \sqrt {{a^2} + {a^2}} \\\\ = \sqrt 2 a\\\end{array}$

The force on charge 4 due to charge 2 is,

${F_{24}} = \frac{{k{q_2}{q_4}}}{{{r_{24}}^2}}$

Substitute $\sqrt 2 a$ for ${r_{24}}$ and q for ${q_2}$ and ${q_4}$ in the above equation.

${F_{24}} = \frac{{k{q^2}}}{{2{a^2}}}$

Resolve the force ${F_{24}}$ into horizontal and vertical component.

The horizontal component of force ${F_{24}}$ is,

${F_{24x}} = {F_{24}}\cos \theta$

The vertical component of force ${F_{24}}$ is,

${F_{24y}} = {F_{24}}\sin \theta$

The angle $\theta$ is given as,

$\begin{array}{c}\\\theta = {\tan ^{ - 1}}\left( {\frac{a}{a}} \right)\\\\ = {45^{\rm{o}}}\\\end{array}$

Refer the above figure and determine the net force in the horizontal direction.

${F_x} = {F_{34}} + {F_{24}}\cos \theta$

Substitute $\frac{{k{q^2}}}{{{a^2}}}$ for ${F_{34}}$ , $\frac{{k{q^2}}}{{2{a^2}}}$ for ${F_{24}}$ , and ${45^{\rm{o}}}$ for $\theta$ in the above equation.

$\begin{array}{c}\\{F_x} = \frac{{k{q^2}}}{{{a^2}}} + \frac{{k{q^2}}}{{2{a^2}}}\cos {45^{\rm{o}}}\\\\ = \frac{{k{q^2}}}{{{a^2}}}\left( {1 + \frac{1}{{2\sqrt 2 }}} \right)\\\end{array}$

Determine the net force in the vertical direction.

${F_y} = {F_{14}} + {F_{24}}\sin \theta$

Substitute $\frac{{k{q^2}}}{{{a^2}}}$ for ${F_{14}}$ , $\frac{{k{q^2}}}{{2{a^2}}}$ for ${F_{24}}$ , and ${45^{\rm{o}}}$ for $\theta$ in the above equation.

$\begin{array}{c}\\{F_y} = \frac{{k{q^2}}}{{{a^2}}} + \frac{{k{q^2}}}{{2{a^2}}}\sin {45^{\rm{o}}}\\\\ = \frac{{k{q^2}}}{{{a^2}}}\left( {1 + \frac{1}{{2\sqrt 2 }}} \right)\\\end{array}$

The magnitude of the net force on charge 4 is given as,

${F_{{\rm{net}}}} = \sqrt {{F_x}^2 + {F_y}^2}$

Substitute $\frac{{k{q^2}}}{{{a^2}}}\left( {1 + \frac{1}{{2\sqrt 2 }}} \right)$ for ${F_x}$ and ${F_y}$ in the above equation.

$\begin{array}{c}\\{F_{{\rm{net}}}} = \sqrt {{{\left( {\frac{{k{q^2}}}{{{a^2}}}\left( {1 + \frac{1}{{2\sqrt 2 }}} \right)} \right)}^2} + {{\left( {\frac{{k{q^2}}}{{{a^2}}}\left( {1 + \frac{1}{{2\sqrt 2 }}} \right)} \right)}^2}} \\\\ = \frac{{k{q^2}}}{{{a^2}}}\left( {1 + \frac{1}{{2\sqrt 2 }}} \right)\sqrt 2 \\\end{array}$

Substitute $8.99 \times {10^9}{\rm{ N}} \cdot {{\rm{m}}^2}{\rm{/}}{{\rm{C}}^{\rm{2}}}$ for k, 0.180 m for a, and 6.04 mC for q in the above equation.

$\begin{array}{c}\\{F_{{\rm{net}}}} = \frac{{\left( {8.99 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right){{\left( {6.04{\rm{ mC}}\left( {\frac{{{{10}^{ - 3}}{\rm{ C}}}}{{1{\rm{ mC}}}}} \right)} \right)}^2}}}{{{{\left( {0.180{\rm{ m}}} \right)}^2}}}\left( {1 + \frac{1}{{2\sqrt 2 }}} \right)\sqrt 2 \\\\ = 1.94 \times {10^7}{\rm{ N}}\\\end{array}$

As all the charges are equal, the net force on each charge will be same.

Ans:

The magnitude of net force on each charge is $1.94 \times {10^7}{\rm{ N}}$ .