Question

Using the symmetry of the arrangement, determine the direction of the force on q in the figure below, given that qa = qb = 7.55 μC, qc = qd = –7.55 μC, and q is positive. qa is top left, qb is top right, qc is bottom left, qd is bottom right, amd q is in the middle for the figure.

Calculate the magnitude of the force on the charge q, given that the square is 11.00 cm on a side and q = 2.15 μC. (In Newtons)

Answer 1

Concepts and reason

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

Firstly, calculate the distance between the charges at the corner of the square and the charge at the center of the square. Then, find the forces on the central charge due to each of the charges at the corner of the square. Then, find the direction of the net force acting on the charge placed at the center of the square. Finally, determine the net force acting on the charge at the central charge.

Fundamentals

Electrostatic force: It is the force between two charges separated by a distance $r$ .

It is given by,

${F_{AB}} = \frac{{k{q_A}{q_B}}}{{{r_{BA}}^2}}$

Here, $k$ is the coulomb’s constant, ${q_A}{\rm{ and }}{q_B}$ are the charges and ${r_{AB}}$ is the distance between two charges ${q_A}{\rm{ and }}{q_B}$ .

The magnitude of resultant vector by using the parallelogram law of vector addition is given as,

$R = \sqrt {{P^2} + {Q^2} + 2PQ\cos \theta }$

Here, P and Q are the vector at the adjacent sides of the parallelogram and $\theta$ is the angle between vector P and Q.

(a)

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

The forces are equal and opposite so that they cancel out. The net force has no value such that it has no direction.

The resultant of OC and OA is F₁ and resultant of OB and OD is F₂. The force F₁ can be resolved along horizontal direction and downward direction. Similarly, the force F₂ can also be resolved along the negative horizontal direction and the downward direction.

The horizontal forces cancel out as they are equal in magnitude and opposite in direction. The components along the downward direction gets added up as they are in same direction.

The resulting force is in downward direction.

Refer figure 1 and determine the distance between the charges at the corner of the square and the charge at the center of the square.

In $\Delta AOC$ ,

$A{O^2} + O{C^2} = A{C^2}$

The distances AO and OC are same and are denoted by r.

Substitute r for AO and OD and a for AD in the above expression.

$\begin{array}{c}\\{r^2} + {r^2} = {a^2}\\\\{r^2} = \frac{{{a^2}}}{2}\\\end{array}$

The electrostatic force on the charge q due to charge at point A is,

${F_{{\rm{OA}}}} = \frac{{k{q_{\rm{a}}}q}}{{{r^2}}}$

Substitute Q for ${q_{\rm{a}}}$ and $\frac{{{a^2}}}{2}$ for r² in the above expression.

${F_{{\rm{OA}}}} = \frac{{2kQq}}{{{a^2}}}$

Similarly, the electrostatic force on the charge q due to charge at point D is,

${F_{{\rm{OD}}}} = \frac{{k{q_{\rm{d}}}q}}{{{r^2}}}$

Substitute Q for ${q_{\rm{d}}}$ and $\frac{{{a^2}}}{2}$ for r² in the above expression.

${F_{{\rm{OD}}}} = \frac{{2kQq}}{{{a^2}}}$

The direction of forces acting on the charge at the center is along the line joining the points A and D. The magnitudes of both forces are equal but one with negative sign.

The net force along the line AD acting on the center charge is the sum of both forces.

${F_1} = {F_{{\rm{OA}}}} + {F_{{\rm{OD}}}}$

Substitute $\frac{{2kQq}}{{{a^2}}}$ for ${F_{OA}}$ and $\frac{{2kQq}}{{{a^2}}}$ for ${F_{OC}}$ in the above expression.

$\begin{array}{c}\\{F_1} = \frac{{2kQq}}{{{a^2}}} + \left( {\frac{{2kQq}}{{{a^2}}}} \right)\\\\ = \frac{{4kqQ}}{{{a^2}}}\\\end{array}$

Similarly, the electrostatic force on charge q due to the charge at point B is,

${F_{{\rm{OB}}}} = \frac{{k{q_{\rm{b}}}q}}{{{r^2}}}$

Substitute Q for ${q_{\rm{b}}}$ and $\frac{{{a^2}}}{2}$ for r² in the above expression.

${F_{{\rm{OB}}}} = \frac{{2kQq}}{{{a^2}}}$

Similarly, the electrostatic force on charge at Q due to charge at point C is,

${F_{{\rm{OC}}}} = \frac{{k{q_{\rm{c}}}q}}{{{r^2}}}$

Substitute Q for ${q_{\rm{c}}}$ and $\frac{{{a^2}}}{2}$ for r² in the above expression.

${F_{{\rm{OC}}}} = \frac{{2kQq}}{{{a^2}}}$

The direction of forces acting on the charge at the center is along the line joining the points B and C. The magnitudes of both forces are equal but one with negative sign.

The net force along the line BC acting on the center charge is the sum of both forces.

${F_2} = {F_{{\rm{OB}}}} + {F_{{\rm{OC}}}}$

Substitute $\frac{{2kqQ}}{{{a^2}}}$ for ${F_{{\rm{OB}}}}$ and $\frac{{2kqQ}}{{{a^2}}}$ for ${F_{{\rm{OC}}}}$ in the above expression.

$\begin{array}{c}\\{F_2} = \frac{{2kqQ}}{{{a^2}}} + \left( {\frac{{2kqQ}}{{{a^2}}}} \right)\\\\ = \frac{{4kqQ}}{{{a^2}}}\\\end{array}$

The angle between the forces ${F_1}$ and ${F_2}$ is 90 degrees. The net force can be calculated by using the following expression.

$F = \sqrt {{F_1}^2 + {F_2}^2 + 2{F_1}{F_2}\cos \theta }$

Substitute $\frac{{4kqQ}}{{{a^2}}}$ for ${F_1}{\rm{ and }}{F_2}$ and ${90^{\rm{o}}}$ for $\theta$ in the above expression.

$\begin{array}{c}\\F = \sqrt {{{\left( {\frac{{4kqQ}}{{{a^2}}}} \right)}^2} + {{\left( {\frac{{4kqQ}}{{{a^2}}}} \right)}^2} + 2\left( {\frac{{4kqQ}}{{{a^2}}}} \right)\left( {\frac{{4kqQ}}{{{a^2}}}} \right)\cos {{90}^{\rm{o}}}} \\\\ = \frac{{4\sqrt 2 kqQ}}{{{a^2}}}\\\end{array}$

Substitute $7.55\;\mu {\rm{C}}$ for Q, $2.15{\rm{ }}\mu {\rm{C}}$ for q, 11.0 cm for a, and $8.99 \times {10^9}{\rm{ N}} \cdot {{\rm{m}}^2}{\rm{/}}{{\rm{C}}^2}$ for k in the equation $F = \frac{{4\sqrt 2 kqQ}}{{{a^2}}}$ .

$\begin{array}{c}\\F = \frac{{4\sqrt 2 \left( {8.99 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2}{\rm{/}}{{\rm{C}}^2}} \right)\left( {2.15{\rm{ }}\mu {\rm{C}}\left( {\frac{{{{10}^{ - 6}}{\rm{ C}}}}{{1{\rm{ }}\mu {\rm{C}}}}} \right)} \right)\left( {7.55\;\mu {\rm{C}}\left( {\frac{{{{10}^{ - 6}}{\rm{ C}}}}{{1{\rm{ }}\mu {\rm{C}}}}} \right)} \right)}}{{{{\left( {11.0{\rm{ cm}}\left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1{\rm{ cm}}}}} \right)} \right)}^2}}}\\\\ = 68.2{\rm{ N}}\\\end{array}$

Ans:

The net force on the charge q is in downward direction and the magnitude of the net force is 68.2 N