Question

Four identical charges Q are placed at the corners of a square of side L.

1.Find the magnitude total force exerted on one charge by the other three charges.

 |F| =  ?

Answer 1

Concepts and reason

‎The concepts that are required to solve this problem are electric force between two point charges, resolution of a vector in to rectangular components, and the net force on an object due to multiple forces acting on it simultaneously.

Initially, call four identical point charges as A, B, C, and D. Place the four charges at four corners of a square of side length L in clockwise order. Later, calculate the electric force on the charge B due to charges A, C, and D in vector form. Later, add these three forces on charge B in vector form. This force vector is called net force vector on the charge B due to other three charges. Later, calculate the magnitude of this force vector.

Fundamentals

The magnitude of the electric force F on a charge ${q_1}$ due to other charge ${q_2}$ can be calculated using the following formula:

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

Here, k is the Coulomb’s constant and r is the distance between the two charges.

From the super position principle, the net electric force ${\vec F_{{\rm{net,B}}}}$ on a charge B due to other three charges A, C, and D is equal to the vector sum of the electric force on the charge B due to these three charges.

${\vec F_{{\rm{net,B}}}} = {\vec F_{\rm{A}}} + {\vec F_{\rm{C}}} + {\vec F_{\rm{D}}}$

Here, ${\vec F_{\rm{A}}}$ is the vector form of electric force on charge B due to charge A, ${\vec F_{\rm{C}}}$ is the vector form of electric force on charge B due to charge C, and ${\vec F_{\rm{D}}}$ is the vector form of electric force on charge B due to charge D.

From the Pythagoras theorem, in a right angle triangle, the square of the hypotenuse is equal to the sum of the square of each side of the triangle.

Magnitude of a two dimensional vector $\vec s = {s_x}\hat i + {s_y}\hat j$ can be calculated using the following formula:

$\left| {\vec s} \right| = \sqrt {s_x^2 + s_y^2}$

Here, ${s_x}$ is the x-component of the vector and ${s_y}$ is the y-component of the vector.

The figure given below represents the arrangement of four charges at corners of a square and the direction of all the electric forces acting on a charge due to other three charges:

Substitute Q for ${q_1}$ and ${q_2}$ , L for r in the equation $F = \frac{{k{q_1}{q_2}}}{{{r^2}}}$ , and calculate the magnitude of the force ${F_{\rm{A}}}$ on charge B due to charge A.

$\begin{array}{c}\\{F_{\rm{A}}} = \frac{{k\left( Q \right)\left( Q \right)}}{{{L^2}}}\\\\ = \frac{{k{Q^2}}}{{{L^2}}}\\\end{array}$

The vector form of the force on charge B due to charge A is,

${\vec F_{\rm{A}}} = \frac{{k{Q^2}}}{{{L^2}}}\hat i$

Substitute Q for ${q_1}$ and ${q_2}$ , L for r in the equation $F = \frac{{k{q_1}{q_2}}}{{{r^2}}}$ , and calculate the magnitude of the force ${F_{\rm{C}}}$ on charge B due to charge C.

$\begin{array}{c}\\{F_{\rm{C}}} = \frac{{k\left( Q \right)\left( Q \right)}}{{{L^2}}}\\\\ = \frac{{k{Q^2}}}{{{L^2}}}\\\end{array}$

The vector form of the force on charge B due to charge C is,

${\vec F_{\rm{C}}} = \frac{{k{Q^2}}}{{{L^2}}}\hat j$

Substitute Q for ${q_1}$ and ${q_2}$ , $\sqrt 2 L$ for r in the equation $F = \frac{{k{q_1}{q_2}}}{{{r^2}}}$ , and calculate the magnitude of the force ${F_{\rm{D}}}$ on charge B due to charge D.

$\begin{array}{c}\\{F_{\rm{D}}} = \frac{{k\left( Q \right)\left( Q \right)}}{{{{\left( {\sqrt 2 L} \right)}^2}}}\\\\ = \frac{{k{Q^2}}}{{2{L^2}}}\\\end{array}$

The vector form of the force on charge B due to charge D is,

${\vec F_{\rm{D}}} = \frac{{k{Q^2}}}{{2{L^2}}}\left( {\cos 45^\circ \hat i + \sin 45^\circ \hat j} \right)$

Substitute $\frac{{k{Q^2}}}{{{L^2}}}\hat i$ for ${\vec F_{\rm{A}}}$ and $\frac{{k{Q^2}}}{{{L^2}}}\hat j$ for ${\vec F_{\rm{C}}}$ , and $\frac{{k{Q^2}}}{{2{L^2}}}\left( {\cos 45^\circ \hat i + \sin 45^\circ \hat j} \right)$ for ${\vec F_{\rm{D}}}$ in the equation ${\vec F_{{\rm{net,B}}}} = {\vec F_{\rm{A}}} + {\vec F_{\rm{C}}} + {\vec F_{\rm{D}}}$ .

$\begin{array}{c}\\{{\vec F}_{{\rm{net,B}}}} = \frac{{k{Q^2}}}{{{L^2}}}\hat i + \frac{{k{Q^2}}}{{{L^2}}}\hat j + \frac{{k{Q^2}}}{{2{L^2}}}\left( {\cos 45^\circ \hat i + \sin 45^\circ \hat j} \right)\\\\ = \left( {\frac{{k{Q^2}}}{{{L^2}}} + \frac{{k{Q^2}}}{{2\sqrt 2 {L^2}}}} \right)\hat i + \left( {\frac{{k{Q^2}}}{{{L^2}}} + \frac{{k{Q^2}}}{{2\sqrt 2 {L^2}}}} \right)\hat j\\\end{array}$

The magnitude of the vector ${\vec F_{{\rm{net,B}}}}$ is,

$\begin{array}{c}\\\left| {{{\vec F}_{{\rm{net,B}}}}} \right| = \sqrt {{{\left( {\frac{{k{Q^2}}}{{{L^2}}} + \frac{{k{Q^2}}}{{2\sqrt 2 {L^2}}}} \right)}^2} + {{\left( {\frac{{k{Q^2}}}{{{L^2}}} + \frac{{k{Q^2}}}{{2\sqrt 2 {L^2}}}} \right)}^2}} \\\\ = \sqrt 2 \left( {\frac{{k{Q^2}}}{{{L^2}}} + \frac{{k{Q^2}}}{{2\sqrt 2 {L^2}}}} \right)\\\\ = \left( {\frac{{\sqrt 2 k{Q^2}}}{{{L^2}}} + \frac{{k{Q^2}}}{{2{L^2}}}} \right)\\\\ = \frac{{k{Q^2}}}{{{L^2}}}\left( {\sqrt 2 + \frac{1}{2}} \right)\\\end{array}$

Ans:

The magnitude of the net force acting on a charge due to other three charges when four identical charges are placed at four corners of a square is $\left| {\vec F} \right| = \frac{{k{Q^2}}}{{{L^2}}}\left( {\sqrt 2 + \frac{1}{2}} \right)$ .