Chapter 23, Problem 23.85CP

Eight charged panicles, each of magnitude q, are located on the corners of a cube of edge s as shown in Figure P22.48. (a) Determine the x, y, and z components of the total force exerted by the other charges on the charge located at point A. What are (b) the magnitude and (c) the direction of this total force?

Figure P22.48

To determine

The $x$, $y$ and $z$ components of the force.

Answer to Problem 23.85CP

The $x$, $y$ and $z$ components of the force are, ${\overrightarrow{F}}_{\text{x}\left(\text{net}\right)}=\left[\frac{k{q}^2}{s^2}\left(1+\frac{1}{\sqrt{2}}+\frac{1}{3\sqrt{3}}\right)\right]\widehat{\text{i}}$, ${\overrightarrow{F}}_{\text{y}\left(\text{net}\right)}=\left[\frac{k{q}^2}{s^2}\left(1+\frac{1}{\sqrt{2}}+\frac{1}{3\sqrt{3}}\right)\right]\widehat{\text{j}}$ and ${\overrightarrow{F}}_{z\left(\text{net}\right)}=\left[\frac{k{q}^2}{s^2}\left(1+\frac{1}{\sqrt{2}}+\frac{1}{3\sqrt{3}}\right)\right]\widehat{\text{k}}$ respectively.

Explanation of Solution

The side of the each cube is $s$ and the magnitude of each charge is $q$.

Write the expression for the coulomb’s force law between the two charges

    $F=\frac{k{q}^2}{r^2}$                                                                                                        (1)

Here, $k$ is the coulomb force constant and $r$ is the distance between the charges.

The charge configuration is as shown in figure below.

Figure (1)

The charges present at the sides are distance $s$ from point A. The charges at the face diagonals are distance $\sqrt{2}s$ from point A. The charge at the body diagonal is $\sqrt{3s}$ distance from the point A.

For the net force in $y$ direction due to the symmetry the charges at point 3, 4 and 7 do not participate.

For force due to charge 6 is only in the y-direction

Substitute $s$ for $r$ in equation (1).

    $F_{6\left(\text{y}\right)}=\frac{k{q}^2}{s^2}$

The force due to charge 5 and 2 makes an angle $45°$ with the $y$ axis so the force due to charge 5 and 2

    $F_{5\text{(y)}}=F_{\text{2(y)}}=F\cos \theta$

Substitute $45°$ for $\theta$ and $\frac{k{q}^2}{r^2}$ for $F$ in the above equation.

    $F_{5\text{(y)}}=F_{\text{2(y)}}=\frac{k{q}^2}{r^2}\cos 45°$

Substitute $\sqrt{2s}$ for $r$ in the above equation.

    $\begin{array}{c}{F}_{5\text{(y)}}=F_{\text{2(y)}}=\frac{k{q}^2}{{\left(\sqrt{2s}\right)}^2}\cos 45°\\ =\frac{k{q}^2}{2\sqrt{2s^2}}\end{array}$

The charge 1 is $\sqrt{3s}$ distance away from the point A and force due to the charge 1 makes angle $\varphi$ with the y axis.

Write the expression for the cosine value of the angle

    $\begin{array}{c}\cos \varphi =\frac{\sqrt{2s}}{\sqrt{3s}}\\ =\sqrt{\frac{2}{3}}\end{array}$

Write the expression for the force due to charge 1

    $F_{\text{1}\left(\text{y}\right)}=F\cos \theta \cos \varphi$

Substitute $\sqrt{\frac{2}{3}}$ for $\cos \varphi$, $45°$ for $\theta$ and $\frac{k{q}^2}{r^2}$ for $F$ in the above equation.

    $F_{\text{1(y)}}=\frac{k{q}^2}{r^2}\left(\sqrt{\frac{2}{3}}\right)\frac{1}{\sqrt{2}}$

Substitute $\sqrt{3s}$ for $r$ in the above equation.

    $\begin{array}{c}{F}_{\text{1(y)}}=\frac{k{q}^2}{{\left(\sqrt{3s}\right)}^2}\left(\sqrt{\frac{2}{3}}\right)\sqrt{\frac{1}{2}}\\ =\frac{1}{3\sqrt{3}}\frac{k{q}^2}{s^2}\end{array}$

Write the expression for the net force in $y$ direction

    $F_{\text{y}\left(\text{net}\right)}=F_{1\left(\text{y}\right)}+F_{2\left(\text{y}\right)}+F_{5\left(\text{y}\right)}+F_{6\left(\text{y}\right)}$

Substitute $\frac{k{q}^2}{s^2}$ for $F_{6\left(\text{y}\right)}$, $\frac{k{q}^2}{2\sqrt{2s^2}}$ for $F_{2\left(\text{y}\right)}$ and $F_{5\left(\text{y}\right)}$ and $\frac{1}{3\sqrt{3}}\frac{k{q}^2}{s^2}$ for $F_{6\left(\text{y}\right)}$ in the above equation.

    $\begin{array}{c}{F}_{\text{y}\left(\text{net}\right)}=\frac{k{q}^2}{s^2}+\frac{k{q}^2}{2\sqrt{2s^2}}+\frac{k{q}^2}{2\sqrt{2s^2}}+\frac{k{q}^2}{3\sqrt{3s^2}}\\ =\frac{k{q}^2}{s^2}\left(1+\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{2}}+\frac{1}{3\sqrt{3}}\right)\\ =\frac{k{q}^2}{s^2}\left(1+\frac{1}{\sqrt{2}}+\frac{1}{3\sqrt{3}}\right)\end{array}$

Thus, net force in $y$ direction is ${\overrightarrow{F}}_{\text{y}\left(\text{net}\right)}=\left[\frac{k{q}^2}{s^2}\left(1+\frac{1}{\sqrt{2}}+\frac{1}{3\sqrt{3}}\right)\right]\widehat{\text{j}}$

From the cubical symmetry the net force in $x$ and $z$ direction is also same.

Thus, net force in $x$ direction

    ${\overrightarrow{F}}_{\text{x}\left(\text{net}\right)}=\left[\frac{k{q}^2}{s^2}\left(1+\frac{1}{\sqrt{2}}+\frac{1}{3\sqrt{3}}\right)\right]\widehat{\text{i}}$

And the net force in $z$ direction

    ${\overrightarrow{F}}_{z\left(\text{net}\right)}=\left[\frac{k{q}^2}{s^2}\left(1+\frac{1}{\sqrt{2}}+\frac{1}{3\sqrt{3}}\right)\right]\widehat{\text{k}}$

Conclusion:

Therefore, the $x$, $y$ and $z$ components of the force are, ${\overrightarrow{F}}_{\text{x}\left(\text{net}\right)}=\left[\frac{k{q}^2}{s^2}\left(1+\frac{1}{\sqrt{2}}+\frac{1}{3\sqrt{3}}\right)\right]\widehat{\text{i}}$, ${\overrightarrow{F}}_{\text{y}\left(\text{net}\right)}=\left[\frac{k{q}^2}{s^2}\left(1+\frac{1}{\sqrt{2}}+\frac{1}{3\sqrt{3}}\right)\right]\widehat{\text{j}}$ and ${\overrightarrow{F}}_{z\left(\text{net}\right)}=\left[\frac{k{q}^2}{s^2}\left(1+\frac{1}{\sqrt{2}}+\frac{1}{3\sqrt{3}}\right)\right]\widehat{\text{k}}$ respectively.

To determine

The magnitude of the net force.

Answer to Problem 23.85CP

The magnitude of the net force is $3.29\frac{k{q}^2}{s^2}$.

Explanation of Solution

The side of the each cube is $s$ and the magnitude of each charge is $q$.

Write the expression for the magnitude of the net force

    $F_{\text{net}}=\sqrt{{\left(\left|{\overrightarrow{F}}_{\text{x}\left(\text{net}\right)}\right|\right)}^2+{\left(\left|{\overrightarrow{F}}_{\text{y}\left(\text{net}\right)}\right|\right)}^2+{\left(\left|{\overrightarrow{F}}_{\text{z}\left(\text{net}\right)}\right|\right)}^2}$

Substitute $\left[\frac{k{q}^2}{s^2}\left(1+\frac{1}{\sqrt{2}}+\frac{1}{3\sqrt{3}}\right)\right]$ for $\left(\left|{\overrightarrow{F}}_{\text{x}\left(\text{net}\right)}\right|\right)$, $\left(\left|{\overrightarrow{F}}_{\text{y}\left(\text{net}\right)}\right|\right)$ and $\left(\left|{\overrightarrow{F}}_{\text{z}\left(\text{net}\right)}\right|\right)$ respectively in the above equation.

    $\begin{array}{c}{F}_{\text{net}}=\sqrt{{\left[\frac{k{q}^2}{s^2}\left(1+\frac{1}{\sqrt{2}}+\frac{1}{3\sqrt{3}}\right)\right]}^2+{\left[\frac{k{q}^2}{s^2}\left(1+\frac{1}{\sqrt{2}}+\frac{1}{3\sqrt{3}}\right)\right]}^2+{\left[\frac{k{q}^2}{s^2}\left(1+\frac{1}{\sqrt{2}}+\frac{1}{3\sqrt{3}}\right)\right]}^2}\\ =\sqrt{{\left[\frac{k{q}^2}{s^2}\left(1.8995\right)\right]}^2+{\left[\frac{k{q}^2}{s^2}\left(1.8995\right)\right]}^2+{\left[\frac{k{q}^2}{s^2}\left(1.8995\right)\right]}^2}\\ =3.29\frac{k{q}^2}{s^2}\end{array}$

Conclusion:

Therefore, the magnitude of the net force is $3.29\frac{k{q}^2}{s^2}$.

To determine

The direction of the net force.

Answer to Problem 23.85CP

The direction of the net force is away from the origin.

Explanation of Solution

There is a force component in $x$, $y$ and $z$ directions and the magnitude of all three components is equal so no single component dominant.

Hence, the net force isn’t along any particular direction rather it has components in all three directions with equal magnitudes and it is pointed away from the origin

Conclusion:

Therefore, the net direction is away from the origin making equal angle with all the three axes.