Chapter 23, Problem 75PQ

Eight small conducting spheres with identical charge q = −2.00 μC are placed at the corners of a cube of side d = 0.500 m (Fig. P23.75). What is the total force on the sphere at the origin (sphere A) due to the other seven spheres?

Figure P23.75

To determine

The total force on the sphere A due to seven spheres.

Answer to Problem 75PQ

The total force on the sphere A due to seven spheres is $\overrightarrow{F}=\left(-0.273\widehat{i}-0.273\widehat{j}-0.273\widehat{k}\right)\text{N}$.

Explanation of Solution

Write the expression for Coulomb’s law.

  $\overrightarrow{F}=\frac{k{q}_1{q}_2}{r^2}\widehat{r}$

Here, $F$ is the electrostatic force between the particles, $k$ is the Coulomb’s constant, $q_1$ is the charge of particle 1, $q_2$ is the charge of particle 2, $r$ is the distance between the centers of the particles.

The forces produced by other spheres on A.

Charge	force
B	${\overrightarrow{F}}_{\text{BA}}=-\frac{k{q}^2}{d^2}\widehat{j}$
C	$\begin{array}{c}{\overrightarrow{F}}_{\text{CA}}=-\frac{k{q}^2}{d^2+d^2}\frac{\widehat{i}+\widehat{j}}{\sqrt{2}}\\ =-\left(\frac{k{q}^2}{2\sqrt{2}{d}^2}\right)\widehat{i}-\left(\frac{k{q}^2}{2\sqrt{2}{d}^2}\right)\widehat{j}\end{array}$
D	${\overrightarrow{F}}_{\text{DA}}=-\frac{k{q}^2}{d^2}\widehat{i}$
E	${\overrightarrow{F}}_{\text{EA}}=-\frac{k{q}^2}{d^2}\widehat{k}$
F	$\begin{array}{c}{\overrightarrow{F}}_{\text{FA}}=-\frac{k{q}^2}{d^2+d^2}\frac{\widehat{i}+\widehat{k}}{\sqrt{2}}\\ =-\left(\frac{k{q}^2}{2\sqrt{2}{d}^2}\right)\widehat{i}-\left(\frac{k{q}^2}{2\sqrt{2}{d}^2}\right)\widehat{k}\end{array}$
G	$\begin{array}{c}{\overrightarrow{F}}_{\text{GA}}=-\frac{k{q}^2}{d^2+d^2+d^2}\frac{\widehat{i}+\widehat{j}+\widehat{k}}{\sqrt{3}}\\ {\overrightarrow{F}}_{\text{GA}}=-\left(\frac{k{q}^2}{3\sqrt{3}{d}^2}\right)\widehat{i}-\left(\frac{k{q}^2}{3\sqrt{3}{d}^2}\right)\widehat{j}-\left(\frac{k{q}^2}{3\sqrt{3}{d}^2}\right)\widehat{k}\end{array}$
H	$\begin{array}{c}{\overrightarrow{F}}_{\text{HA}}=-\frac{k{q}^2}{d^2+d^2}\frac{\widehat{i}+\widehat{k}}{\sqrt{2}}\\ =-\left(\frac{k{q}^2}{2\sqrt{2}{d}^2}\right)\widehat{i}-\left(\frac{k{q}^2}{2\sqrt{2}{d}^2}\right)\widehat{k}\end{array}$

Write the expression to find the total force on the sphere A due to seven spheres.

  $\overrightarrow{F}={\overrightarrow{F}}_{\text{BA}}+{\overrightarrow{F}}_{\text{CA}}+{\overrightarrow{F}}_{\text{DA}}+{\overrightarrow{F}}_{\text{EA}}+{\overrightarrow{F}}_{\text{FA}}+{\overrightarrow{F}}_{\text{GA}}+{\overrightarrow{F}}_{\text{HA}}$

Here, ${\overrightarrow{F}}_{\text{BA}}$ is force on A due to B, ${\overrightarrow{F}}_{\text{CA}}$ is force on A due to C, ${\overrightarrow{F}}_{\text{DA}}$ is force on A due to D, ${\overrightarrow{F}}_{\text{EA}}$ is force on A due to E, ${\overrightarrow{F}}_{\text{FA}}$ is force on A due to F, ${\overrightarrow{F}}_{\text{GA}}$ is force on A due to G, ${\overrightarrow{F}}_{\text{HA}}$ is force on A due to H.

Substitute $-\frac{k{q}^2}{d^2}\widehat{j}$ for ${\overrightarrow{F}}_{\text{BA}}$, $-\left(\frac{k{q}^2}{2\sqrt{2}{d}^2}\right)\widehat{i}-\left(\frac{k{q}^2}{2\sqrt{2}{d}^2}\right)\widehat{j}$ for ${\overrightarrow{F}}_{\text{CA}}$, $-\frac{k{q}^2}{d^2}\widehat{i}$ for ${\overrightarrow{F}}_{\text{DA}}$, $-\frac{k{q}^2}{d^2}\widehat{k}$ for ${\overrightarrow{F}}_{\text{EA}}$, $-\left(\frac{k{q}^2}{2\sqrt{2}{d}^2}\right)\widehat{i}-\left(\frac{k{q}^2}{2\sqrt{2}{d}^2}\right)\widehat{k}$ for ${\overrightarrow{F}}_{\text{FA}}$, $-\left(\frac{k{q}^2}{3\sqrt{3}{d}^2}\right)\widehat{i}-\left(\frac{k{q}^2}{3\sqrt{3}{d}^2}\right)\widehat{j}-\left(\frac{k{q}^2}{3\sqrt{3}{d}^2}\right)\widehat{k}$ for ${\overrightarrow{F}}_{\text{GA}}$, $-\left(\frac{k{q}^2}{2\sqrt{2}{d}^2}\right)\widehat{i}-\left(\frac{k{q}^2}{2\sqrt{2}{d}^2}\right)\widehat{k}$ for ${\overrightarrow{F}}_{\text{HA}}$ to find the total force on the sphere A due to seven spheres.

  $\begin{array}{c}\overrightarrow{F}=\left(-\frac{k{q}^2}{d^2}\widehat{j}\right)+\left(-\left(\frac{k{q}^2}{2\sqrt{2}{d}^2}\right)\widehat{i}-\left(\frac{k{q}^2}{2\sqrt{2}{d}^2}\right)\widehat{j}\right)\\ +\left(-\frac{k{q}^2}{d^2}\widehat{i}\right)+\left(-\frac{k{q}^2}{d^2}\widehat{k}\right)+\left(-\left(\frac{k{q}^2}{2\sqrt{2}{d}^2}\right)\widehat{i}-\left(\frac{k{q}^2}{2\sqrt{2}{d}^2}\right)\widehat{k}\right)\\ +\left(-\left(\frac{k{q}^2}{3\sqrt{3}{d}^2}\right)\widehat{i}-\left(\frac{k{q}^2}{3\sqrt{3}{d}^2}\right)\widehat{j}-\left(\frac{k{q}^2}{3\sqrt{3}{d}^2}\right)\widehat{k}\right)+\\ \left(-\left(\frac{k{q}^2}{2\sqrt{2}{d}^2}\right)\widehat{i}-\left(\frac{k{q}^2}{2\sqrt{2}{d}^2}\right)\widehat{k}\right)\\ =-\frac{k{q}^2}{d^2}\left(1+\frac{2}{2\sqrt{2}}+\frac{1}{3\sqrt{3}}\right)\left(\widehat{i}+\widehat{j}+k\right)\end{array}$

Conclusion:

Substitute $8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$, $2.00\times {10}^{-6}\text{C}$ for $q$, $0.500\text{m}$ for $d$ to find the total force on the sphere A due to seven spheres.

  $\begin{array}{c}\overrightarrow{F}=-\frac{\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right){\left(2.00\times {10}^{-6}\text{C}\right)}^2}{{\left(0.500\text{m}\right)}^2}\left(1.90\right)\left(\widehat{i}+\widehat{j}+k\right)\\ =\left(-0.273\widehat{i}-0.273\widehat{j}-0.273\widehat{k}\right)\text{N}\end{array}$

Therefore, the total force on the sphere A due to seven spheres is $\overrightarrow{F}=\left(-0.273\widehat{i}-0.273\widehat{j}-0.273\widehat{k}\right)\text{N}$.