Chapter 23, Problem 40PQ

Three charged metal spheres are arrayed in the xy plane so that they form an equilateral triangle (Fig. P23.40). What is the net electrostatic force on the sphere at the origin?

Figure P23.40

To determine

The net electrostatic force on the sphere at the origin.

Answer to Problem 40PQ

The net electrostatic force on the sphere at the origin is $\overrightarrow{F}=\left(6.01\times {10}^{-6}\text{N}\right)\widehat{i}-\left(7.43\times {10}^{-6}\text{N}\right)\widehat{j}$.

Explanation of Solution

The diagram for the forces on charge on the origin.

Write the expression for Coulomb’s law.

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

Here, $F_E$ is the electrostatic force between the spheres, $k$ is the Coulomb’s constant, $q_1$ is the charge of sphere 1, $q_2$ is the charge of sphere 2, $r$ is the distance between the centers of the spheres.

Substitute $8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$, $-5.50\text{nC}$ for $q_1$, $3.00\text{nC}$ for $q_2$ and $12.0\text{cm}$ for $r$ to find the magnitude of the force on charge $-5.50\text{nC}$ due to $3.00\text{nC}$.

  $\begin{array}{c}{F}_1=\frac{\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left|\left(-5.50\text{nC}\right)\left(3.00\text{nC}\right)\right|}{{\left(12.0\text{cm}\right)}^2}\\ =\frac{\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left|\left(-5.50\text{nC}\frac{{10}^{-9}\text{C}}{1.0\text{nC}}\right)\left(3.00\text{nC}\frac{{10}^{-9}\text{C}}{1.0\text{nC}}\right)\right|}{{\left(12.0\text{cm}\frac{{10}^{-2}\text{m}}{1.0\text{cm}}\right)}^2}\\ =1.03\times {10}^{-5}\text{N}\end{array}$

The magnitude of the force on charge $-5.50\text{nC}$ due to $3.00\text{nC}$ is in the $+x$ direction.

Substitute $8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$, $-5.50\text{nC}$ for $q_1$, $-2.50\text{nC}$ for $q_2$ and $12.0\text{cm}$ for $r$ to find the magnitude of the force on charge $-5.50\text{nC}$ due to $-2.50\text{nC}$.

  $\begin{array}{c}{F}_2=\frac{\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left|\left(-5.50\text{nC}\right)\left(-2.50\text{nC}\right)\right|}{{\left(12.0\text{cm}\right)}^2}\\ =\frac{\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left|\left(-5.50\text{nC}\frac{{10}^{-9}\text{C}}{1.0\text{nC}}\right)\left(-2.50\text{nC}\frac{{10}^{-9}\text{C}}{1.0\text{nC}}\right)\right|}{{\left(12.0\text{cm}\frac{{10}^{-2}\text{m}}{1.0\text{cm}}\right)}^2}\\ =8.58\times {10}^{-6}\text{N}\end{array}$

The magnitude of the force on charge $-5.50\text{nC}$ due to $-2.50\text{nC}$ is at an angle $60°$ below to the $-x$ direction.

Write the expression to find the net electrostatic force on sphere at origin.

  $\overrightarrow{F}=F_x\widehat{i}+F_y\widehat{j}$                                                                                              (II)

Here, $F_x$ is the x-component of net electrostatic force on sphere at origin, $F_y$ is the y-component of net electrostatic force on sphere at origin, $\overrightarrow{F}$ is the net electrostatic force on sphere at origin.

Write the expression to find the x-component of net electrostatic force on sphere at origin.

  $F_x=-F_2\cos \theta +F_1$

Substitute $1.03\times {10}^{-5}\text{N}$ for $F_1$, $8.58\times {10}^{-6}\text{N}$ for $F_2$ and $60°$ for $\theta$ to find the x-component of net electrostatic force on sphere at origin.

  $\begin{array}{c}{F}_x=\left(-8.58\times {10}^{-6}\text{N}\right)\cos 60°+\left(1.03\times {10}^{-5}\text{N}\right)\\ =6.01\times {10}^{-6}\text{N}\end{array}$

Write the expression to find the y-component of net electrostatic force on sphere at origin.

  $F_y=-F_2\sin \theta$

Substitute $8.58\times {10}^{-6}\text{N}$ for $F_2$ and $60°$ for $\theta$ to find the y-component of net electrostatic force on sphere at origin.

  $\begin{array}{c}{F}_y=\left(-8.58\times {10}^{-6}\text{N}\right)\cos 60°\\ =-7.43\times {10}^{-6}\text{N}\end{array}$

Conclusion:

Substitute $6.01\times {10}^{-6}\text{N}$ for $F_x$ and $-7.43\times {10}^{-6}\text{N}$ for $F_y$ in equation (II) to find the net electrostatic force on sphere at origin.

  $\begin{array}{c}\overrightarrow{F}=\left(6.01\times {10}^{-6}\text{N}\right)\widehat{i}+\left(-7.43\times {10}^{-6}\text{N}\right)\widehat{j}\\ =\left(6.01\times {10}^{-6}\text{N}\right)\widehat{i}-\left(7.43\times {10}^{-6}\text{N}\right)\widehat{j}\end{array}$

Thus, the net electrostatic force on the sphere at the origin is $\overrightarrow{F}=\left(6.01\times {10}^{-6}\text{N}\right)\widehat{i}-\left(7.43\times {10}^{-6}\text{N}\right)\widehat{j}$.