Chapter 24, Problem 51PQ

Figure P24.51 shows four small charged spheres arranged at the corners of a square with side d = 25.0 cm.

a. What is the electric field at the location of the sphere with charge +2.00 nC?

b. What is the total electric force exerted on the sphere with charge +2.00 nC by the other three spheres?

FIGURE P24.51

To determine

The electric field at the location of the sphere with charge $+2.00\text{nC}$.

Answer to Problem 51PQ

The electric field at the location of the sphere with charge $+2.00\text{nC}$ is $\left(-347\widehat{i}-635\widehat{j}\right)\text{N/C}$.

Explanation of Solution

Sketch the diagram showing the electric fields and position vectors of the charges.

Write the equation for the electric field of a charged particle.

$\overrightarrow{E}=\frac{kq}{r^2}\widehat{r}$ (I)

Here, $\overrightarrow{E}$$E$ is the electric field of a charged particle, $k$ is the Coulomb’s constant, $r$ is the distance and $\widehat{r}$ is the surface unit vector in the direction for $r$.

Use equation (I) to find the equation for the sum of the electric field at the location of the sphere with charge $+2.00\text{nC}$.

$\overrightarrow{E}=\frac{k{q}_1}{r_1^2}{\widehat{r}}_1+\frac{k{q}_3}{r_3^2}{\widehat{r}}_3+\frac{k{q}_4}{r_4^2}{\widehat{r}}_4$ (II)

Here, $q_1$ is the is the first charge, $q_4$ is the is the third charge, $q_3$ is the is the third charge, $r_1$ is the distance of the first charge to the $+2.00\text{nC}$ charge, $r_3$ is the distance of the third charge to the $+2.00\text{nC}$ charge, $r_4$ is the distance of the fourth charge to the $+2.00\text{nC}$ charge, ${\widehat{r}}_1$ is the unit vector in the direction of $r_1$, ${\widehat{r}}_3$ is the unit vector in the direction of $r_3$ and ${\widehat{r}}_4$ is the unit vector in the direction of $r_4$.

Conclusion:

Substitute $8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$ for $k$, $1.00\text{nC}$ for $q_1$, $4.00\text{nC}$ for $q_3$ and $3.00\text{nC}$ for $q_4$, $d$ for $r_1$, $\sqrt{2}d$ for $r_3$, $d$ for $r_4$, $\widehat{i}$ for ${\widehat{r}}_1$, $\left(\cos 45.0\widehat{i}+\sin 45.0\widehat{j}\right)$ for ${\widehat{r}}_3$ and $\widehat{j}$ for ${\widehat{r}}_4$ in equation (II) to find $\overrightarrow{E}$.

$\begin{array}{c}\overrightarrow{E}=-k\left[\frac{1.00\text{nC}}{d^2}\widehat{i}-\frac{4.00\text{nC}}{{\left(\sqrt{2}d\right)}^2}\left(\cos 45.0\widehat{i}+\sin 45.0\widehat{j}\right)-\frac{3.00\text{nC}}{d^2}\widehat{j}\right]\\ =-k\left[\begin{array}{l}\frac{\left(1.00\text{nC}\right)\left(\frac{{10}^{-9}\text{C}}{1\text{nC}}\right)}{d^2}\widehat{i}-\frac{\left(4.00\text{nC}\right)\left(\frac{{10}^{-9}\text{C}}{1\text{nC}}\right)}{{\left(\sqrt{2}d\right)}^2}\left(\cos 45.0\widehat{i}+\sin 45.0\widehat{j}\right)\\ -\frac{\left(3.00\text{nC}\right)\left(\frac{{10}^{-9}\text{C}}{1\text{nC}}\right)}{d^2}\widehat{j}\end{array}\right]\\ =\frac{8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}}{{\left(\left(25.0\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}^2}\left({10}^{-9}\text{C}\right)\left[\left(-1-2\cos 45.0°\right)\widehat{i}+\widehat{j}\right]\\ =\left(-347\widehat{i}-635\widehat{j}\right)\text{N/C}\end{array}$

Thus, the electric field at the location of the sphere with charge $+2.00\text{nC}$ is $\left(-347\widehat{i}-635\widehat{j}\right)\text{N/C}$.

To determine

The total electric forceon the sphere with charge $+2.00\text{nC}$.

Answer to Problem 51PQ

The total electric force on the sphere with charge $+2.00\text{nC}$ is $\left(-6.94\times {10}^{-7}\widehat{i}-1.27\times {10}^{-6}\widehat{j}\right)\text{N}$

Explanation of Solution

Write the equation for the electric force.

$\overrightarrow{F}=q\overrightarrow{E}$ (III)

Here, $\overrightarrow{F}$ is the electric force.

Conclusion:

Substitute $2.00\text{nC}$ for $q$ and $\left(-347\widehat{i}-635\widehat{j}\right)\text{N/C}$ for $\overrightarrow{E}$ in equation (III) to find $\overrightarrow{F}$.

$\begin{array}{c}\overrightarrow{F}=\left(2.00\text{nC}\right)\left(\left(-347\widehat{i}-635\widehat{j}\right)\text{N/C}\right)\\ =\left(\left(2.00\text{nC}\right)\left(\frac{{10}^{-9}\text{C}}{1\text{nC}}\right)\right)\left(\left(-347\widehat{i}-635\widehat{j}\right)\text{N/C}\right)\\ =\left(-6.94\times {10}^{-7}\widehat{i}-1.27\times {10}^{-6}\widehat{j}\right)\text{N}\end{array}$

Thus, the total electric force on the sphere with charge $+2.00\text{nC}$ is $\left(-6.94\times {10}^{-7}\widehat{i}-1.27\times {10}^{-6}\widehat{j}\right)\text{N}$.