Chapter 16, Problem 45P

To determine

The electric force on the charge placed at origin.

Answer to Problem 45P

The net electric force on the charge at origin is $1.61\text{N }\widehat{x}$.

Explanation of Solution

The charges are $3.00\text{μC}$, $-5.00\text{μC}$, and $8.00\text{μC}$ whose positions are given in the figure.

The charges are $3.00\text{μC}$, $-5.00\text{μC}$ and $8.00\text{μC}$ which are kept at a distance $3.0\text{cm}$.

Write the expression for net electric force experienced by a charge due to two other charges

${\displaystyle \sum \overrightarrow{F}}=\frac{k\left|q_1\right|\left|q_2\right|}{r_{12}^2}{\widehat{r}}_{12}+\frac{k\left|q_1\right|\left|q_3\right|}{r_{13}^2}{\widehat{r}}_{13}$                                                            (I)

Here, ${\displaystyle \sum \overrightarrow{F}}$ is the net electric force, $k$ is the Coulomb’s constant, $q_1$, $q_2$, $q_3$ are the three point charges, $r_{12}$ is the distance between $q_1$ and $q_2$, $r_{13}$ is the distance between $q_1$ and $q_3$, ${\widehat{r}}_{12}$, ${\widehat{r}}_{13}$ are the corresponding direction vectors along the line joining the point charges.

Since $q_2$ is negative with respect to $q_1$, ${\widehat{r}}_{12}$ is in the positive x-direction i.e. $\widehat{x}$ whereas $q_3$ is positive with respect to $q_1$, and hence ${\widehat{r}}_{13}$ is in the negative x-direction, $-\widehat{x}$.

Substitute $8.988\times {10}^9{\text{Nm}}^2{\text{C}}^{-2}$ for $k$, $3.00\text{μC}$ for $q_1$, $-5.00\text{μC}$ for $q_2$, $8.00\text{μC}$ for $q_3$, $20.0\text{cm}$ for $r_{12}$, $35.0\text{cm}$ for $r_{13}$, $\widehat{x}$ for ${\widehat{r}}_{12}$ and $-\widehat{x}$ for ${\widehat{r}}_{13}$ in (I) to ${\displaystyle \sum \overrightarrow{F}}$.

$\begin{array}{c}{\displaystyle \sum \overrightarrow{F}}=\{\begin{array}{l}\frac{\left(8.988\times {10}^9{\text{Nm}}^2{\text{C}}^{-2}\right)\left(\left|3.00\text{μC}\right|\right)\left(\left|-5.00\text{μC}\right|\right)}{{\left(20.0\text{ cm}\right)}^2}\left(\widehat{x}\right)\\ +\frac{\left(8.988\times {10}^9{\text{Nm}}^2{\text{C}}^{-2}\right)\left(\left|3.00\text{μC}\right|\right)\left(\left|8.00\text{μC}\right|\right)}{{\left(35.0\text{ cm}\right)}^2}\left(-\widehat{x}\right)\end{array}\\ =\left(8.988\times {10}^9{\text{Nm}}^2{\text{C}}^{-2}\right)\left(3.00\text{μC}\right)\left[\frac{5.00\text{μC}}{{\left(20.0\text{ cm}\right)}^2}-\frac{8.00\text{μC}}{{\left(35.0\text{ cm}\right)}^2}\right]\\ =\left(8.988\times {10}^9{\text{Nm}}^2{\text{C}}^{-2}\right)\left(3.00\text{μC}\right)\left[\frac{5.00\text{C}}{{\left(20.0\text{ m}\right)}^2}-\frac{8.00\text{C}}{{\left(35.0\text{ m}\right)}^2}\right]\times \frac{{10}^{-6}}{{10}^{-4}}\end{array}$

${\displaystyle \sum \overrightarrow{F}}=1.61\text{N }\widehat{x}$

Thus, the net electric force is $1.61\text{N }\widehat{x}$.