In Fig. 21-26, particle 1 of charge $q_1=-80.0\text{μC}$and particle 2 of charge $q_2=+40.0\text{μC}$are held at separation $L=20.0\text{\hspace{0.17em}cm}$on an xaxis. In unit-vector notation, what is the net electrostatic force on particle 3, of charge $q_3=20.0\text{\hspace{0.17em}μC}$, if particle 3 is placed at (a)$x=40.0$cm and (b) $x=80.0\text{\hspace{0.17em}cm}$? What should be the (c) xand (d) ycoordinates of particle 3 if the net electrostatic force on it due to particles1 and 2 is zero?

$q_1=-80.0\text{\hspace{0.17em}μC}$;$q_2=+40.0\text{μC}$And$L=20.0\text{cm}$

$q_3=20.0\text{μC}$at$x=40.0\text{\hspace{0.17em}cm}$ and$x=80.0\text{\hspace{0.17em}cm}$

Coulomb’s law states that the force of attraction or repulsion between the two charges is directly proportional to the product of the charges and inversely proportional to the square of a distance between two charges.

Charge$q_1=-80.0\times {10}^{-6}\text{C}$is at the origin, and charge$q_2=40.0\times {10}^{-6}\text{C}$is at x = 0.20 m. The force on$q_3=20.0\times {10}^{-6}\text{C}$is due to the attractive and repulsive forces from q1 and q2, respectively.

In symbols ${\overrightarrow{F}}_{3net}={\overrightarrow{F}}_{32}+{\overrightarrow{F}}_{31}$

$F_{31}=k\frac{q_3\left|q_1\right|}{r_{31}^2},F_{32}=k\frac{q_3\left|q_2\right|}{r_{32}^2}$

In this case= 0.40 m and= 0.20 m, with$F_{31}$directed toward –x and

$F_{32}$directed in the +x direction. Using the value of k

$\begin{array}{c}{\overrightarrow{F}}_{3net}=-\left|{\overrightarrow{F}}_{32}\right|\widehat{i}+\left|{\overrightarrow{F}}_{31}\right|\widehat{i}\\ =(-k\frac{q_3\left|q_1\right|}{r_{31}^2}+k\frac{q_3\left|q_2\right|}{r_{32}^2})\widehat{i}\\ =k{q}_3(-\frac{\left|q_1\right|}{r_{31}^2}+\frac{q_2}{r_{32}^2})\widehat{i}\\ =(9.0\times {10}^9{\text{Nm}}^{\text{2}}{\text{/C}}^{\text{2}})(20\times {10}^{-6}\text{C})(\frac{-80.0\times {10}^{-6}\text{C}}{{(0.40\text{m})}^2}+\frac{40.0\times {10}^{-6}\text{C}}{{(0.20\text{m})}^2})\widehat{i}\\ =(90\text{N})\widehat{i}\end{array}$

In this case= 0.80 m and= 0.60 m, with$F_{31}$directed toward –x and $F_{32}$

toward +x. Now we obtain,

$\begin{array}{c}{F}_{3net}=-\left|F_{32}\right|\widehat{i}+\left|F_{31}\right|\widehat{i}=(-k\frac{q_3\left|q_1\right|}{r_{31}^2}+k\frac{q_3\left|q_2\right|}{r_{32}^2})\widehat{i}\\ =k{q}_3(-\frac{\left|q_1\right|}{r_{31}^2}+\frac{q_2}{r_{32}^2})\widehat{i}\\ =(9.0\times {10}^9{\text{Nm}}^{\text{2}}{\text{/C}}^{\text{2}})(20\times {10}^{-6}\text{C})(\frac{-80.0\times {10}^{-6}\text{C}}{{(0.80\text{m})}^2}+\frac{40.0\times {10}^{-6}\text{C}}{{\left(0.60\text{m}\right)}^2})\widehat{i}\\ =(-2.50\text{N})\widehat{i}\end{array}$

Between the locations treated in parts (a) and (b), there must be one where$F_{3net}=0$ . Writing $r_{31}$= x and $r_{32}$= x – 0.20 m, we equate $F_{31}$ and $F_{32}$ , and after cancelling common factors, arrive at