Question

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?

Short answer

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Step-by-step solution

Given

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

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

Understanding the concept

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.

(a) Calculate the net electrostatic force on particle 3, of charge q3 = 20.0 μC, if particle 3 is placed at x = 40.0 cm

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

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

(b) Calculate the net electrostatic force on particle 3, of charge q3 = 20.0 μC, if particle 3 is placed at x = 80.0 cm

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

toward +x. Now we obtain,

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

Calculate the (c) x and (d) y coordinates of particle 3 if the net electrostatic force on it due to particles 1 and 2 is zero

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