Question

Three point charges are positioned on the x axis. If the charges and corresponding positions are +32 µC at x = 0, +20 µC at x = 40 cm, and –60 µC at x = 60 cm, what is the magnitude of the electrostatic force on the +32-µC charge?

Answer 1

Answer:

The magnitude of the electrostatic force on $\mathrm{+32\,\mu C}$ is:   $\mathbf{12\;N}$

Explanation:

Consider

• $q_1=+32\;\mu\text{C}$ is at position $x_1=0\;\text{m}$
• $q_2=+20\;\mu\text{C}$ is at position $x_2=40\;\text{cm}$
• $q_3=-60\;\mu\text{C}$ is at position $x_3=60\;\text{cm}$

The sum of all horizontal forces on $q_1$ is given as

  $\sum F_x=F_{12}+F_{13}$

where

• $F_{12}$ is the force exerted by $q_2$
• $F_{13}$ is the force exerted by $q_3$

The force on $q_1$ exerted by $q_2$ is repulsive, so the direction of the force $F_{12}$ is to the left (negative direction). Thus

  $F_{12}=-\frac{K\,|q_1|\,|q_2|}{r_{12}^2}\\F_{12}=-\frac{K\,|q_1|\,|q_2|}{(x_2\;-\;x_1)^2}\\F_{12}=\mathrm{-\frac{(8.988\times10^9\;\frac{Nm^2}{C^2})\,|+32\;\mu C|\,|+20\;\mu C|}{(40\;cm\;-\;0\;cm)^2}}\\F_{12}=\mathrm{-\frac{(8.988\times10^9\;\frac{Nm^2}{C^2})\,(32\;\mu C)\,(20\;\mu C)}{(40\;cm)^2}}\\F_{12}=\mathrm{-\frac{(8.988\times10^9\;\frac{Nm^2}{C^2})\,(32\times10^{-6}\;C)\,(20\times10^{-6}\;C)}{(0.40\;m)^2}}\\F_{12}=\mathrm{-36\;N}$

The force on $q_1$ exerted by $q_3$ is attractive, so the direction of the force $F_{13}$ is to the right (positive direction). Thus

  $F_{13}=+\frac{K\,|q_1|\,|q_3|}{r_{13}^2}\\F_{13}=+\frac{K\,|q_1|\,|q_3|}{(x_3\;-\;x_1)^2}\\F_{13}=\mathrm{+\frac{(9\times10^9\;\frac{Nm^2}{C^2})\,|+32\;\mu C|\,|-60\;\mu C|}{(60\;cm\;-\;0\;cm)^2}}\\F_{13}=\mathrm{+\frac{(9\times10^9\;\frac{Nm^2}{C^2})\,(32\;\mu C)\,(60\;\mu C)}{(60\;cm)^2}}\\F_{13}=\mathrm{+\frac{(9\times10^9\;\frac{Nm^2}{C^2})\,(32\times10^{-6}\;C)\,(60\times10^{-6}\;C)}{(0.60\;m)^2}}\\F_{13}=\mathrm{+48\;N}$

Therefore

  $\sum F_x=\mathrm{-36\;N+48\;N} \;\;\;\;\;\Rightarrow\;\;\;\;\; \mathbf{\sum F_x=+12\;N}$

the electrostatic force on $q_1$ is to the right and has a magnitude of $\mathbf{12\;N}$