Answer :
Answer:
The magnitude of the electrostatic force on [tex]\mathrm{+32\,\mu C}[/tex] is: [tex]\mathbf{12\;N}[/tex]
Explanation:
Consider
- [tex]q_1=+32\;\mu\text{C}[/tex] is at position [tex]x_1=0\;\text{m}[/tex]
- [tex]q_2=+20\;\mu\text{C}[/tex] is at position [tex]x_2=40\;\text{cm}[/tex]
- [tex]q_3=-60\;\mu\text{C}[/tex] is at position [tex]x_3=60\;\text{cm}[/tex]
The sum of all horizontal forces on [tex]q_1[/tex] is given as
[tex]\sum F_x=F_{12}+F_{13}[/tex]
where
- [tex]F_{12}[/tex] is the force exerted by [tex]q_2[/tex]
- [tex]F_{13}[/tex] is the force exerted by [tex]q_3[/tex]
The force on [tex]q_1[/tex] exerted by [tex]q_2[/tex] is repulsive, so the direction of the force [tex]F_{12}[/tex] is to the left (negative direction). Thus
[tex]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}[/tex]
The force on [tex]q_1[/tex] exerted by [tex]q_3[/tex] is attractive, so the direction of the force [tex]F_{13}[/tex] is to the right (positive direction). Thus
[tex]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}[/tex]
Therefore
[tex]\sum F_x=\mathrm{-36\;N+48\;N} \;\;\;\;\;\Rightarrow\;\;\;\;\; \mathbf{\sum F_x=+12\;N}[/tex]
the electrostatic force on [tex]q_1[/tex] is to the right and has a magnitude of [tex]\mathbf{12\;N}[/tex]
