Answer:
The electric force on q₁ is [tex]4.5\times10^{-3}\ N[/tex].
Explanation:
Given that,
Charge on [tex]q_{1}=-6.0\ \mu C[/tex]
Distance [tex]x= -3.0\ m[/tex]
Charge on [tex]q_{2}=1.0\ \mu C[/tex] at origin
Distance [tex]x= 3.0\ m[/tex]
Charge on [tex]q_{3}=-1.0\ \mu C[/tex]
We need to calculate the electric force on q₁
Using formula of electric force
[tex]F_{12}=\dfrac{kq_{1}q_{2}}{r^2}[/tex]
Put the value into the formula
[tex]F_{12}=\dfrac{9\times10^{9}\times(-6.0\times10^{-6})\times1.0\times10^{-6}}{(3.0)^2}[/tex]
[tex]F_{12}=-0.006\ N[/tex]
Negative sign shows the attraction force.
We need to calculate the electric force F₁₃
[tex]F_{13}=\dfrac{9\times10^{9}\times(-6.0\times10^{-6})\times(-1.0\times10^{-6})}{(6.0)^2}[/tex]
[tex]F_{13}=0.0015\ N[/tex]
Positive sign shows the repulsive force.
We need to calculate the net electric force
[tex]F=F_{12}+F_{13}[/tex]
[tex]F=-0.006+0.0015[/tex]
[tex]F=0.0045\ N[/tex]
[tex]F=4.5\times10^{-3}\ N[/tex]
Hence, The electric force on q₁ is [tex]4.5\times10^{-3}\ N[/tex].
