Given that there are three charges forming the right triangle such that
[tex]q1\text{ = -3 }\mu C=-3\times10^{-6\text{ }}C[/tex]
is placed 4 cm = 0.04 m from the charge
[tex]q2=\text{ 6 }\mu C=6\times10^{-6\text{ }}C[/tex]
Another charge
[tex]q3=\text{ 2}\mu C=2\times10^{-6}\text{ C}[/tex]
is placed 3 cm = 0.03 m right from the charge q3.
Here, Coulomb's law can be used to calculate force,
[tex]F=k\frac{q1q2}{r^2}[/tex]
Here, F is the force, q1 and q2 are the charges, r is the distance between q1 and q2 and k is a constant whose value is
[tex]9\times10^9Nm^2/C^2[/tex]
Substituting values for the force on q2 due to q1, we get
[tex]\begin{gathered} F_{12}=\frac{9\times10^9\times(3\times10^{-6})\times6\times10^{-6}}{(0.04)^2} \\ =\text{ }1.01\text{ N} \end{gathered}[/tex]
Substituting values for the force on q2 due to q3, we get
[tex]\begin{gathered} F_{32}=\frac{9\times10^9\times(2\times10^{-6})\times6\times10^{-6}}{(0.03)^2} \\ =\text{ 120 N} \end{gathered}[/tex]
Hence the total force is
[tex]\begin{gathered} F_2=F_{12}+F_{32} \\ =120\text{ +1.01} \\ =\text{ }121.01\text{ N} \end{gathered}[/tex]
Here, A is the attractive force due to q1 and B is the repulsive force due to q3
R is the resultant due to both of them.
R=A+B
Please note there are arrows above them.
Thus R is the resultant direction.
