Example 3.1: Electrostatic force between three colinear point charges

Question:  A particle of charge $q_1=+6.0\,\mu{\rm C}$ is located on the $x$-axis at coordinate $x_1=5.1\,{\rm cm}$. A second particle of charge $q_2=-5.0\,\mu{\rm C}$ is placed on the $x$-axis at $x_2=-3.4\,{\rm cm}$. What is the magnitude and direction of the total electrostatic force acting on a third particle of charge $q_3=+2.0\,\mu{\rm C}$ placed at the origin ($x=0$)?
 
Solution: The force $f$ acting between charges 1 and 3 is given by

$$f = k_e\, \frac{q_1 \,q_3}{x_1^{~2}} = (8.988\times 10^9)\,\... ...,(2\times 10^{-6})}{(5.1\times 10^{-2})^2} = +41.68\,{\rm N}.$$

Since $f>0$, the force is repulsive. This means that the force $f_{13}$ exerted by charge 1 on charge 3 is directed along the $-x$-axis (i.e., from charge 1 towards charge 3), and is of magnitude $\vert f\vert$. Thus, $f_{13} = - 41.69\,{\rm N}$. Here, we adopt the convention that forces directed along the $+x$-axis are positive, and vice versa. The force $f'$ acting between charges 2 and 3 is given by

$$f' = k_e\, \frac{q_2 \,q_3}{\vert x_2\vert^{2}} = (8.988\tim... ...,(2\times 10^{-6})}{(3.4\times 10^{-2})^2} = -77.75\,{\rm N}.$$

Since $f'<0$, the force is attractive. This means that the force $f_{23}$ exerted by charge 2 on charge 3 is directed along the $-x$-axis (i.e., from charge 3 towards charge 2), and is of magnitude $\vert f'\vert$. Thus, $f_{23} = - 77.75\,{\rm N}$.

The resultant force $f_3$ acting on charge 3 is the algebraic sum of the forces exerted by charges 1 and 2 separately (the sum is algebraic because all the forces act along the $x$-axis). It follows that

$$f_3 = f_{13} + f_{23} = - 41.69 - 77.75 =-119.22\,{\rm N}.$$

Thus, the magnitude of the total force acting on charge 3 is $119.22\,{\rm N}$, and the force is directed along the $-x$-axis (since $f_3<0$).