Example 3.2: Electrostatic force between three non-colinear point charges

Question: Suppose that three point charges, $q_a$, $q_b$, and $q_c$, are arranged at the vertices of a right-angled triangle, as shown in the diagram. What is the magnitude and direction of the electrostatic force acting on the third charge if $q_a=-6.0\,\mu{\rm C}$, $q_b=+4.0\,\mu{\rm C}$, $q_c = +2.0\,\mu{\rm C}$, $a=4.0$m, and $b=3.0$m?
 
Solution: The magnitude $f_{ac}$ of the force ${\bf f}_{ac}$ exerted by charge $q_a$ on charge $q_c$ is given by

$$f_{ac} = k_e\,\frac{\vert q_a\vert\, q_c}{c^2}= (8.988\times... ...2\times 10^{-6})} {(4^2+3^2)} = 4.31 \times 10^{-3} \,{\rm N},$$

where use has been made of the Pythagorean theorem. The force is attractive (since charges $q_a$ and $q_c$ are of opposite sign). Hence, the force is directed from charge $q_c$ towards charge $q_a$, as shown in the diagram. The magnitude $f_{bc}$ of the force ${\bf f}_{bc}$ exerted by charge $q_b$ on charge $q_c$ is given by

$$f_{bc} = k_e\,\frac{q_b q_c}{b^2}= (8.988\times 10^9)\,\frac... ...)\,(2\times 10^{-6})} {(3^2)} = 7.99 \times 10^{-3} \,{\rm N}.$$

The force is repulsive (since charges $q_b$ and $q_c$ are of the same sign). Hence, the force is directed from charge $q_b$ towards charge $q_c$, as shown in the diagram. Now, the net force acting on charge $q_c$ is the sum of ${\bf f}_{ac}$ and ${\bf f}_{bc}$. Unfortunately, since ${\bf f}_{ab}$ and ${\bf f}_{bc}$ are vectors pointing in different directions, they cannot be added together algebraically. Fortunately, however, their components along the $x$- and $y$-axes can be added algebraically. Now, it is clear, from the diagram, that ${\bf f}_{bc}$ is directed along the $+x$-axis. If follows that

$$f_{bc\,x} = f_{bc} = 7.99 \times 10^{-3} \,{\rm N},$$

$$f_{bc\,y} = 0.$$

It is also clear, from the diagram, that ${\bf f}_{ac}$ subtends an angle

$$\theta =\tan^{-1} (a/b) = \tan^{-1}(4/3) = 53.1^\circ$$

with the $-x$-axis, and an angle $90^\circ-\theta$ with the $+y$-axis. It follows from the conventional laws of vector projection that

$$f_{ac\,x} = -f_{ac}\,\cos\theta = -(4.31 \times 10^{-3})\,(0.6)=-2.59\times 10^{-3}\,{\rm N},$$

$$f_{ac\,y} = f_{ac}\,\cos(90^\circ -\theta) = f_{ac} \,\sin\theta = (4.31 \times 10^{-3})\,(0.8) = 3.45 \times 10^{-3}\,{\rm N}.$$

The $x$- and $y$-components of the resultant force ${\bf f}_c$ acting on charge $q_c$ are given by

$$f_{c\,x} = f_{ac\,x} + f_{bc\,x} = -2.59\times 10^{-3} + 7.99 \times 10^{-3} =5.40\times 10^{-3}\,{\rm N},$$

$$f_{c\,y} = f_{ac\,y} + f_{bc\,y} = 3.45 \times 10^{-3}\,{\rm N}.$$

Thus, from the Pythagorean theorem, the magnitude of the resultant force is

$$f_c = \sqrt{(f_{c\,x})^2 + (f_{c\,y})^2} = 6.4\times 10^{-3}\,{\rm N}.$$

Furthermore, the resultant force subtends an angle

$$\phi = \tan^{-1} (f_{c\,y}/f_{c\,x}) = 32.6^\circ$$

with the $+x$-axis, and an angle $90^\circ-\phi = 57.4^\circ$ with the $+y$-axis.