91Ó°ÊÓ

Coulomb's law states that the magnitude F of the force exerted by two-point charges${\mathit{q}}_{\mathbf{1}}$ and${\mathit{q}}_{\mathbf{2}}$ separated by r is directly proportional to the product of the charges$\left(q_1×q_2\right)$ and inversely proportional to the square of the distance between them.

$F=k\frac{q_1{q}_2}{r^2}$

Here; F is the force on each point charge exerting on each other, k is the proportionality constant, r is the distance between the charges$q_1$ and $q_2$.

Consider the given data as below.

The charge,$q_2=-2.00\mathrm{μ}C=-2.00×{10}^{-6}C$

The distance between two point charges, r = 0..5 m

The Coulomb’s constant, localid="1665130019862" $k=9×{10}^{9}Nâ‹\dots m^2/C^2$

In example 21.4, the net force on Q is a repulsive force, but now Q has both repulsive and attractive forces. Therefore, the charge,

$Q=4.00\mathrm{μ}C=4.00×{10}^{-6}C$

Because $q_1$exerted repulsive forces in the negative direction of the y-axis with angle $\mathrm{Î\pm }$with the x-axis, the negative charge $q_2$exerted an attractive force towards itself, so the y-axis direction will be;

Because the two forces acting on Q have the same magnitude, the net force in the new case will be;

$F_{net}=F_{1Q}+F_{2Q}$

Here, localid="1665130545351" $F_{1Q}=F_{2Q}$. Therefore,

localid="1665130595651" $$\begin{gathered} F_{net}=2F_{2Q} \\ =2\left(k\frac{\left|q_{2}Q\right|}{r^2}\right)\sin \mathrm{Î\pm } \end{gathered}$$

Substitute known values in the above expression, and you have

$$\begin{gathered} F_{net}=2\left(9×{10}^{9}N·m^2/C^2×\frac{\left|-3.00×{10}^{-6}C\right|\left|4.00×{10}^{-6}C\right|}{{\left(0.5m\right)}^2}\right)\left(\frac{0.3m}{0.5m}\right) \\ =345.6×{10}^{-3}N \\ =0.35N \end{gathered}$$

Hence, the net force is 0.35 N and its direction is in the negative y-direction.