Question

In Example 21.4, suppose the point charge on the y-axis at $\mathit{y}\mathbf{=}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{30}\mathit{m}$ has negative charge $\mathbf{-}\mathbf{2}\mathit{\mu }\mathit{C}$, and the other charges remain the same. Find the magnitude and direction of the net force on Q. How does your answer differ from that in Example 21.4? Explain the differences.

Short answer

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

Step-by-step solution

Coulomb’s law

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$\left(q_1\times q_2\right)$ 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 $q_2$.

Magnitude and direction of the net force on Q

Consider the given data as below.

The charge,$q_2=-2.00\mu C=-2.00\times {10}^{-6}C$

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

The Coulomb’s constant, localid="1665130019862" $k=9\times {10}^{9}N\cdot m^2/C^2$

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,

$Q=4.00\mu C=4.00\times {10}^{-6}C$

Because $q_1$exerted repulsive forces in the negative direction of the y-axis with angle $\alpha$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" $$\begin{gathered} F_{net}=2F_{2Q} \\ =2\left(k\frac{\left|q_{2}Q\right|}{r^2}\right)\sin \alpha \end{gathered}$$

Substitute known values in the above expression, and you have

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

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