The situation depicted in Fig. $24-5$ is that of two tiny charged spheres separated by $10.0 \mathrm{~cm}$ in air. Find $(a)$ the electric field $E$ at point $P,(b)$ the force on a $-4.0 \times 10^{-8} \mathrm{C}$ charge placed at $P$, and $(c)$ where in the region the electric field would be zero (in the absence of the $-4.0 \times 10^{-8}$ C charge).
(a) A positive test charge placed at $P$ will be repelled to the right by the positive charge $q_{1}$ and attracted to the right by the negative charge $q_{2} .$ Because $\overrightarrow{\mathbf{E}}_{1}$ and $\overrightarrow{\mathbf{E}}_{2}$ have the same direction, we can add their magnitudes to obtain the magnitude of the resultant field:
$$E=E_{1}+E_{2}=k_{0} \frac{\left|q_{1}\right|}{r_{1}^{2}}+k_{0} \frac{\left|q_{2}\right|}{r_{2}^{2}}=\frac{k_{0}}{r_{1}^{2}}\left(\left|q_{1}\right|+\left|q_{2}\right|\right)$$
where $r_{1}=r_{2}=0.05 \mathrm{~m}$, and $\left|q_{1}\right|$ and $\left|q_{2}\right|$ are the absolute values of $q_{1}$ and $q_{2}$. Hence,
$$E=\frac{9.0 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}}{(0.050 \mathrm{~m})^{2}}\left(25 \times 10^{-8} \mathrm{C}\right)=9.0 \times 10^{5} \mathrm{~N} / \mathrm{C}$$
directed toward the right.
(b) A charge $q$ placed at $P$ will experience a force $E q .$ Therefore,
$$F_{E}=E q=\left(9.0 \times 10^{5} \mathrm{~N} / \mathrm{C}\right)\left(-4.0 \times 10^{-8} \mathrm{C}\right)=-0.036 \mathrm{~N}$$
The negative sign tells us the force is directed toward the left. This is correct because the electric field represents the force on a positive charge. The force on a negative charge is opposite in direction to
the field.
(c) Reasoning as in Problem $24.9$, we conclude that the field will be zero somewhere to the right of the $-5.0 \times 10^{-8} \mathrm{C}$ charge. Represent the distance to that point from the $-5.0 \times 10^{-8} \mathrm{C}$ charge by $d$ At that point,
$$E_{1}-E_{2}=0$$
because the field due to the positive charge is to the right, while the field due to the negative charge is to the left. Thus,
$$k_{0}\left(\frac{\left|q_{1}\right|}{r_{1}^{2}}-\frac{\left|q_{2}\right|}{r_{2}^{2}}\right)=\left(9.0 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}\right)\left[\frac{20 \times 10^{-8} \mathrm{C}}{(d+0.10 \mathrm{~m})^{2}}-\frac{5.0 \times 10^{-8} \mathrm{C}}{d^{2}}\right]=0$$
Simplifying, we obtain
$$3 d^{2}-0.2 d-0.01=0$$
The quadratic formula yields $d=0.10 \mathrm{~m}$ and $-0.03 \mathrm{~m}$. Only the plus sign has meaning here, and therefore $d=0.10 \mathrm{~m}$. The point in question is $10 \mathrm{~cm}$ to the right of the negative charge.