Chapter 23, Problem 26P

(a) Because of the inverse square nature of the electric field, any location where the field is zero must be closer to the weaker charge (q₂). Also, in between the two charges, the fields due to the two charges are parallel to each other (both to the left) and cannot cancel. Thus the only places where the field can be zero are closer to the weaker charge, but not between them. In the diagram, this is the point to the left of q₂. Take rightward as the positive direction.

$\begin{array}{c}E=\frac{1}{4\pi {\epsilon }_0}\frac{\left|q_2\right|}{x^2}-\frac{1}{4\pi {\epsilon }_0}\frac{q_1}{{(d+x)}^2}=0\to \left|q_2\right|{(d+x)}^2=q_1{x}^2\to \\ x=\frac{\sqrt{\left|q_2\right|}}{\sqrt{q_1}-\sqrt{\left|q_2\right|}}d=\frac{\sqrt{2.0\times {10}^{-6}\text{C}}}{\sqrt{3.4\times {10}^{-6}\text{C}}-\sqrt{2.0\times {10}^{-6}\text{C}}}(5.0\text{cm})=\boxed{16\text{cm}\text{left}\text{of}{q}_2}\end{array}$

(b) The potential due to the positive charge is positive everywhere, and the potential due to the negative charge is negative everywhere. Since the negative charge is smaller in magnitude than the positive charge, any point where the potential is zero must be closer to the negative charge. So consider locations between the charges (position x₁) and to the left of the negative charge (position x₂) as shown in the diagram.

$\begin{array}{c}{V}_{\text{location 1}}=\frac{1}{4\pi {\epsilon }_0}\left[\frac{q_1}{(d-x_1)}+\frac{q_2}{x_1}\right]=0\to x_1=\frac{q_{2}d}{(q_2-q_1)}=\frac{\left(-2.0\times {10}^{-6}\text{C}\right)(5.0\text{cm})}{\left(-5.4\times {10}^{-6}\text{C}\right)}=1.852\text{cm}\\ V_{\text{location 2}}=\frac{1}{4\pi {\epsilon }_0}\left[\frac{q_1}{(d+x_1)}+\frac{q_2}{x_2}\right]=0\to \\ V_{\text{location 2}}=\frac{1}{4\pi {\epsilon }_0}\left[\frac{q_1}{(d+x_1)}+\frac{q_2}{x_2}\right]=0\to \\ x_2=-\frac{q_{2}d}{(q_1+q_2)}=\frac{\left(-2.0\times {10}^{-6}\text{C}\right)(5.0\text{cm})}{\left(1.4\times {10}^{-6}\text{C}\right)}=7.143\text{cm}\\ \begin{array}{l}\text{So the two locations where the potential is zero are }\\ \boxed{\begin{array}{l}1.9\text{ cm from the negative charge towards the positive}\\ \text{ charge, and 7}\text{.1 cm from the negative charge away from the positive charge}\text{.}\end{array}}\end{array}\end{array}$

(II) Two point charges, 3.4 μC and −2.0μC, are placed 5.0 cm apart on the x axis. At what points along the x axis is (a) the electric field zero and (b) the potential zero? Let V = 0 at r = ∞.