Irodov 3.4: A Negative Charge Holds Two Apart – Amit & Deepali Sinha Foundation

Problem Irodov 3.4

Positive point charges $q_1$ and $q_2$ sit at the points with position vectors $\vect{r}_1$ and $\vect{r}_2$ . A negative point charge $q_3$ is to be placed at a point with position vector $\vect{r}_3$ so that the net force on every one of the three charges is zero. Find $q_3$ and $\vect{r}_3$ .

The two positive charges q_1 and q_2 at the ends of their position vectors r_1 and r_2 from the origin O. — The two positive charges $q_1$ and $q_2$ at the ends of their position vectors $\vect{r}_1$ and $\vect{r}_2$ from the origin $O$ .

Given

$q_1 > 0$ at $\vect{r}_1$
$q_2 > 0$ at $\vect{r}_2$
$q_3 < 0$

Find

q_3

and

\vect{r}_3

for zero force on all three charges

Solution

Step 1Where the third charge must sit

On $q_1$ the repulsion of $q_2$ acts along the line through the two charges, so the attraction of $q_3$ can cancel it only if $q_3$ lies on that line. On $q_3$ both forces are attractions, toward $q_1$ and toward $q_2$ ; they point opposite ways only if $q_3$ lies between the two. Call $d = \lvert\vect{r}_2 - \vect{r}_1\rvert$ the separation of the given charges, and $d_1$ , $d_2$ the distances of $q_3$ from $q_1$ and $q_2$ :

d_1 + d_2 = d \tag{1}

The three charges on one line, drawn for q_2 = 2q_1: the negative q_3 between the two positive ones, pulled equally toward each by F_31 and F_32. — The three charges on one line, drawn for $q_2 = 2q_1$ : the negative $q_3$ between the two positive ones, pulled equally toward each by $\vect{F}_{31}$ and $\vect{F}_{32}$ .

The two attractions on $q_3$ must have equal magnitudes. The factor $\lvert q_3\rvert/(4\pi\varepsilon_0)$ cancels:

\begin{aligned} \frac{q_1\lvert q_3\rvert}{4\pi\varepsilon_0 d_1^2} &= \frac{q_2\lvert q_3\rvert}{4\pi\varepsilon_0 d_2^2}\\ \frac{d_1}{d_2} &= \sqrt{\frac{q_1}{q_2}} \end{aligned}

With (1), the distance from $q_1$ is

d_1 = \frac{\sqrt{q_1}}{\sqrt{q_1} + \sqrt{q_2}}\,d \tag{2}

Step 2The position vector

The point $\vect{r}_3$ lies on the segment from $\vect{r}_1$ to $\vect{r}_2$ , the fraction $d_1/d$ of the way along it. With (2):

\begin{aligned} \vect{r}_3 &= \vect{r}_1 + \frac{d_1}{d}\,(\vect{r}_2 - \vect{r}_1)\\ &= \vect{r}_1 + \frac{\sqrt{q_1}}{\sqrt{q_1}+\sqrt{q_2}}\,(\vect{r}_2 - \vect{r}_1) \end{aligned}

Collecting the terms in $\vect{r}_1$ over the common denominator:

\vect{r}_3 = \frac{\vect{r}_1\sqrt{q_2} + \vect{r}_2\sqrt{q_1}}{\sqrt{q_1} + \sqrt{q_2}}

(3)

Step 3The size of the third charge

Now $q_1$ must be in equilibrium too: the attraction of $q_3$ at distance $d_1$ must balance the repulsion of $q_2$ at distance $d$ . The factor $q_1/(4\pi\varepsilon_0)$ cancels, and (2) gives $d_1/d$ :

\begin{aligned} \frac{q_1\lvert q_3\rvert}{4\pi\varepsilon_0 d_1^2} &= \frac{q_1 q_2}{4\pi\varepsilon_0 d^2}\\ \lvert q_3\rvert &= q_2\,\frac{d_1^2}{d^2} = q_2\,\frac{q_1}{(\sqrt{q_1}+\sqrt{q_2})^2} \end{aligned}

The charge is negative, so

q_3 = -\frac{q_1 q_2}{(\sqrt{q_1} + \sqrt{q_2})^2}

(4)

Check

The third condition, on $q_2$ , was never imposed, and it must hold by itself, because the three internal forces sum to zero by Newton’s third law. Directly: the pull of $q_3$ on $q_2$ must balance the push of $q_1$ , which needs $\lvert q_3\rvert/d_2^2 = q_1/d^2$ . With $d_2 = d - d_1$ from (1) and (2), and $\lvert q_3\rvert$ from the box,

\begin{aligned} d_2 &= \frac{\sqrt{q_2}}{\sqrt{q_1}+\sqrt{q_2}}\,d\\ \frac{\lvert q_3\rvert}{d_2^2} &= \frac{q_1 q_2}{(\sqrt{q_1}+\sqrt{q_2})^2}\cdot\frac{(\sqrt{q_1}+\sqrt{q_2})^2}{q_2\,d^2} = \frac{q_1}{d^2} \end{aligned}

as required. Special case: for $q_1 = q_2 = q$ the charge $-q/4$ sits at the midpoint, the familiar result.

Remark

The equilibrium cannot be stable. Move $q_3$ a little toward $q_1$ along the line: the pull of $q_1$ grows and that of $q_2$ weakens, so $q_3$ runs on toward $q_1$ . This is Earnshaw’s theorem at work: no arrangement of charges held only by electrostatic forces can be in stable equilibrium, because the potential energy of a charge in a static field has no minimum in empty space.