Two fixed charges -2Q and Q are located at the points with coordinates (-3a, 0) and (+3a, 0) respect

Q: Why do we use absolute values for distance when calculating potential on the $x$-axis?

Distance is a scalar quantity and must always be positive. Depending on where you are on the $x$-axis relative to the charges, the algebraic difference in coordinates might be negative, so the absolute value ensures we are using the correct physical distance.

Q: How did we know the locus would be a circle?

When you equate the distances from two fixed points in a constant ratio (other than 1:1), the resulting locus is always a circle. This is a classic geometric property known as the Circle of Apollonius!

Q: Why does the potential go to negative infinity near $-2Q$ and positive infinity near $Q$?

The potential is inversely proportional to the distance from the charge. As you get infinitely close to a charge, the distance approaches zero, causing the potential to blow up. The sign of the infinity matches the sign of the source charge.

Q: Why does the particle move outwards from the center?

Positive charges naturally tend to move from regions of higher potential to lower potential to minimize their potential energy. Since the center is at a positive potential and the circle is at zero potential, the particle is pushed outwards.

Q: Could the particle stop before reaching the circle?

No, because there are no other forces acting on it, and the potential continuously decreases from the center to the circle. It will keep accelerating, converting its potential energy entirely into kinetic energy by the time it reaches the zero-potential boundary.

Question

Two fixed charges $- 2Q$ and $Q$ are located at the points with coordinates $(- 3a, 0)$ and $(+ 3a, 0)$ respectively in the $x-y$ plane. (1991)
(a) Show that all points in the $x-y$ plane where the electric potential due to the two charges is zero, lie on a circle. Find its radius and the location of its centre.
(b) Give the expression $V(x)$ at a general point on the $x$-axis and sketch the function $V(x)$ on the whole $x$-axis.
(c) If a particle of charge $+q$ starts form rest at the centre of the circle, show by a short quantitative argument that the particle eventually crosses the circle. Find its speed when it does so.

Sigma Percentile · Accepted Answer

### Visualizing the Electrostatic Landscape

Imagine a vast, empty $x-y$ plane. We anchor two charges onto this canvas: a heavy, negative charge $-2Q$ at $x = -3a$, and a lighter, positive charge $Q$ at $x = +3a$. These two charges create an invisible landscape of electric potential around them. Our mission is to map out the regions where this potential perfectly balances out to zero, and then explore the journey of a test charge placed within this field.

### The Locus of Zero Potential

To find the points where the net electric potential is zero, we consider a general point $P(x, y)$. The total potential at $P$ is simply the algebraic sum of the potentials due to each individual charge.

$$V = \frac{1}{4\pi\varepsilon_0} \left[ \frac{Q}{\sqrt{(3a-x)^2 + y^2}} + \frac{-2Q}{\sqrt{(3a+x)^2 + y^2}} ight]$$

Setting this potential to zero means the magnitudes of the potentials from both charges must be exactly equal.

$$\frac{Q}{\sqrt{(3a-x)^2 + y^2}} = \frac{2Q}{\sqrt{(3a+x)^2 + y^2}}$$

To eliminate the square roots, we square both sides. This gives us a purely algebraic equation:

$$4[(3a-x)^2 + y^2] = (3a+x)^2 + y^2$$

Expanding the squares and grouping the terms carefully, we get:

$$4(9a^2 - 6ax + x^2 + y^2) = 9a^2 + 6ax + x^2 + y^2$$
$$3x^2 - 30ax + 3y^2 + 27a^2 = 0$$

Dividing the entire equation by 3 and completing the square for the $x$ terms reveals a beautiful geometric truth:

$$x^2 - 10ax + y^2 + 9a^2 = 0$$
$$(x-5a)^2 + y^2 = (4a)^2$$

This is the standard equation of a circle! The locus of all points with zero potential forms a perfect circle with its center at $(5a, 0)$ and a radius of $4a$. This fascinating result is a classic example of the Circle of Apollonius.

### Analyzing the Potential Along the Axis

Now, let's restrict our view strictly to the $x$-axis, where $y = 0$. The potential expression simplifies, but we must use absolute values to ensure distances remain positive.

$$V(x) = \frac{Q}{4\pi\varepsilon_0} \left[ \frac{1}{|x-3a|} - \frac{2}{|x+3a|} ight]$$

This function has vertical asymptotes at the locations of the charges. As we approach $x = -3a$, the potential plunges to $-\infty$. As we approach $x = 3a$, it skyrockets to $+\infty$.

The potential crosses zero at exactly two points on the axis: $x = a$ and $x = 9a$. These are precisely the points where our zero-potential circle intersects the $x$-axis! Between $x = a$ and $x = 3a$, the potential is positive, dominated by the closer $+Q$ charge. Beyond $x = 9a$, the $-2Q$ charge takes over, and the potential becomes negative again.

### The Journey of the Test Charge

For the grand finale, imagine placing a positive test charge $+q$ at the center of our zero-potential circle, $C(5a, 0)$. Let's calculate the potential at this exact spot:

$$V_C = \frac{Q}{4\pi\varepsilon_0} \left[ \frac{1}{2a} - \frac{2}{8a} ight] = \frac{Q}{16\pi\varepsilon_0 a}$$

The potential at the center is strictly positive. Since the boundary of the circle is at zero potential, our positive charge $+q$ is sitting on a "potential hill." Nature dictates that positive charges will spontaneously roll down to regions of lower potential to minimize their potential energy. Therefore, the particle will accelerate outwards and inevitably cross the circle!

To find its speed as it crosses the boundary, we invoke the principle of conservation of energy. The gain in kinetic energy must equal the loss in potential energy:

$$\frac{1}{2}mv^2 = q(V_C - V_{	ext{circle}})$$

Since $V_{	ext{circle}} = 0$, we simply plug in our value for $V_C$:

$$\frac{1}{2}mv^2 = q \left( \frac{Q}{16\pi\varepsilon_0 a} ight)$$

Solving for $v$, we arrive at the final speed of the particle:

**$$v = \sqrt{\frac{Qq}{8\pi\varepsilon_0 ma}}$$**

And there we have it—a complete journey from abstract potential fields to the dynamic motion of a particle!

Visualized Solution

$Visualizing the Setup$

$Potential at a General Point P (x, y)$

$Locus of Zero Potential$

$Equation of the Circle$

$Potential on the x -axis$

$Graph of V (x)$

$Motion of the Particle$

$Speed of the Particle$

The Sigma Insight: Electric Potential and Potential Difference

Visualizing the Electrostatic Landscape

The Locus of Zero Potential

Analyzing the Potential Along the Axis

The Journey of the Test Charge

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Potential at a General Point P(x,y)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Locus of Zero Potential

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equation of the Circle

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Potential on the x-axis

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Graph of V(x)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Motion of the Particle

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Speed of the Particle