Animated Solution for Physics - Electrostatics: Comprehension Passage

The problem of finding the locus of points where the electric potential is zero due to two unequal charges is a classic and beautiful application of coordinate geometry in physics. It perfectly demonstrates how algebraic manipulation can reveal hidden geometric structures—in this case, the famous Apollonius circle.

Imagine you are looking at a flat 2D plane. We have two point charges placed on the x-axis. At the origin $(0,0)$, we place a negative charge $-Q$. A little further down the x-axis, at the point $(2,0)$, we place a positive charge $+\frac{Q}{\sqrt{3}}$.

The question tells us that there is a set of points in this plane where the total electric potential is exactly zero, and this set forms a circle. Our goal is to find the radius $R$ and the center $(b,0)$ of this circle.

Let's pick a random point $P(x,y)$ on this equipotential circle. The total electric potential at $P$ is simply the scalar sum of the potentials created by each individual charge.

Here, $r_1$ is the distance from the origin to $P$, and $r_2$ is the distance from $(2,0)$ to $P$. Because the total potential is zero, we can move the negative term to the other side of the equation.

Notice how beautifully the constants $k$ and $Q$ cancel out! This tells us that the shape of the zero-potential surface doesn't depend on the actual magnitude of $Q$, but only on the ratio of the charges.

Cross-multiplying gives us a simple geometric constraint:

Now, we need to express these distances in terms of our coordinates $x$ and $y$ using the standard distance formula.

Substituting these into our constraint equation, we get:

To get rid of those intimidating square roots, we simply square both sides. This is a safe operation because distances are always positive.

Now comes the execution phase. We need to expand the squared term on the right side.

Distributing the $3$ across the terms inside the parenthesis:

To see the true shape of this equation, let's bring all the terms to one side. Subtracting $x^2$ and $y^2$ from the right side yields:

This is starting to look like a circle! To make it standard, let's divide the entire equation by $2$.

To find the center and radius, we must mold this equation into the standard circle form: $(x-h{)}^2+(y-k{)}^2=R^2$. We do this by completing the square for the $x$ terms.

Take the $x$ terms: $x^2-6x$. Half of $-6$ is $-3$, and $(-3{)}^2$ is $9$. So, we add and subtract $9$.

This perfectly collapses into:

Moving the constant to the right side gives us our final, pristine equation:

We can rewrite the right side as a square to explicitly see the radius.

Comparing this to the standard equation $(x-b{)}^2+y^2=R^2$, the parameters of our equipotential circle are crystal clear.

This problem is a fantastic reminder that physical constraints (like zero potential) often translate into elegant geometric shapes. The locus of a point whose distance from two fixed points is in a constant ratio (other than 1) is always a circle, known mathematically as the Circle of Apollonius!