Animated Solution for Physics - Electrostatics: Two electric dipoles, $A,B$ with respective dipole moments ${\mathbf{d}}_A=-4qa\hat{\mathbf{i}}$ and ${\mathbf{d}}_B=-2qa\hat{\mathbf{i}}$ are placed on the X-axis with a separation $R$, as shown in the figure. The distance from $A$ at which both of them produce the same potential is

The journey to mastering electrostatics is paved with problems that test not just your memory of formulas, but your physical intuition. This problem is a beautiful example of how a simple setup can lead to an elegant algebraic dance. Let's dive into the world of electric dipoles and uncover the hidden symmetry in this system.

Imagine you are standing on the X-axis, looking at two electric dipoles, $A$ and $B$. They are separated by a distance $R$.

The problem tells us their dipole moments are ${\mathbf{d}}_A=-4qa\hat{\mathbf{i}}$ and ${\mathbf{d}}_B=-2qa\hat{\mathbf{i}}$. Notice the negative sign and the $\hat{\mathbf{i}}$ vector? This means both dipoles are pointing in the exact same direction—towards the negative X-axis.

This alignment is crucial. It tells us that the electric potential they create in the space between them will have a consistent sign behavior. Our mission is to find a specific point where the potential from dipole $A$ perfectly matches the potential from dipole $B$.

Before we crunch the numbers, let's recall the tool we need. The electric potential $V$ due to a short dipole at a distance $r$ on its axial line is given by:

where $k$ is Coulomb's constant and $p$ is the magnitude of the dipole moment.

Because both dipoles point in the same direction, if we pick a point $P$ between them, the potential from both dipoles will be positive (or negative, depending on your exact convention, but they will be the same sign). This means we can simply equate their magnitudes to find our magical point.

Let's place our point $P$ somewhere between $A$ and $B$. We don't know exactly where it is yet, so let's assign a variable.

Let the distance from dipole $B$ to point $P$ be $x$. Since the total distance between the dipoles is $R$, the distance from dipole $A$ to point $P$ must naturally be $R-x$.

Now, we translate our physical condition into mathematics. We want the potential from $A$ to equal the potential from $B$:

Substituting our dipole formula, we get:

I know this equation might look a bit heavy, but let's take a breath and simplify it. Physics problems often have a way of cleaning themselves up beautifully.

First, let's cancel out the common terms on both sides: $k$, $q$, and $a$. We can also divide both sides by 2. This leaves us with a much cleaner relation:

Now, let's cross-multiply to get rid of the fractions:

This is a quadratic equation, but we don't need the quadratic formula! Since we are dealing with physical distances, we can simply take the positive square root of both sides:

Let's group the $x$ terms together by moving $-x$ to the left side:

Solving for $x$, we find:

We found $x$! Are we done? Absolutely not.

This is where JEE loves to set traps. Look back at our setup. We defined $x$ as the distance from dipole $B$. But the question specifically asks for the distance from dipole $A$.

The distance from $A$ is $R-x$. Let's substitute our value of $x$ into this expression:

To subtract these, we need a common denominator:

The $+R$ and $-R$ cancel out perfectly, leaving us with our final, elegant answer:

And there it is! By carefully setting up our geometry, trusting the algebra, and staying vigilant at the very end, we've conquered this problem. Always remember to double-check what the question is actually asking for before you celebrate!