List-I shows four configurations, each consisting of a pair of ideal electric dipoles. Each dipole h

Q: Why is the equatorial field opposite to the dipole moment?

Think about the electric field lines originating from the positive charge and terminating at the negative charge. At any point on the equatorial plane, the horizontal components of the field from the two charges add up, pointing exactly anti-parallel to the dipole moment vector (which points from negative to positive).

Q: Why is the axial field twice as strong as the equatorial field at the same distance?

This is a fundamental property of the dipole field's geometry. On the axis, the point is closer to one charge than the other, leading to a stronger net field. On the equator, the point is equidistant from both charges, and only the horizontal components survive, resulting in a weaker field. Mathematically, it emerges from the binomial expansion of the exact field equations.

Q: In Case R, why doesn't the right dipole produce a field in the y-direction?

The right dipole in Case R is aligned along the x-axis, and point X is also on the x-axis. This means X is strictly an axial point for this dipole. The electric field on the axis of a dipole is always purely along the axis itself. There is no perpendicular (y-direction) component.

Q: What if the distance to X was different for the two dipoles?

If the distances were different, say $r_1$ and $r_2$, you would simply substitute those specific distances into the formulas: $E \propto 1/r_1^3$ and $E \propto 1/r_2^3$. The superposition principle still holds perfectly; you just add the resulting vectors.

Q: Why do we use $k = 1/(4\pi\epsilon_0)$ instead of keeping it as $k$?

The options in List-II are given in terms of $\epsilon_0$. To match our derived expressions with the given options, we must expand the Coulomb constant $k$. It also helps in seeing how factors like 2 and 4 cancel out, as seen in Case S.

Question

List-I shows four configurations, each consisting of a pair of ideal electric dipoles. Each dipole has a dipole moment of magnitude $p$, oriented as marked by arrows in the figures. In all the configurations the dipoles are fixed such that they are at a distance $2r$ apart along the $x$ direction. The midpoint of the line joining the two dipoles is $X$. The possible resultant electric fields $\vec{E}$ at $X$ are given in List-II. Choose the option that describes the correct match between the entries in List-I to those in List-II.

List-I:

P.

Q.

R.

S.

List-II:

1. $\vec{E} = 0$

2. $\vec{E} = -\frac{p}{2\pi\epsilon_0 r^3} \hat{j}$

3. $\vec{E} = -\frac{p}{4\pi\epsilon_0 r^3} (\hat{i} - \hat{j})$

4. $\vec{E} = \frac{p}{4\pi\epsilon_0 r^3} (2\hat{i} - \hat{j})$

5. $\vec{E} = \frac{p}{\pi\epsilon_0 r^3} \hat{i}$

Sigma Percentile · Accepted Answer

### The Power of Superposition

When dealing with multiple electric dipoles, the principle of superposition is our greatest ally. Instead of getting overwhelmed by the combined system, we can isolate each dipole, calculate its individual electric field at the point of interest, and then simply add the resulting vectors together.

In this problem, we are tasked with finding the net electric field at a midpoint $X$ between two dipoles in four different configurations. The key to unlocking this is recognizing the geometric relationship between the point $X$ and each dipole.

### The Master Tools: Axial and Equatorial Fields

Before diving into the cases, let's arm ourselves with the two fundamental formulas for the electric field of a short dipole.

If a point lies on the **axis** of the dipole at a distance $r$, the electric field is given by:

$$ \vec{E}_{	ext{axial}} = \frac{2kp}{r^3} \hat{p} $$

Notice that the axial field points in the exact same direction as the dipole moment vector $\hat{p}$.

If the point lies on the **equatorial plane** (the perpendicular bisector) at a distance $r$, the electric field is:

$$ \vec{E}_{	ext{equatorial}} = -\frac{kp}{r^3} \hat{p} $$

Here, the negative sign is crucial! It tells us that the equatorial field points in the exact opposite direction to the dipole moment. Also, note that the axial field is twice as strong as the equatorial field for the same distance $r$. We will use $k = \frac{1}{4\pi\epsilon_0}$ throughout our calculations.

### Case P: The Parallel Equators

Let's analyze configuration P. Both dipoles are oriented vertically upwards (in the $+\hat{j}$ direction). The point $X$ lies on the horizontal line connecting them.

For the left dipole, $X$ is on its equatorial line. Therefore, its field at $X$ will point downwards:

$$ \vec{E}_1 = -\frac{kp}{r^3}\hat{j} $$

Similarly, for the right dipole, $X$ is also on its equatorial line. Its field will also point downwards:

$$ \vec{E}_2 = -\frac{kp}{r^3}\hat{j} $$

Adding these two parallel vectors gives us the net field:

$$ \vec{E}_P = -\frac{2kp}{r^3}\hat{j} = -\frac{p}{2\pi\epsilon_0 r^3}\hat{j} $$

This perfectly matches option (2) from List-II.

### Case Q: The Perfect Cancellation

In configuration Q, the left dipole still points upwards, producing a downward equatorial field $\vec{E}_1 = -\frac{kp}{r^3}\hat{j}$.

However, the right dipole is now flipped, pointing downwards ($-\hat{j}$). Since $X$ is still on its equatorial line, the field it produces will be opposite to its dipole moment, meaning it will point upwards:

$$ \vec{E}_2 = -\frac{k(-p\hat{j})}{r^3} = +\frac{kp}{r^3}\hat{j} $$

When we superimpose these two fields, we find they are equal in magnitude but opposite in direction. They perfectly cancel each other out!

$$ \vec{E}_Q = \vec{E}_1 + \vec{E}_2 = 0 $$

This matches option (1).

### Case R: The Orthogonal Dance

Configuration R introduces a twist. The left dipole is still vertical, giving our familiar downward equatorial field $\vec{E}_1 = -\frac{kp}{r^3}\hat{j}$.

But look at the right dipole! It is oriented horizontally along the x-axis ($+\hat{i}$). The point $X$ lies directly on the axis of this dipole. Therefore, we must use the axial field formula. The field will point in the same direction as the dipole moment:

$$ \vec{E}_2 = \frac{2kp}{r^3}\hat{i} $$

The net field is the vector sum of these two orthogonal components:

$$ \vec{E}_R = \frac{2kp}{r^3}\hat{i} - \frac{kp}{r^3}\hat{j} = \frac{kp}{r^3}(2\hat{i} - \hat{j}) $$

Substituting $k = \frac{1}{4\pi\epsilon_0}$, we get:

$$ \vec{E}_R = \frac{p}{4\pi\epsilon_0 r^3}(2\hat{i} - \hat{j}) $$

This is an exact match for option (4).

### Case S: The Axial Reinforcement

Finally, in configuration S, both dipoles are lying flat along the x-axis, pointing to the right ($+\hat{i}$). The point $X$ lies on the axial line for both of them.

Both dipoles will produce an axial field pointing to the right:

$$ \vec{E}_1 = \frac{2kp}{r^3}\hat{i} $$
$$ \vec{E}_2 = \frac{2kp}{r^3}\hat{i} $$

Adding them together, they reinforce each other to create a strong net field:

$$ \vec{E}_S = \frac{4kp}{r^3}\hat{i} $$

When we substitute the value of $k$, the factor of 4 beautifully cancels out with the $4\pi$ in the denominator:

$$ \vec{E}_S = \frac{4p}{4\pi\epsilon_0 r^3}\hat{i} = \frac{p}{\pi\epsilon_0 r^3}\hat{i} $$

This matches option (5).

### The Final Verdict

By systematically applying the axial and equatorial field formulas and respecting vector addition, we have successfully mapped all configurations. The final matching is P $ightarrow$ 2, Q $ightarrow$ 1, R $ightarrow$ 4, and S $ightarrow$ 5. This problem beautifully illustrates how complex electrostatic setups can be dismantled into simple, manageable atomic steps.

List-I

List-II

Select Matching Pairs:

Visualized Solution

\text{Dipole Electric Field Formulas}

\text{Case P: Both Dipoles Equatorial}

\text{Case Q: Anti-parallel Equatorial}

\text{Case R: Equatorial and Axial}

\text{Case S: Both Dipoles Axial}

\text{Final Matrix Match}

The Sigma Insight: Electric Dipole

The Power of Superposition

The Master Tools

Case P

Case Q

Case R

Case S

The Final Verdict

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