Charges -q and +q located at A and B, respectively, constitute an electric dipole. Distance AB = 2a,

Q: Why do we ignore $a^2$ in the denominator?

Because $y \gg 2a$, $y^2$ is massively larger than $a^2$. Adding a tiny number to a huge number doesn't change it significantly, making the math much cleaner without losing accuracy.

Q: Does the sign of the charge $Q$ matter here?

The sign of $Q$ determines the *direction* of the force, but the question only asks for the *magnitude* of the new force relative to the old one. The scaling factor (27) remains the same regardless of the sign.

Q: What if the problem didn't say $y/3 \gg 2a$?

If $y/3$ wasn't much larger than $2a$, we couldn't use the short dipole approximation for the second case. We would have to use the exact formula, and the ratio wouldn't be a neat integer like 27.

Q: Why is the electric field proportional to $1/r^3$ and not $1/r^2$ like a point charge?

A dipole consists of two opposite charges. From far away, their fields almost cancel each other out. The remaining field drops off faster ($1/r^3$) than the field of a single isolated charge ($1/r^2$).

Q: Would the answer change if the charge was on the axial line?

No! The electric field on the axial line is also proportional to $1/r^3$ (specifically $2kP/r^3$). The factor of 2 cancels out when taking the ratio, so the force would still increase by a factor of 27.

Question

Charges $-q$ and $+q$ located at $A$ and $B$, respectively, constitute an electric dipole. Distance $AB = 2a$, $O$ is the mid point of the dipole and $OP$ is perpendicular to $AB$. A charge $Q$ is placed at $P$, where $OP = y$ and $y \gg 2a$. The charge $Q$ experiences an electrostatic force $F$.
If $Q$ is now moved along the equatorial line to $P'$ such that $OP' = \left(\frac{y}{3}\right)$, the force on $Q$ will be close to $\left(\frac{y}{3} \gg 2a\right)$

Options:

(A) $\frac{F}{3}$

(B) $3F$

(C) $9F$

(D) $27F$

Sigma Percentile · Accepted Answer

### The Setup: A Dipole and a Charge

Imagine an electric dipole as a tiny, rigid barbell with a positive charge on one end and a negative charge on the other. This arrangement creates a unique electric field around it. In our problem, we place a test charge $Q$ on the equatorial line—a line that runs perfectly perpendicular to the dipole, right through its center.

The electric field $E$ created by a dipole on its equatorial line at a distance $r$ is given by the exact formula:

$$E = \frac{kP}{(r^2 + a^2)^{3/2}}$$

Here, $P = q(2a)$ is the dipole moment, representing the strength of the dipole.

### The Short Dipole Approximation

The magic happens when we move far away from the dipole. The problem explicitly states that the distance $y$ is much, much greater than the dipole's length $2a$ ($y \gg 2a$).

Because $y$ is so massive compared to $a$, the term $a^2$ becomes completely negligible when added to $y^2$. It's like adding a drop of water to the ocean! We can safely approximate the denominator:

$$y^2 + a^2 \approx y^2$$

Substituting this back into our electric field equation, the square and the square root cancel out, leaving us with a beautifully simple inverse-cube relationship:

$$E_1 = \frac{kP}{(y^2)^{3/2}} = \frac{kP}{y^3}$$

Consequently, the initial electrostatic force $F$ experienced by the charge $Q$ is simply the charge multiplied by this field:

$$F = Q E_1 = \frac{kPQ}{y^3}$$

### The Inverse Cube Magic

Now, we bring the charge $Q$ much closer to the dipole, to a new distance $r' = \frac{y}{3}$. The problem assures us that even at this closer distance, the short dipole approximation still holds true ($\frac{y}{3} \gg 2a$).

Let's calculate the new electric field $E_2$ at this closer point by plugging our new distance into the approximated formula:

$$E_2 = \frac{kP}{(y/3)^3}$$

When we cube the fraction in the denominator, we get $y^3 / 27$. The $27$ flips up to the numerator, drastically amplifying the field:

$$E_2 = 27 \frac{kP}{y^3}$$

### The Final Verdict

With the new electric field calculated, finding the new force $F'$ is a breeze. We multiply the charge $Q$ by the new field $E_2$:

$$F' = Q E_2 = 27 \left(\frac{kPQ}{y^3}\right)$$

Notice that the term inside the parentheses is exactly our initial force $F$!

$$F' = 27F$$

By moving the charge three times closer, the force didn't just triple, and it didn't just increase by a factor of nine (as it would for a single point charge). Because of the dipole's unique $1/r^3$ field geometry, the force skyrocketed by a factor of **27**! This highlights the rapid decay and equally rapid intensification of dipole fields.

Select Answer:

Visualized Solution

Visualizing the Setup

Electric Field on Equatorial Line

Applying the Short Dipole Approximation

Initial Force on Charge $Q$

Moving to the New Position $P^{'}$

Electric Field at $P^{'}$

Final Force Calculation

The Way Forward

The Sigma Insight: Electric Dipole

The Setup

The Short Dipole Approximation

The Inverse Cube Magic

The Final Verdict

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