Four point charges -q, +q, +q and -q are placed on Y-axis at y = -2d, y = -d, y = +d and y = +2d, re

Q: Why do the Y-components of the electric field cancel out?

Because the charge distribution is perfectly symmetric about the X-axis. For every charge at $+y$, there is an identical charge at $-y$. Their Y-components are equal and opposite, thus summing to zero.

Q: Why can't we just substitute $d=0$ right at the beginning since $D \gg d$?

If you substitute $d=0$, all four charges collapse to the origin. You'd have $+2q$ and $-2q$ at the exact same point, giving a net charge of zero and an electric field of exactly zero. The binomial expansion helps us find the *first non-zero* term, which tells us how the field behaves at large, but not infinite, distances.

Q: What exactly is a quadrupole?

A quadrupole is an arrangement of four charges whose total charge is zero and whose dipole moment is also zero. It's the next level of complexity after a dipole. While a dipole field drops off as $1/r^3$, a quadrupole field drops off even faster, as $1/r^4$.

Q: How do we know when to use the binomial expansion?

Whenever you see a condition like $D \gg d$ or $x \ll 1$ in a physics problem involving fractions or roots, it's a massive flashing sign to use the binomial expansion: $(1+x)^n \approx 1+nx$.

Q: What would happen if the charges were $+q, -q, +q, -q$ instead?

That would break the symmetry! The Y-components wouldn't perfectly cancel in the same way, and the system would have a non-zero dipole moment. The dominant field at large distances would then drop off as $1/D^3$ instead of $1/D^4$.

Question

Four point charges $-q, +q, +q$ and $-q$ are placed on Y-axis at $y = -2d$, $y = -d$, $y = +d$ and $y = +2d$, respectively. The magnitude of the electric field $E$ at a point on the X-axis at $x = D$, with $D >> d$, will behave as

Options:

(A) $E \propto \frac{1}{D}$

(B) $E \propto \frac{1}{D^3}$

(C) $E \propto \frac{1}{D^2}$

(D) $E \propto \frac{1}{D^4}$

Sigma Percentile · Accepted Answer

### The Magic of Multipoles

Imagine you are standing far away from a complex arrangement of charges. If you are far enough, the fine details blur, and you only see the dominant behavior. This is the essence of multipole expansion!

In this problem, we are dealing with a highly symmetric arrangement of four charges. This isn't just any random setup; it's a **linear quadrupole**.

Let's break down how we find the electric field for this fascinating system.

### Analyzing the Setup and Symmetry

We have four charges placed along the Y-axis: $+q$ at $y = \pm d$, and $-q$ at $y = \pm 2d$. We want to find the electric field at a point $P$ on the X-axis, at a distance $D$.

The first thing to notice is the beautiful symmetry. For every charge above the X-axis, there is an identical charge at the exact same distance below the X-axis.

Because of this, the Y-components of the electric fields produced by these pairs will be equal and opposite. They perfectly cancel each other out!

Therefore, we only need to worry about the X-components of the electric fields.

### The Master Equation

Let's write down the net electric field along the X-axis. The field from the positive charges points away from them, while the field from the negative charges points towards them.

Using Coulomb's law and resolving the vectors along the X-axis, we get:

$$ E_{net} = 2E_{+q}\cos	heta_1 - 2E_{-q}\cos	heta_2 $$

Substituting the expressions for the electric field and the cosine of the angles from the geometry of the setup:

$$ E_{net} = 2 \left( \frac{kq}{D^2+d^2} ight) \left( \frac{D}{\sqrt{D^2+d^2}} ight) - 2 \left( \frac{kq}{D^2+(2d)^2} ight) \left( \frac{D}{\sqrt{D^2+(2d)^2}} ight) $$

This simplifies to:

$$ E_{net} = \frac{2kqD}{(D^2+d^2)^{3/2}} - \frac{2kqD}{(D^2+4d^2)^{3/2}} $$

### The Art of Approximation

This equation looks terrifying, but let's take a breath. The problem gives us a crucial piece of information: $D \gg d$.

This is a massive flashing sign telling us to use the **binomial expansion**. To do this, we need to factor out $D^2$ from the denominators so we can create terms that are much smaller than 1.

$$ E_{net} = \frac{2kq}{D^2} \left[ \left(1+\frac{d^2}{D^2}ight)^{-3/2} - \left(1+\frac{4d^2}{D^2}ight)^{-3/2} ight] $$

Now, we apply the binomial approximation $(1+x)^n \approx 1+nx$:

$$ \left(1+\frac{d^2}{D^2}ight)^{-3/2} \approx 1 - \frac{3d^2}{2D^2} $$

$$ \left(1+\frac{4d^2}{D^2}ight)^{-3/2} \approx 1 - \frac{12d^2}{2D^2} $$

### Final Calculation

Let's substitute these approximations back into our master equation:

$$ E_{net} \approx \frac{2kq}{D^2} \left[ \left(1 - \frac{3d^2}{2D^2}ight) - \left(1 - \frac{12d^2}{2D^2}ight) ight] $$

Notice how the $1$s perfectly cancel out! This is why we couldn't just set $d=0$ at the beginning. If we did, the entire field would be zero. The binomial expansion reveals the first non-zero term.

$$ E_{net} \approx \frac{2kq}{D^2} \left[ \frac{9d^2}{2D^2} ight] $$

$$ E_{net} = \frac{9kqd^2}{D^4} $$

### The Grand Conclusion

We have arrived at our final result! The magnitude of the electric field is:

$$ E \propto \frac{1}{D^4} $$

This is the hallmark of a **quadrupole**. A single charge (monopole) has a field that drops as $1/r^2$. A dipole drops as $1/r^3$. And as we've just beautifully proven, a quadrupole drops as $1/r^4$.

The further you go in the multipole expansion, the faster the field fades away!

Animated Solution for Physics - Electrostatics: Four point charges $- q, + q, + q$ and $- q$ are placed on Y-axis at $y = - 2 d$ , $y = - d$ , $y = + d$ and $y = + 2 d$ , respectively. The magnitude of the electric field $E$ at a point on the X-axis at $x = D$ , with $D >> d$ , will behave as

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Visualized Solution

$Visualizing the Setup$

$Electric Field Contributions$

$Symmetry and Superposition$

$Net Electric Field Equation$

$Simplifying the Expression$

$Binomial Approximation$

$Final Calculation$

$Conclusion$

$The Linear Quadrupole$

The Sigma Insight: Electric Field

The Magic of Multipoles

Analyzing the Setup and Symmetry

The Master Equation

The Art of Approximation

Final Calculation

The Grand Conclusion

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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; } Electric Field Contributions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Symmetry and Superposition

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Net Electric Field Equation

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

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Linear Quadrupole