Determine the electric dipole moment of the system of three charges, placed on the vertices of an eq

Q: Why can we just split the $-2q$ charge into two $-q$ charges?

This is due to the principle of superposition. A charge of $-2q$ is physically and mathematically equivalent to two $-q$ charges placed at the exact same location. By treating them as separate entities, we can pair them up with other charges to form standard dipoles, making the calculation much simpler.

Q: Why does the dipole moment vector point from negative to positive?

By convention in physics, the electric dipole moment vector $\vec{p}$ is defined to point from the negative charge to the positive charge. This aligns with the definition $\vec{p} = q\vec{d}$, where $\vec{d}$ is the displacement vector from $-q$ to $+q$.

Q: How did we know the angle between the two dipole moments is $60^\circ$?

The charges are placed at the vertices of an equilateral triangle, which has internal angles of $60^\circ$. The dipole moment vectors lie along the sides of this triangle, starting from the top vertex and pointing towards the base vertices. Therefore, the angle between these two vectors is exactly the vertex angle of the triangle, which is $60^\circ$.

Q: Why do the horizontal components cancel out?

The system is perfectly symmetric about the y-axis. The two dipole moments have equal magnitudes and are mirrored across the y-axis. Their horizontal components are equal in magnitude but point in opposite directions (one to the left, one to the right), so their sum is exactly zero.

Q: What if the triangle wasn't equilateral?

If the triangle were not equilateral, the magnitudes of the dipole moments might be different (if the sides are unequal), and the angle between them would not be $60^\circ$. You would still use the same method—splitting the top charge and forming two dipoles—but you'd have to use the specific side lengths and angles of that triangle for your vector addition.

Question

Determine the electric dipole moment of the system of three charges, placed on the vertices of an equilateral triangle as shown in the figure.

Options:

(A) $\sqrt{3} ql \frac{\hat{j} - \hat{i}}{\sqrt{2}}$

(B) $2ql \hat{j}$

(C) $-\sqrt{3} ql \hat{j}$

(D) $(ql) \frac{\hat{i} + \hat{j}}{\sqrt{2}}$

Sigma Percentile · Accepted Answer

### The Setup: A Triangle of Charges

Imagine an equilateral triangle with charges placed at its vertices. We have a $+q$ at the origin, another $+q$ at a distance $l$ on the x-axis, and a $-2q$ at the top vertex.

Our goal is to find the net electric dipole moment of this entire system. At first glance, this might seem tricky because a standard dipole consists of only two charges: one positive and one negative.

Here, we have three charges, and one of them is $-2q$. How do we proceed?

### The Master Stroke: Splitting the Charge

Look closely at the top vertex. We have a $-2q$ charge. What if we cleverly split this $-2q$ into two separate $-q$ charges?

This is perfectly valid due to the **Principle of Superposition**. A charge of $-2q$ is physically and mathematically equivalent to two $-q$ charges placed at the exact same location.

By treating them as separate entities, we can pair them up with the other charges to form standard dipoles. This simple trick will make our lives so much easier.

### Constructing the Dipoles

By splitting the charge, we can pair one $-q$ with the $+q$ at the origin to form our first dipole, $\vec{p}_1$.

Then, we pair the second $-q$ with the other $+q$ to form our second dipole, $\vec{p}_2$.

By convention, the electric dipole moment vector points from the negative charge to the positive charge. Both dipoles have a magnitude of $p = ql$. Because it is an equilateral triangle, the angle between these two dipole vectors is exactly $60^\circ$.

### The Vector Addition: Finding the Magnitude

Now it is just vector addition! To find the resultant of two equal vectors at $60^\circ$, we use the parallelogram law of vector addition.

The magnitude of the resultant dipole moment is given by:
$$p_{net} = \sqrt{p_1^2 + p_2^2 + 2p_1 p_2 \cos 60^\circ}$$

Substituting $p_1 = p_2 = ql$ and $\cos 60^\circ = 1/2$, we get:
$$p_{net} = \sqrt{(ql)^2 + (ql)^2 + 2(ql)^2 \left(\frac{1}{2}\right)}$$

The math simplifies beautifully:
$$p_{net} = \sqrt{3(ql)^2} = \sqrt{3}ql$$

### Symmetry and Direction: The Final Piece

But what about the direction? Look at the symmetry of the system.

The horizontal components of $\vec{p}_1$ and $\vec{p}_2$ are equal and opposite. One points to the left and the other points to the right, so they cancel out perfectly.

The vertical components, however, both point downwards. They add up, pointing straight down along the negative y-axis.

Therefore, our final vector answer is:
$$\vec{p}_{net} = -\sqrt{3} ql \hat{j}$$

### The Takeaway: Power of Symmetry

This method of splitting charges to form dipoles is a very powerful tool in electrostatics.

You can use it for squares, hexagons, or any symmetric charge distribution. Always look for hidden symmetries; they are the key to solving complex physics problems effortlessly.

Animated Solution for Physics - Electrostatics: Determine the electric dipole moment of the system of three charges, placed on the vertices of an equilateral triangle as shown in the figure.

Select Answer:

Visualized Solution

$System Setup$

$Splitting the Charge$

$Forming Dipoles$

$Resultant Magnitude$

$Resultant Direction$

$Conclusion$

The Sigma Insight: Electric Dipole

The Setup

The Master Stroke

Constructing the Dipoles

The Vector Addition

Symmetry and Direction

The Takeaway

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