An electric dipole is formed by two charges +q and -q located in xy-plane at (0, 2) mm and (0, -2) m

Q: Why can we use the short dipole approximation here?

The distance between the charges is around $4$ mm, while the distance to point P is $\sqrt{100^2 + 100^2} \approx 141$ mm. Since $r \gg l$, the short dipole formula $V = \frac{K \vec{P} \cdot \vec{r}}{r^3}$ is highly accurate and simplifies the calculation immensely.

Q: Does the direction of the dipole moment matter?

Absolutely! The dipole moment vector $\vec{P}$ is always directed from the negative charge to the positive charge. Reversing this would flip the sign of your potential.

Q: Can I solve this without using vectors, just by finding distances?

Yes, you could calculate the exact distance from each charge to point P and use $V = \frac{Kq}{r_1} - \frac{Kq}{r_2}$. However, the algebra becomes incredibly messy with square roots. Vectors make it a one-line calculation!

Q: Why did the $x$-component of the new dipole moment decrease the potential?

The dot product $\vec{P} \cdot \vec{r}$ measures how much of the dipole moment aligns with the position vector. The new dipole has a negative $x$-component ($-2\hat{i}$), which opposes the positive $x$-direction of $\vec{r}$ ($100\hat{i}$), reducing the overall dot product and thus the potential.

Q: What if the charges were not equal and opposite?

If the charges don't sum to zero, it's no longer a pure dipole. You would have a monopole (point charge) contribution as well, and you'd have to calculate the potential using the superposition principle for each charge individually.

Question

An electric dipole is formed by two charges +q and -q located in xy-plane at (0, 2) mm and (0, -2) mm, respectively, as shown in the figure. The electric potential at point P(100, 100) mm due to the dipole is $V_0$. The charges +q and -q are then moved to the points (-1, 2) mm and (1, -2) mm, respectively. What is the value of electric potential at P due to the new dipole?

Options:

(A) $V_0 / 4$

(B) $V_0 / 2$

(C) $V_0 / \sqrt{2}$

(D) $3V_0 / 4$

Sigma Percentile · Accepted Answer

### The Power of Vectors in Electrostatics

Imagine you are standing far away from a tiny pair of charges—an electric dipole. When you are at a distance much larger than the separation between the charges, the intricate details of their individual positions blur together. Instead, they act as a single entity characterized by a powerful vector: the **dipole moment** ($\vec{P}$).

In this problem, we are asked to find the electric potential at a distant point $P(100, 100)$ mm. Because the distance to $P$ ($100\sqrt{2} \approx 141$ mm) is vastly greater than the dipole length ($\approx 4$ mm), we can confidently unleash the short dipole approximation. The potential $V$ at a position vector $\vec{r}$ is elegantly given by the dot product:

$$V = \frac{K \vec{P} \cdot \vec{r}}{r^3}$$

This single equation is our master key. It transforms a messy algebraic calculation involving square roots and inverse distances into a clean, swift vector operation.

### Analyzing the Initial Setup

Let's look at the initial state. We have a positive charge $+q$ at $(0, 2)$ and a negative charge $-q$ at $(0, -2)$.

The dipole moment vector $\vec{P}$ always points from the negative charge to the positive charge. Here, the displacement is purely vertical, moving $4$ units up the y-axis. Therefore, our initial dipole moment is:

$$\vec{P}_1 = q(4\hat{j}) = 4q\hat{j}$$

The point $P$ is located at $(100, 100)$, giving us the position vector:

$$\vec{r} = 100\hat{i} + 100\hat{j}$$

Now, we substitute these into our master equation to find the initial potential, $V_0$. When we take the dot product of $\vec{P}_1$ and $\vec{r}$, the $\hat{i}$ component vanishes because the initial dipole has no horizontal spread.

$$V_0 = \frac{K (4q\hat{j}) \cdot (100\hat{i} + 100\hat{j})}{r^3} = \frac{K(400q)}{r^3}$$

Keep this result safe; it is our baseline for comparison.

### The Shift: A New Dipole Emerges

Suddenly, the charges are shifted. The positive charge moves to $(-1, 2)$ and the negative charge moves to $(1, -2)$. The dipole has been tilted and stretched!

To find the new potential, we don't need to panic or draw complex geometry. We just need the new dipole moment, $\vec{P}_2$. We calculate the displacement vector from the new position of $-q$ to the new position of $+q$:

$$\vec{d} = (-1 - 1)\hat{i} + (2 - (-2))\hat{j} = -2\hat{i} + 4\hat{j}$$

Multiplying by the charge $q$, we get our new dipole moment:

$$\vec{P}_2 = q(-2\hat{i} + 4\hat{j})$$

Notice how the y-component remains the same, but a new, negative x-component has appeared.

### The Final Calculation: Dot Product Magic

We are ready for the final strike. We substitute $\vec{P}_2$ and our unchanged position vector $\vec{r}$ into the potential formula:

$$V = \frac{K [q(-2\hat{i} + 4\hat{j})] \cdot (100\hat{i} + 100\hat{j})}{r^3}$$

Let's execute the dot product. We multiply the $\hat{i}$ components and the $\hat{j}$ components:

$$V = \frac{K q [(-2)(100) + (4)(100)]}{r^3}$$

$$V = \frac{K q (-200 + 400)}{r^3} = \frac{K(200q)}{r^3}$$

The magic happens when we compare this new potential $V$ to our initial potential $V_0$.

$$V_0 = \frac{K(400q)}{r^3}$$
$$V = \frac{K(200q)}{r^3}$$

It is crystal clear: the new potential is exactly half of the original potential!

$$V = \frac{V_0}{2}$$

By trusting the vector dot product, we bypassed tedious trigonometry and arrived at the exact answer with absolute mathematical rigor.

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Visualized Solution

$Visualizing the Initial Setup$

$Initial Dipole Moment$

$Potential due to a Dipole$

$Calculating Initial Potential V_{0}$

$The New Dipole Setup$

$New Dipole Moment$

$Calculating New Potential V$

$Comparing V and V_{0}$

$The Way Forward$

The Sigma Insight: Electric Dipole

The Power of Vectors in Electrostatics

Analyzing the Initial Setup

The Shift

The Final Calculation

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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 Initial Setup

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Potential due to a Dipole

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Initial Potential V0​

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Comparing V and V0​

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