A positive point charge q is fixed at origin. A dipole with a dipole moment p is placed along the x-

Q: Why is the initial energy zero?

Because the dipole is initially "far away" (at infinity), where the electric field and potential due to the point charge are zero.

Q: Why does the dipole move towards the origin?

The electric field of the point charge is non-uniform. It is stronger closer to the origin. The negative charge of the dipole is closer to the origin than the positive charge, so the attractive force on the negative charge is stronger than the repulsive force on the positive charge, resulting in a net attractive force.

Q: Can we find the force on the dipole directly?

Yes, the force on a dipole in a non-uniform electric field is given by $\mathbf{F} = (\mathbf{p} \cdot \nabla)\mathbf{E}$. This will give the exact same magnitude but opposite direction to the force on the point charge.

Q: Why is the electric field of the dipole at the origin in the $+x$ direction?

The origin lies on the axial line of the dipole. For a point on the axial line, the electric field is always in the direction of the dipole moment $\mathbf{p}$, regardless of whether the point is "in front" or "behind" the dipole.

Q: What if the dipole was pointing in the $-x$ direction?

If $\mathbf{p}$ was in the $-x$ direction, the potential energy would be positive, meaning it would need external work to bring it closer. The force on the point charge would be in the $-x$ direction (attractive).

Question

A positive point charge $q$ is fixed at origin. A dipole with a dipole moment $\mathbf{p}$ is placed along the $x$-axis far away from the origin with $\mathbf{p}$ pointing along positive $x$-axis. Find : (a) the kinetic energy of the dipole when it reaches a distance $d$ from the origin, and (b) the force experienced by the charge $q$ at this moment.

Sigma Percentile · Accepted Answer

The Dance of the Dipole and the Point Charge

Imagine a vast, empty space. Right at the center, the origin, sits a solitary positive point charge, $q$. It acts like a beacon, radiating an electric field in all directions. Now, far away on the $x$-axis, a tiny electric dipole with moment $\mathbf{p}$ is placed. It's pointing directly away from the origin, in the positive $x$-direction.

Because the electric field from the point charge is non-uniform—getting stronger as you get closer to the origin—the dipole feels a net force. The negative end of the dipole is slightly closer to the positive charge $q$ than the positive end is. This means the attractive force wins over the repulsive force, and the dipole begins to accelerate towards the origin!

Our goal is to find the kinetic energy of this dipole when it reaches a distance $d$ from the origin, and the exact force the point charge $q$ experiences at that very moment.

### Analyzing the Setup and Energy Conservation

Let's tackle the first part. The dipole starts its journey from "far away," which in physics translates to infinity. At infinity, the electric field from the point charge is practically zero, meaning the dipole has zero potential energy. Since it's released from rest, its initial kinetic energy is also zero.

We can use the powerful principle of conservation of mechanical energy. The total energy of the system must remain constant.

$$K_i + U_i = K_f + U_f$$

$$0 + 0 = K_f + U_f$$

This tells us that the final kinetic energy $K_f$ will simply be the negative of its final potential energy $U_f$.

### The Potential Energy of the Dipole

To find $U_f$, we need the formula for the potential energy of a dipole in an external electric field:

$$U = -\mathbf{p} \cdot \mathbf{E}$$

Here, $\mathbf{E}$ is the electric field created by our point charge $q$ at the location of the dipole (distance $d$). The electric field of a point charge is given by:

$$\mathbf{E}_q = \frac{1}{4\pi\varepsilon_0} \frac{q}{d^2} \hat{\mathbf{i}}$$

Both the dipole moment $\mathbf{p}$ and the electric field $\mathbf{E}_q$ are pointing in the positive $x$-direction ($\hat{\mathbf{i}}$). When we take their dot product, we get a positive value.

$$U_f = -(p\hat{\mathbf{i}}) \cdot \left( \frac{1}{4\pi\varepsilon_0} \frac{q}{d^2} \hat{\mathbf{i}} \right) = -\frac{pq}{4\pi\varepsilon_0 d^2}$$

Now, plugging this back into our energy conservation equation:

$$K_f = -U_f = \frac{pq}{4\pi\varepsilon_0 d^2}$$

**This is the kinetic energy of the dipole at distance $d$.**

### The Force on the Point Charge

Now for part (b). We need to find the force experienced by the point charge $q$. We could use Newton's Third Law (the force on $q$ by the dipole is equal and opposite to the force on the dipole by $q$), but let's calculate it directly using the electric field produced by the dipole.

The point charge $q$ is located at the origin, which lies on the *axial line* of the dipole. The dipole is at distance $d$, pointing away from the origin.

The formula for the electric field due to a dipole on its axial line at a distance $d$ is:

$$\mathbf{E}_p = \frac{1}{4\pi\varepsilon_0} \frac{2p}{d^3}$$

Crucially, the direction of the electric field on the axial line is *always* in the same direction as the dipole moment $\mathbf{p}$. Since $\mathbf{p}$ points in the $+\hat{\mathbf{i}}$ direction, the electric field at the origin is also in the $+\hat{\mathbf{i}}$ direction!

$$\mathbf{E}_p = \frac{1}{4\pi\varepsilon_0} \frac{2p}{d^3} \hat{\mathbf{i}}$$

Finally, the force on the charge $q$ is simply the charge multiplied by this electric field:

$$\mathbf{F} = q\mathbf{E}_p = q \left( \frac{1}{4\pi\varepsilon_0} \frac{2p}{d^3} \hat{\mathbf{i}} \right)$$

Simplifying this, we get our final answer:

**$$\mathbf{F} = \frac{pq}{2\pi\varepsilon_0 d^3} \hat{\mathbf{i}}$$**

It's a beautiful demonstration of how fields interact. The dipole accelerates towards the charge, and the charge is pulled towards the dipole. Physics in perfect harmony!

Visualized Solution

$Visualizing the Setup$

$Conservation of Energy$

$Potential Energy of Dipole$

$Electric Field due to Point Charge$

$Calculating Kinetic Energy$

$Force on the Point Charge$

$Electric Field of Dipole$

$Calculating the Force$

$Food for Thought$

The Sigma Insight: Electric Dipole

Analyzing the Setup and Energy Conservation

The Potential Energy of the Dipole

The Force on the Point Charge

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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; } Conservation of Energy

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Electric Field due to Point Charge

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Force on the Point Charge

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

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

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