A system of three charges are placed as shown in the figure:

If $$D >> d$$, the potential energy of the system is best given by:

Problem Statement

A system of three charges is placed as shown in the figure. If $$D \gg d$$, the potential energy of the system is best given by:

• (1) $$\frac{1}{4\pi \epsilon_0} \left[ +\frac{q^2}{d} + \frac{qQd}{D^2} \right]$$
• (2) $$\frac{1}{4\pi \epsilon_0} \left[ -\frac{q^2}{d} - \frac{qQd}{D^2} \right]$$
• (3) $$\frac{1}{4\pi \epsilon_0} \left[ -\frac{q^2}{d} + \frac{2qQd}{D^2} \right]$$
• (4) $$\frac{1}{4\pi \epsilon_0} \left[ -\frac{q^2}{d} - \frac{qQd}{2D^2} \right]$$
• The distance between $$+q$$ and $$-q$$ is $$d$$. Therefore, $$+q$$ is at $$x = -\frac{d}{2}$$ and $$-q$$ is at $$x = +\frac{d}{2}$$.
• The distance $$D$$ is measured from the midpoint of the dipole to the charge $$Q$$. Therefore, the charge $$Q$$ is located at $$x = D$$.
• Between $$+q$$ and $$-q$$: $$r_{12} = d$$
• Between $$+q$$ and $$Q$$: $$r_{13} = D - \left(-\frac{d}{2}\right) = D + \frac{d}{2}$$
• Between $$-q$$ and $$Q$$: $$r_{23} = D - \frac{d}{2}$$

Step-by-Step Solution

1. Identify the Exact Positions of the Charges

Let the midpoint between the charges $$+q$$ and $$-q$$ be the origin ($$x = 0$$).

The distances between the pairs of interacting charges are:

2. Set Up the Total Electrostatic Potential Energy Equation

The total potential energy $$U$$ of a three-charge system is the sum of the potential energies of all unique pairs:

$$U = \frac{1}{4\pi\epsilon_0} \left[ \frac{(+q)(-q)}{r_{12}} + \frac{(+q)(Q)}{r_{13}} + \frac{(-q)(Q)}{r_{23}} \right]$$

Substitute the distances into the equation:

$$U = \frac{1}{4\pi\epsilon_0} \left[ -\frac{q^2}{d} + \frac{qQ}{D + \frac{d}{2}} - \frac{qQ}{D - \frac{d}{2}} \right]$$

3. Simplify using Binomial Approximation

Take out $$D$$ from the denominators of the second and third terms to utilize the condition $$D \gg d$$:

$$U = \frac{1}{4\pi\epsilon_0} \left[ -\frac{q^2}{d} + \frac{qQ}{D}\left(1 + \frac{d}{2D}\right)^{-1} - \frac{qQ}{D}\left(1 - \frac{d}{2D}\right)^{-1} \right]$$

Since $$\frac{d}{2D} \ll 1$$, we apply the binomial expansion $$(1 \pm x)^{-1} \approx 1 \mp x$$:

$$U \approx \frac{1}{4\pi\epsilon_0} \left[ -\frac{q^2}{d} + \frac{qQ}{D}\left(1 - \frac{d}{2D}\right) - \frac{qQ}{D}\left(1 + \frac{d}{2D}\right) \right]$$

Expand the terms inside the bracket:

$$U \approx \frac{1}{4\pi\epsilon_0} \left[ -\frac{q^2}{d} + \frac{qQ}{D} - \frac{qQd}{2D^2} - \frac{qQ}{D} - \frac{qQd}{2D^2} \right]$$

The $$\frac{qQ}{D}$$ and $$-\frac{qQ}{D}$$ terms cancel each other out:

$$U \approx \frac{1}{4\pi\epsilon_0} \left[ -\frac{q^2}{d} - \frac{2qQd}{2D^2} \right]$$

$$U \approx \frac{1}{4\pi\epsilon_0} \left[ -\frac{q^2}{d} - \frac{qQd}{D^2} \right]$$

Correct Answer

The correct option is (1).