Animated Solution for Physics - Electrostatics: A system of three charges are placed as shown in the figure If $D\gg d$, the potential energy of the system is best given by

Welcome to a beautiful problem in electrostatics that perfectly blends the principle of superposition with mathematical approximation. We are presented with a one-dimensional system consisting of three point charges: $+q$, $-q$, and $Q$, arranged sequentially on a straight line.

The geometry is clearly defined: the distance between the first two charges ($+q$ and $-q$) is $d$, and the distance from the first charge ($+q$) to the distant charge ($Q$) is $D$. Our objective is to find the total electrostatic potential energy of this entire system, given the crucial constraint that $D\gg d$.

The total potential energy $U$ of a system of discrete point charges is the total work required to assemble the system by bringing the charges in from infinity one by one. Mathematically, this is the sum of the potential energies of all unique pairs of charges in the system:

For our three-charge system, there are exactly three unique pairs to consider: 1. The pair $(+q,-q)$ separated by distance $d$. 2. The pair $(+q,Q)$ separated by distance $D$. 3. The pair $(-q,Q)$. To find their separation, we look at the geometry: the total distance from $+q$ to $Q$ is $D$, and $-q$ is at a distance $d$ from $+q$. Therefore, the distance between $-q$ and $Q$ is simply $D-d$.

Let's write down the potential energy contribution for each of these three pairs:

Simplifying the signs, we get:

Notice that the first term is entirely independent of $Q$ and $D$. It represents the "self-energy" of the $+q$ and $-q$ pair. The next two terms represent the interaction of this pair with the distant charge $Q$. Let's factor out the common terms to make the algebra cleaner:

Now, we focus on the expression inside the parentheses. Let's combine the fractions by finding a common denominator:

Substituting this back into our master equation yields:

This is the exact potential energy of the system. However, the problem provides a powerful constraint: $D\gg d$. This means that $d$ is negligibly small compared to $D$.

A Word of Caution: A common mistake is to set $d=0$ everywhere. If you do that in the numerator, the entire interaction term vanishes! We only apply the approximation to additive or subtractive terms where the small quantity is overwhelmed by a much larger one.

In the denominator, we have the term $(D-d)$. Since $D$ is massive compared to $d$, we can safely approximate $(D-d)\approx D$.

Applying this approximation to the denominator, we get:

Our potential energy equation now beautifully simplifies to:

Finally, replacing the Coulomb constant $k$ with its standard form $\frac{1}{4\pi {ϵ}_0}$, we arrive at the final expression:

This perfectly matches option (d).

There is a deeply elegant, alternative way to view this problem. Notice that the pair $+q$ and $-q$ separated by a small distance $d$ forms an electric dipole. The dipole moment is $p=qd$, pointing from $-q$ to $+q$ (which is the $-x$ direction in our setup).

The total energy of the system can be viewed as the internal self-energy of the dipole plus the interaction energy of the dipole with the external electric field created by $Q$.

1. Self-Energy: The energy required to assemble the dipole itself is simply $-\frac{k{q}^2}{d}$. 2. Interaction Energy: The energy of a dipole in an external electric field is given by $U_{\text{int}}=-\vec{p}\cdot \vec{E}$. The electric field ${\vec{E}}_Q$ produced by $Q$ at the location of the dipole points to the left ($-x$ direction) with a magnitude of $\frac{kQ}{D^2}$. Since both $\vec{p}$ and ${\vec{E}}_Q$ point in the same direction, their dot product is positive, leaving the interaction energy negative:

Adding these two components together yields the exact same result instantly! Mastering multiple perspectives like this is what transforms a good physics student into a great one.