Question

Obtain an expression for the work done to dissociate the system of three charges $q$, $-4q$ and $2q$ placed at the vertices A, B and C respectively of an equilateral triangle of side '$a$'.

Hint:

Always remember that "work done to assemble" equals $+U$, while "work done to dissociate" equals $-U$.
A bound system always has a negative potential energy, meaning positive work must be aggressively injected to break it apart completely.

Answer 1

Concept:
• The total electrostatic potential energy ($U$) of a discrete system of charges is precisely the total amount of external work that was originally done to assemble them from infinity to their current configuration.
• To completely "dissociate" the system, an external agent must do work exactly equal to the negative of this initial potential energy to tear them completely apart and push them back out to infinity.
• The potential energy for a continuous system of multiple charges is the algebraic sum of the potential energies of all unique pairs of charges.

Step 1: Formulate the total potential energy of the system
For a carefully defined system consisting of three point charges $q_1$, $q_2$, and $q_3$ separated by pairwise distances $r_{12}$, $r_{23}$, and $r_{13}$, the formula is:
\[ U = \frac{1}{4\pi\epsilon_0} \left[ \frac{q_1 q_2}{r_{12}} + \frac{q_2 q_3}{r_{23}} + \frac{q_1 q_3}{r_{13}} \right] \]

Step 2: Substitute the provided parameters
From the problem description, the charges are situated on the vertices of an equilateral triangle.
This guarantees that all pairwise separation distances are perfectly equal: $r_{12} = r_{23} = r_{13} = a$.
The specific charges are defined as: $q_1 = q$, $q_2 = -4q$, and $q_3 = 2q$.
Inserting these into the potential energy equation:
\[ U = \frac{1}{4\pi\epsilon_0} \left[ \frac{(q)(-4q)}{a} + \frac{(-4q)(2q)}{a} + \frac{(q)(2q)}{a} \right] \]

Step 3: Simplify the mathematical expression
Multiply the charges carefully keeping track of the negative signs:
\[ U = \frac{1}{4\pi\epsilon_0 a} \left[ -4q^2 - 8q^2 + 2q^2 \right] \]
Combine the algebraic terms strictly inside the bracket:
\[ U = \frac{1}{4\pi\epsilon_0 a} \left[ -10q^2 \right] \]
\[ U = -\frac{10q^2}{4\pi\epsilon_0 a} \]
This negative energy definitively indicates that the system is in an attractive, bound state.

Step 4: Calculate the final work done to dissociate
The total external work required to dissociate the system is completely given by the difference between the final potential energy (at infinity) and the initial potential energy.
Since the potential energy at infinite separation is exactly zero ($U_f = 0$):
\[ W_{dissociate} = U_f - U_i \]
\[ W_{dissociate} = 0 - \left( -\frac{10q^2}{4\pi\epsilon_0 a} \right) \]
\[ W_{dissociate} = \frac{10q^2}{4\pi\epsilon_0 a} \]