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 \(a\).
Solution and Explanation
Given
Charges at the vertices of an equilateral triangle of side \(a\):
- At A: \(\;q\)
- At B: \(\;-4q\)
- At C: \(\;2q\)
Step 1: Electrostatic Potential Energy of the System
The total electrostatic potential energy of the system is the sum of the potential energies of all pairs of charges.
\[ U=\frac{1}{4\pi\varepsilon_0} \left( \frac{q_1q_2}{a} +\frac{q_2q_3}{a} +\frac{q_3q_1}{a} \right) \]
Step 2: Calculate the Potential Energy of Each Pair
Between A and B:
\[ U_{AB} =\frac{1}{4\pi\varepsilon_0} \left(\frac{q(-4q)}{a}\right) = -\frac{4q^2}{4\pi\varepsilon_0a} \]
Between B and C:
\[ U_{BC} =\frac{1}{4\pi\varepsilon_0} \left(\frac{(-4q)(2q)}{a}\right) = -\frac{8q^2}{4\pi\varepsilon_0a} \]
Between C and A:
\[ U_{CA} =\frac{1}{4\pi\varepsilon_0} \left(\frac{(2q)(q)}{a}\right) = \frac{2q^2}{4\pi\varepsilon_0a} \]
Step 3: Total Potential Energy
Adding the three contributions,
\[ U = \frac{1}{4\pi\varepsilon_0a} \left(-4q^2-8q^2+2q^2\right) \]
\[ U = -\frac{10q^2}{4\pi\varepsilon_0a} \]
Step 4: Work Done to Dissociate the System
To dissociate the system, all the charges are taken to infinity where the potential energy becomes zero.
Hence,
\[ W=U_{\infty}-U_{\text{initial}} \]
\[ W=0-\left(-\frac{10q^2}{4\pi\varepsilon_0a}\right) \]
\[ \boxed{ W=\frac{10q^2}{4\pi\varepsilon_0a} } \]
Final Answer
The work required to dissociate the three-charge system is
\[ \boxed{ W=\frac{10q^2}{4\pi\varepsilon_0a} } \]
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