Question:
Three point charges \(+q,-2q\) and \(-2q\) are placed at the vertices of an equilateral triangle of side a. The work done by some external force to increase their separation to '\(2a\)' will be \(\left(K=\frac{1}{4\pi \varepsilon _{0}}\right)\)
Three point charges \(+q,-2q\) and \(-2q\) are placed at the vertices of an equilateral triangle of side a. The work done by some external force to increase their separation to '\(2a\)' will be \(\left(K=\frac{1}{4\pi \varepsilon _{0}}\right)\)
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Compute the total potential energy of the system at separation a and at 2a.
Updated On: Nov 12, 2026
- \(\frac{2q^{2}K}{a}\)
- \(\frac{q^{2}K}{2a}\)
- \(\frac{qK}{a^{2}}\)
- zero
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept
Work done by the external force equals the change in electric potential energy of the system, \(W=U_f-U_i\).
Step 2: Initial Energy
\(U_i=K\left[\frac{(q)(-2q)}{a}+\frac{(q)(-2q)}{a}+\frac{(-2q)(-2q)}{a}\right]=\frac{Kq^{2}}{a}(-2-2+4)=0\).
Step 3: Final Energy
The pairwise sum has the same charge products with every distance \(2a\): \(U_f=\frac{Kq^{2}}{2a}(-2-2+4)=0\).
Step 4: Work Done
\(W=0-0=0\), option (D).
Final Answer:
The work done by the external force is zero. \[ \boxed{\text{(D) }\text{zero}} \]
Work done by the external force equals the change in electric potential energy of the system, \(W=U_f-U_i\).
Step 2: Initial Energy
\(U_i=K\left[\frac{(q)(-2q)}{a}+\frac{(q)(-2q)}{a}+\frac{(-2q)(-2q)}{a}\right]=\frac{Kq^{2}}{a}(-2-2+4)=0\).
Step 3: Final Energy
The pairwise sum has the same charge products with every distance \(2a\): \(U_f=\frac{Kq^{2}}{2a}(-2-2+4)=0\).
Step 4: Work Done
\(W=0-0=0\), option (D).
Final Answer:
The work done by the external force is zero. \[ \boxed{\text{(D) }\text{zero}} \]
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