Question

Three equal \((+q)\) charges are placed in air at the vertices of equilateral triangle. If the side of equilateral triangle is \(l\), the net potential energy of the system is \(\left[K=\frac{1}{4\pi \epsilon _{0}}\right]\)

• A. \(\frac{kq^{2}}{l}\)
• B. \(\frac{2kq^{2}}{l}\)
• C. \(\frac{3kq^{2}}{l}\)
• D. \(\frac{4kq^{2}}{l}\)

Hint:

Add the potential energy of every pair of charges.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Potential energy of a system is the sum over all pairs: \(U=\sum\dfrac{kq_{i}q_{j}}{r_{ij}}\).

Step 2: Pairs
Three charges make three pairs, each separated by \(l\).

Step 3: Calculation
\(U=3\times\dfrac{kq^{2}}{l}=\dfrac{3kq^{2}}{l}\).

Final Answer:
The potential energy of the system is 3kq squared over l. \[ \boxed{\text{(C) }\dfrac{3kq^{2}}{l}} \]