Question

Three charges each of magnitude \(3\mu C\), are placed on the vertices of an equilateral triangle of side 6 cm . The net potential energy of the system will be nearly \(\left[ \frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \text{ SI unit} \right]\)}

• A. 1.4 J
• B. 2.7 J
• C. 4.1 J
• D. 8.2 J

Hint:

For $n$ identical charges at distance $r$, total energy is $^nC_2 \frac{kq^2}{r}$.

Answer 1

The Correct Option is C

Step 1: Formula
Potential energy of 3 charges: $U = \frac{1}{4\pi\epsilon_0} \left( \frac{q_1q_2}{r} + \frac{q_2q_3}{r} + \frac{q_1q_3}{r} \right)$.
Since all $q$ and $r$ are equal, $U = 3 \times \frac{kq^2}{r}$.
Step 2: Substitution
$q = 3 \times 10^{-6} \text{ C}$, $r = 0.06 \text{ m}$, $k = 9 \times 10^9$.
$U = 3 \times \frac{9 \times 10^9 \times (3 \times 10^{-6})^2}{0.06} = 3 \times \frac{9 \times 10^9 \times 9 \times 10^{-12}}{0.06}$.
Step 3: Calculation
$U = \frac{243 \times 10^{-3}}{0.06} = \frac{0.243}{0.06} \approx 4.05 \text{ J}$.
Final Answer: (C)