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: Understanding the Concept:
The total electrostatic potential energy of a system of charges is the sum of the potential energies of all possible pairs of charges. For three identical charges at the vertices of an equilateral triangle, there are three identical pairs.
Step 2: Key Formula or Approach:
The potential energy between two charges \(q_1\) and \(q_2\) separated by distance \(r\) is:
\[ U = \frac{1}{4\pi\epsilon_0} \frac{q_1q_2}{r} \]
For three identical charges \(q\) on a triangle of side \(a\):
\[ U_{net} = 3 \times \left( \frac{1}{4\pi\epsilon_0} \frac{q^2}{a} \right) \]
Step 3: Detailed Explanation:
Given values:
\(q = 3 \mu C = 3 \times 10^{-6} \text{ C}\)
\(a = 6 \text{ cm} = 0.06 \text{ m}\)
\(k = \frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \text{ Nm}^2/\text{C}^2\)
Calculating the energy:
\[ U_{net} = 3 \times \frac{9 \times 10^9 \times (3 \times 10^{-6})^2}{0.06} \]
\[ U_{net} = 3 \times \frac{9 \times 10^9 \times 9 \times 10^{-12}}{0.06} \]
\[ U_{net} = \frac{243 \times 10^{-3}}{0.06} \]
\[ U_{net} = \frac{0.243}{0.06} = 4.05 \text{ J} \]
Rounding to the nearest option, we get approximately 4.1 J.
Step 4: Final Answer:
The net potential energy of the system is nearly 4.1 J.