15) Three charges, each of magnitude \(3\,\text{C}\), are placed at the vertices of an equilateral triangle of side \(4.5\,\text{cm}\) in air. The net potential energy of the system will be:
Solution:
The potential energy between two point charges is given by: \[ U = \frac{k \, q_1 q_2}{r}, \] where \(k \approx 9 \times 10^9\, \text{N}\cdot\text{m}^2/\text{C}^2\), \(q_1\) and \(q_2\) are the charges, and \(r\) is the separation between them.
In an equilateral triangle, all sides are equal. Here, each side is \(4.5\,\text{cm} = 0.045\,\text{m}\) and each charge is \(3\,\text{C}\).
The potential energy for one pair of charges is: \[ U_{\text{pair}} = \frac{k \times (3)(3)}{0.045} = \frac{9k}{0.045}. \]
Substituting \(k = 9 \times 10^9\, \text{N}\cdot\text{m}^2/\text{C}^2\): \[ U_{\text{pair}} = \frac{9 \times 9 \times 10^9}{0.045} = \frac{81 \times 10^9}{0.045}. \]
Calculating the value: \[ \frac{81 \times 10^9}{0.045} = 81 \times \frac{10^9}{0.045} = 81 \times (22.22 \times 10^9) \approx 1.8 \times 10^{12}\, \text{J}. \]
Since there are three pairs of charges in the triangle, the total potential energy is: \[ U_{\text{total}} = 3 \times U_{\text{pair}} = 3 \times 1.8 \times 10^{12}\, \text{J} = 5.4 \times 10^{12}\, \text{J}. \]
Thus, the net potential energy of the system is \(5.4 \times 10^{12}\, \text{J}\), which corresponds to option D.
.png)
0 Comments