Question

Three equal charges, $2.0 \times 10^{-6}$ each, are held fixed at the three corners of an equilateral triangle of side 5 cm. Find the Coulomb force experienced by one of the charges due to the rest two.

• A. 15 N
• B. 24.9 N
• C. 34.6 N
• D. 60 N

Answer 1

Answer: (B)

24.9 N

Answer 2

The Coulomb force experienced by one of the charges is calculated as follows:

1. Force Between Two Charges

   Using Coulomb’s law:

   $$F = k\frac{q^2}{r^2}$$

   where $q = 2.0 \times 10^{-6}\,\text{C}$, $r = 0.05\,\text{m}$, and $k \approx 9.0 \times 10^9\,\text{N} \cdot \text{m}^2/\text{C}^2$.

   $$F = 9.0 \times 10^9 \times \frac{(2.0 \times 10^{-6})^2}{(0.05)^2} = 9.0 \times 10^9 \times \frac{4.0 \times 10^{-12}}{2.5 \times 10^{-3}} = 14.4\,\text{N}$$

2. Net Force Calculation

   At one vertex, two forces of 14.4 N act at an angle of $60^\circ$ (internal angle of an equilateral triangle).

   The resultant force magnitude, using vector addition:

   $$F_{\text{net}} = 2F \cos\left(\frac{60^\circ}{2}\right) = 2 \times 14.4 \times \cos(30^\circ)$$ $$\cos(30^\circ) \approx 0.866 \quad \Longrightarrow \quad F_{\text{net}} \approx 28.8 \times 0.866 = 24.9\,\text{N}$$

Therefore, the Coulomb force experienced by one of the charges is approximately 24.9 N.