Answer:
3.456 × 10⁻⁴³ J
Explanation:
The work done in bringing two charges q₁ and q₂ separated by a distance, r together is W = kq₁q₂/r. With the first charge q₁ in place, the work done in bringing the second charge into the vertex of the equilateral at a distance, r is W₁₂ = kq₁q₂/r. Since the first and second charges are in place, the work done in bringing the third charge q₃ in place is work done W₁₃ between q₁ and q₃ = kq₁q₃/r plus work done W₂₃ between q₂ and q₃ = kq₂q₃/r. So the total work done in assembling the atomic nucleus is W = W₁₂ + W₁₃ + W₂₃ = kq₁q₂/r + kq₁q₃/r + kq₂q₃/r. Since q₁ = q₂ = q₃ = q = 1.6 × 10⁻¹⁹ C, W = 3kq²/r. Since r = side of equilateral triangle = 2.0 × 10⁻¹⁵ m. W = 3 × 9 × 10⁹ × (1.6 × 10⁻¹⁹)²/2.0 × 10⁻¹⁵ = 3.456 × 10⁻⁴³ J ≅ 3.46 × 10⁻⁴³ J
