How much energy is necessary to place three +2.0-µC point charges at the vertices of an equilateral triangle of side 2.0 cm if they started out extremely far away? (k = 1/4πε₀ = 9.0 × 10⁹ N ∙ m²/C²)

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Quizplus · Accepted Answer

The energy required to assemble a system of point charges is given by the potential energy formula $U = k \frac{q_1 q_2}{r}$, where $k$ is Coulomb's constant ($9.0 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2$), $q_1$ and $q_2$ are the charges, and $r$ is the distance between them. For three charges at the vertices of an equilateral triangle, the total energy is the sum of the energies for each pair of charges. Since all charges and distances are the same, this simplifies to $3 \times k \frac{q^2}{r}$. Substituting $q = 2.0 \times 10^{-6} \, \text{C}$, $r = 0.02 \, \text{m}$, and $k = 9.0 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2$, we get $U = 3 \times 9.0 \times 10^9 \times \frac{(2.0 \times 10^{-6})^2}{0.02} = 5.4 \, \text{J}$.

How Much Energy Is Necessary to Place Three +2