A 7.0-μC point charge and a $9.0 - \mu \mathrm { C }$ point charge are initially extremely far apart.How much work does it take to bring the $7.0 - \mu \mathrm { C }$ point charge to the point $x = 3.0 \mathrm {~mm}$ $y = 0.0 \mathrm {~mm}$ ,and the $9.0 - \mu \mathrm { C }$ point charge to the point $x = - 3.0 \mathrm {~mm}$ $y = 0.0 \mathrm {~mm} ?$ (k = 1/4πε₀ = 9.0 × 10⁹ N ∙ m²/C²)

Question

Quizplus · Accepted Answer

The work done is calculated using the formula $ W = k \cdot \frac{q_1 \cdot q_2}{r} $, where $ r $ is the distance between the charges. Here, $ r = 6.0 \, \text{mm} = 0.006 \, \text{m} $. Substituting the given values, $ W = 9.0 \times 10^9 \cdot \frac{7.0 \times 10^{-6} \cdot 9.0 \times 10^{-6}}{0.006} = 94.5 \, \text{J} \approx 95 \, \text{J} $.

A 70-μC Point Charge and A A 9.0 9.0−μC9.0 - \mu \mathrm { C }9.0−μC

A 70-μC Point Charge and A A 9.0 $9.0 - \mu \mathrm { C }$