Two 3.0 μC charges lie on the x-axis,one at the origin and the other at $2.0 \mathrm {~m}$ What is the potential (relative to infinity)due to these charges at a point at $6.0 \mathrm {~m}$ on the x-axis? (k = 1/4πε₀ = 9.0 × 10⁹ N ∙ m²/C²)

Question

Quizplus · Accepted Answer

The potential due to a point charge is given by $V = \frac{kq}{r}$, where $V$ is the potential, $k$ is the Coulomb's constant ($9.0 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2$), $q$ is the charge, and $r$ is the distance from the charge to the point of interest. For two charges, the total potential is the sum of the potentials due to each charge. For the charge at the origin, $r = 6.0 \, \text{m}$, and for the charge at $2.0 \, \text{m}$, $r = 6.0 - 2.0 = 4.0 \, \text{m}$. Thus, the total potential at $6.0 \, \text{m}$ is $V = \frac{(9.0 \times 10^9)(3.0 \times 10^{-6})}{6.0} + \frac{(9.0 \times 10^9)(3.0 \times 10^{-6})}{4.0} = 4500 + 6750 = 11,250 \, \text{V}$, which rounds to $11,000 \, \text{V}$.

Two 30 μC Charges Lie on the X-Axis,one at the Origin