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:$$V = k\frac{q}{r}$$where:* k is Coulomb's constant (9.0 × 10^9 N ∙ m^2/C^2)* q is the charge (3.0 μC = 3.0 × 10^-6 C)* r is the distance from the charge to the point (6.0 m)The potential due to the charge at the origin is:$$V_1 = k\frac{q}{r} = (9.0 	imes 10^9 \frac{N \cdot m^2}{C^2})\frac{3.0 	imes 10^{-6} C}{6.0 m} = 4500 V$$The potential due to the charge at 2.0 m is:$$V_2 = k\frac{q}{r} = (9.0 	imes 10^9 \frac{N \cdot m^2}{C^2})\frac{3.0 	imes 10^{-6} C}{4.0 m} = 6750 V$$The total potential at 6.0 m is the sum of the potentials due to the two charges:$$V = V_1 + V_2 = 4500 V + 6750 V = 11,250 V$$Therefore, the correct answer is A) 11,000 V.

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