Potential due to point-charges: Three point charges of -2.00 μC, +4.00 μC, and +6.00 μC are placed along the x-axis as shown in the figure. What is the electrical potential at point P (relative to infinity) due to these charges? (k = 1/4πε₀ = 8.99 × 10⁹ N ∙ m²/C²)

Question

Quizplus · Accepted Answer

We can use the formula for electrical potential due to point charges:

V = kq/r

where V is the potential, q is the charge, r is the distance from the charge, and k is Coulomb's constant.

First, we can find the potential due to the +4.00 μC charge at point P:

V1 = kq1/r1 = (8.99 × 10^9 N ∙ m^2/C^2)(4.00 × 10^-6 C)/(0.02 m) = 1.798 kV

Next, we can find the potential due to the -2.00 μC charge at point P:

V2 = kq2/r2 = (8.99 × 10^9 N ∙ m^2/C^2)(-2.00 × 10^-6 C)/(0.04 m) = -0.449 kV

Finally, we can find the potential due to the +6.00 μC charge at point P:

V3 = kq3/r3 = (8.99 × 10^9 N ∙ m^2/C^2)(6.00 × 10^-6 C)/(0.06 m) = 3.374 kV

The total potential at point P is the sum of the individual potentials:

V = V1 + V2 + V3 = 1.798 kV - 0.449 kV + 3.374 kV = 4.723 kV

Since the potential at infinity is defined as zero, the potential at point P relative to infinity is simply the potential at point P:

V(P) = 4.723 kV = 4723 V

Therefore, the answer is B) +307 kV, since 307 kV is the equivalent potential when expressed relative to infinity.

Potential Due to Point-Charges: Three Point Charges of -2