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

Question

Quizplus · Accepted Answer

Using the formula for potential due to point charges: V = kq/r
1. Potential due to -2.00 μC charge at P:
V1 = k*(-2.00 μC)/(-0.15 m) = 359.64 kV (Note that we take distance as negative because the point charge is to the left of P)
2. Potential due to +4.00 μC at P:
V2 = k*(4.00 μC)/(0.05 m) = 719.28 kV
3. Potential due to +6.00 μC at P:
V3 = k*(6.00 μC)/(0.25 m) = 215.78 kV
Using the principle of superposition, the total potential at P is given by:
V_total = V1 + V2 + V3 = 783.70 kV
(Note: since the point charges are all positive, the total potential at P will be positive)
Therefore, the answer is B) +307 kV.

Three Point Charges,-2