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

Question

Quizplus · Accepted Answer

Using the formula for electric potential due to a point charge:

V = k*q/r

where V is the electric potential, k is the Coulomb's constant, q is the charge, and r is the distance between the point charge and the point P.

For the -2.00 μC charge, V = (-9x10^9 Nm^2/C^2)*(2x10^-6 C)/0.03 m = -600 kV

For the +4.00 μC charge, V = (9x10^9 Nm^2/C^2)*(4x10^-6 C)/0.04 m = 900 kV

For the +6.00 μC charge, V = (9x10^9 Nm^2/C^2)*(6x10^-6 C)/0.07 m = 557 kV

The total electric potential at point P is the sum of the electric potentials due to each point charge:

V_total = V1 + V2 + V3 = -600 kV + 900 kV + 557 kV = +857 kV

Therefore, the electric potential (relative to infinity) at point P due to these charges is +307 kV.

Three Point Charges,-2