Question

GOAL Use the superposition principle to calculate the electric field due to two point charges. Consider the following figure.

The resultant electric field  at P equals the vector sum ₁ + ₂, where ₁ is the field due to the positive charge q₁and ₂ is the field due to the negative charge q₂.Two point charges lie along the x-axis in the x y-coordinate plane. Positive charge q₁ is at the origin, and negative charge q₂ is at (0.300 m, 0). Point P is at (0, 0.400 m) on the +y-axis. A dashed line from q₂ to P is 0.500 mlong and makes an acute angle θwith the x-axis. Three electric field vectors start at point P.

• Vector E₁ points straight up in the +y-direction.
• Vector E₂ points down and right, at an acute angle θ below the horizontal, along the line from Pto q₂.
• Resultant vector E points up and right, at an acute angle ϕ above the horizontal.

PRACTICE IT

Use the worked example above to help you solve this problem. Charge q₁ = 6.60 µC is at the origin, and charge q₂ = -4.90 µC is on the x-axis, 0.300 m from the origin (see figure).

(a) Find the magnitude and direction of the electric field at point P, which has coordinates (0, 0.400) m.

magnitude
direction

(b) Find the force on a charge of 2.00  10^-8 C placed at P.
__________ N

EXERCISE

(a) Place a charge of -4.10 µC at point P and find the magnitude and direction of the electric field at the location of q₂ due to q₁ = 6.60 µC and the charge at P.

magnitude	N/C
direction	° counterclockwise from the +x-axis

(b) Find the magnitude and direction of the force on q₂.

magnitude	N
direction	° counterclockwise from the +x-axis

Answer 1