Question

Find the direction and magnitude of the net electrostatic force exerted on the point charge q₂ in the figure below. Let q = +2.7 µC and d = 33cm.

Direction

° (from the x-axis, which points to the right

Magnitude

_______  N

Answer 1

q= 2.7*10^-6 C

q1 = q

q2 = -2q

q3 = -3q

q4 = -4q

d = 33 cm = 0.33 m

Force on q2 due to q1, F1 = [Kq1q2/d^2](-j)

=[2Kq^2/d^2](-j)

= - 1.204 j N

Force on q2 due to q3, F3 = [Kq2q3/d^2](i)

=[6Kq^2/d^2](i)

= 3.614 i N

Force on q2 due to q4, F4 = [Kq2q4/(√2 d)^2][cos45 i - sin45 j]

=[8Kq^2/(√2 d)^2][0.707 i - 0.707 j]

= 2.408*0.707[i - j]N

=1.702[ i - j] N

Net force on q2, F =F1 + F3 + F4

= -1.204 j + 3.614 i + 1.702 i - 1.702 j

= 5.316 i - 2.906 j

Magnitude : |F| = 6.058 N

Direction θ = Tan^-1[2.906/5.316]

= 28.66 degrees below the X axis.