Question

Electric charges of - 10 nC, + 10 nC, and - 20 nC are located in the x-y plane at positions of (0, 0.5), (0, -0.5), and (0.5, 0), respectively, where the positions are expressed in m. What is the magnitude of the net force exerted on the - 20 nC charge?

Answer 1

electrostatic force is given by,

F = kq1*q2/d^2

Force will be attractive if both charge have different signs, and force will be repulsive if both charge have same signs. Now

Net force will be

qa = -10 nC, qb = +10 nC & qc = -20 nC

rac = rbc = L = sqrt(0.5^2 + 0.5^2) = 1/sqrt(2)

So,

Fnet = Fac + Fbc

Fac = Force between qa and qc, which will be repulsive at angle below +xaxis

Fbc = Force between qb and qc, which will be attractive at angle below -xaxis

here, angle = arctan(0.5/0.5) = 45 deg

Fac = k*qa*qc/L^2

Fac = (9*10^9)*(10*10^-9)*(20*10^-9)/(1/2)

Fac = 3.6*10^-6 N

Facx = Fac*cos 45 deg = 3.6*10^-6*cos 45 deg = 2.55*10^-6 N

Facy = Fac*sin 45 deg = -3.6*10^-6*sin 45 deg = -2.55*10^-6 N

Now

Fbc = k*qa*qc/L^2

Fbc = (9*10^9)*(10*10^-9)*(20*10^-9)/(1/2)

Fbc = 3.6*10^-6 N

Fbcx = Fbc*cos 45 deg = 3.6*10^-6*cos 45 deg = -2.55*10^-6 N

Fbcy = -Fbc*sin 45 deg = -3.6*10^-6*cos 45 deg = -2.55*10^-6 N

Now Net force will be

Fnet = Fac + Fbc

Fnet = (Facx + Fbcx) i + (Facy + Fbcy) j

F = (2.55*10^-6 - 2.55*10^-6) i + (-2.55*10^-6 - 2.55*10^-6) j

F = (0 i - 5.1*10^-6 j) N

Magnitude of Force will be

|F| = sqrt (0^2 + (-5.1*10^-6)^2)

|F| = 5.1*10^-6 N

|F| = 5.1 N

direction = arctan (0/(-5.1*10^-6)) = towards negative y axis

Comment below if you have any doubt.