In: Physics
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?
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.