Question

3 charges, A = 1.0 µC, B = 2.0 µC, and C = 6.0 µC, are located on three vertices A, B, C of an equilateral triangle with sides 10 cm each.

(a)Calculate the net force on A due to B and C

(b)Calculate the net force on B due to A and C

(c)Calculate the net force on C due to A and B

(d)Another charge q is located at the mid point of the side BC. Calculate q so that net force on the charge at A due to the charges at B, C and q is zero.

Answer 1

force = kq1q2/r^2

a)

force on A due to B

Fab = 9*10^9*1*10^(-6)*2*10^(-6)/(0.1)^2 = 1.8 N

Fac = 9*10^9*1*10^(-6)*6*10^(-6)/(0.1)^2 = 5.4 N

angle between Fab & Fac = 60 degree

Fa = Fab^2 + Fac^2 + 2*Fab*Fac*cos(120) = 1.8^2 + 5.4^2 + 2*0.5*1.8*5.4 = 42.12 N

b)

Fba = 9*10^9*1*10^(-6)*2*10^(-6)/(0.1)^2 = 1.8 N

Fbc = 9*10^9*2*10^(-6)*6*10^(-6)/(0.1)^2 = 10.8 N

Fb = 1.8^2 + 10.8^2 + 2*0.5*1.8*10.8 = 139.32 N

c)

Fca = 9*10^9*1*10^(-6)*6*10^(-6)/(0.1)^2 = 5.4 N

Fcb = 9*10^9*2*10^(-6)*6*10^(-6)/(0.1)^2 = 10.8 N

Fc = 5.4^2 + 10.8^2 + 2*0.5*5.4*10.8 = 204.12 N

d)
net force at A is zero thus q should be negative

net force at A due to B & C = force at A due to q

42.12 = 9*10^9*1*10^(-6)*q/(0.1)^2

q = (42.12/9)*10^(-5) = 46.8 micro C

q = - 46.8 micro C