Question

Consider two point charges located on the x axis: one charge, q1= -19.0 nC , is located at x1 = -1.700 m; the second charge, q2 q= 40.0 nC , is at the origin (x = 0).

What is (F_net3)x the x-component of the net force exerted by these two charges on a third charge q3 = 54.5 nC placed between q₁ and q₂ at x₃= -1.160 m ?

Your answer may be positive or negative, depending on the direction of the force.

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.

Force F1, on q3 due to q1 will be attractive and towards the -ve x-axis

Force F2, on q3 due to q2 will be repulsive and towards the -ve x-axis

So net force on q3 will be

Fnet = -F1 - F2 = -(F1 + F2)

Fnet = -(k*q1*q3/d1^2 + k*q2*q3/d2^2)

d1 = distance between q1 and q3 = -1.160 - (-1.700) = 0.540 m

d2 = distance between q2 and q3 = 0 - (-1.160) = 1.160 m

q1 = -19.0 nC = -19.0*10^-9 C

q2 = 40.0 nC = 40.0*10^-9 C

q3 = 54.5 nC = 54.5*10^-9 C

k = 1/(4*pi*e0) = 1/(4*pi*8.854*10^-12) = 8.99*10^9

So Using these values:

Fnet = -k*q3*[q1/d1^2 + q2/d2^2]

Fnet = -8.99*10^9*54.5*10^-9*[19.0*10^-9/0.540^2 + 40.0*10^-9/1.160^2]

Fnet = -4.65*10^-5 N

(-ve sign means net force will be towards left or -ve x-axis)

Let me know if you've any query.