Question

You arrange two identical charges along the x-axis at positions (-3.0,0) and (3.0,0), with distances measured in centimeters. You've set things up so that the origin of your coordinates lies at the midpoint between the charges. You measure the electric force acting on the charge on the right and find it has a magnitude of 24574.3 N. You want to deduce the electric field vector that is acting on the right-hand charge.

Answer 1

Position of left charge q = (-3.0,0)

Position of right charge q = (3.0, 0)

Distance between charges r = 6 cm = 0.06 m

Electric force F acting on right charge = F = 24574.3 N

It must be acting towards right direction as charges are of same nature and they will repel each other

F = k q^2/r^2

24574.3 = (9 x 10^9)q^2/(0.06)^2

q = 9.9 x 10^-5 C

Electric field vector acting on right hand charge = E = F/q

E = 24574.3/(9.9 x 10^-5) = 2.48 x 10^8 N/C (towards right direction) [ Electric field has the same direction as electric force]