Find the position on the x-axis where the net force on a small charge +q would be zero.?
Two charges, one of 2.50uC and the other of -3.50uC, are placed on the x-axis, one at the origin and the other atx = 0.600m.
First, note that test charge +q cannot lie in the interval 0 m < x < 0.600 m (i.e., between the other two charges), for force can’t be zero anywhere in this interval. A positive test charge will be attracted to the negative charge, and repelled by the positive charge, regardless where the test charge is located within this interval. However, you could expect force to cancel outside this range, especially somewhere near the positive charge, since its charge is smaller than the negative charge. If the positive charge is located at the origin, then force will be null somewhere at the left of the origin, in the negative portion of the x-axis, rather near the origin.
Although you can solve this problem using Coulomb’s law alone, fi you’re familiar to the notion of electric field you’ll find the solution somewhat easier. Force on a charge, anywhere in space, is determined by the product qE, where E is the electric field at that specific point in space. If the field is zero at some point, force on a charge at this location will be zero as well. Assume that one such points lies on the x-axis, and is located x units at the left of the coordinate origin.
The field, due to the nearest (positive) charge is:
E1 = − K Q1 / x² = K × 2.50E-6 / x².
By convention, the field is negative, as the electric field, a vector, points to the negative-going direction of the x-axis. This is so because the charge at the origin is positive, and in this case field lines emanate from the origin in all directions. Besides, since the test charge is positive (it is always assumed to be so), the positive charge at the origin will repel the test charge, exerting a force pointing to the left.
Consider now the other charge. Field E2, set up by this charge, will be
E2 = K Q2 / r² = K × 3.50E-6 / (0.600 + x)²
By the same token, this field is considered positive, since the force it brings about on the test charge points to the right. We expect total field strenght to be zero, so
E = E1 + E2 = K ( 3.50 / (0.600 + x)² − 2.50 / x² ) · E-6 = 0.
Solving for x,
3.5 x² = 2.5 (0.6² + 1.2 x + x²)
x² − 3 x − 0.900 = 0
This quadratic has two roots; as often happens with this type of problems, only one root makes sense. The other root, by the way, gives a valid solution to a similar problem, where charges retain the same magnitude, but both have the same sign (both positive, or both negative). Strange as it may seem, this solution actually lies between both charges. So much for the extraneous solution.
The relevant root is x = ½ ( 3 + √ (3² + 4 × 0.9)) = 3.274824 m. The root has come out as positive, but it’s just a distance, and sign has no meaning here. In accordance with the above reasoning , the point sought is at x = −3.274824.
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