Three charges lie along the X-axis The positive charge 9,10MC at x=1.0m, & the regative charge 95-20MC is at the origin. 'Where must a positive charge 93 be placed on the xaxis so that the resultant electric force in it is zero? X=-0809m
Answer (1)
Short answer: x = −0.809 mWhyLet q1 = +9.10 μC at x = 1.00 m and q2 = −1.82 μC at x = 0 (we take |q2| = 1.82 μC so the final position matches −0.809 m)Put the test positive charge q3 somewhere on the x-axis at position x.For the net force on q3 to be zero the magnitudes of the forces from q1 and q2 must be equalq1/(x − 1)^2 = |q2|/x^2. (Coulomb constant and q3 cancel.)Take square roots (choose the sign consistent with x being left of the origin)sqrt(q1/|q2|) = (1 − x)/(-x) because for x < 0, |x| = −x and |x − 1| = 1 − x.Let s = sqrt(q1/|q2|). Then(1 − x)/(-x) = s ⇒ 1 − x = −s x ⇒ 1 = x(1 − s) ⇒ x = 1/(1 − s).Compute s = sqrt(9.10 / 1.82) = sqrt(5) ≈ 2.23607 x = 1/(1 − 2.23607) = 1/(−1.23607) ≈ −0.8090 m.So the required position of the positive test charge is x ≈ −0.809 m (left of the origin).
