The electric field at the point x=5.0 cm and y=0 points in the positive directio

Question

The electric field at the point x=5.0 cm and y=0 points in the positive direction with a magnitude of 10.0 N/C . At the point x=10 cm and y= 0 the electric field points in the positive direction with a magnitude of 17.0 N/C . Assume this electric field is produced by a single point charge. What is the charge's location, sign of charge, and magnitude of the charge. PLEASE SHOW WORK FOR ALL STEPS FOR LIFESAVER

Explanation / Answer

The electric field is increasing along the positive x-axis. The electric field points in the positive x-direction . Therefore, the single point charge must be located at x >10cm, and it must be NEGATIVE, because the field is pointing TOWARDS it. Let d be the displacement of the point charge from x=10 Then (5+d)^2/d^2 = 16/10 = 1.6 since E field is inversely proportional to d^2 Solve for d: 5+d = d*SQRT(1.6) = d*1.265 d = 5 / 0.265 = 18.9 cm (3 sig figs) So the negative point charge is located at x = 10+d = 28.9 cm Magnitude of the charge can now be found from either of the field strengths and distances. (16 N/C at 18.9 cm, or 10 N/C at 18.9+5 = 23.9 cm). Both should give the same answer using Q = E * 4.pi.Eo.r^2 Q = 16 * 4.pi.Eo.(0.189)^2 = 6.35 * 10^-11 C check: Q = 10 * 4.pi.Eo.(0.239)^2 = 6.35 * 10^-11 C

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