The electric field at the point x = 15.0 cm and y = 0 points in the positive x d

Question

The electric field at the point x = 15.0 cm and y = 0 points in the positive x direction with a magnitude of 5.00 N/C. At the point x = 25.0 cm and y = 0 the electric field points in the positive x direction with a magnitude of 10.0 N/C. Assume that this electric field is produced by a single, point charge
i got 27.02 for the location but cant get find the charge and magnitude

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 >25cm, and it must be NEGATIVE, because the field is pointing TOWARDS it. Let d be the displacement of the point charge from x=25 Then (15+d)^2/d^2 = 10/5 = 2 since E field is inversely proportional to d^2 Solve for d: 15+d = d*SQRT(2) = d*1.414 d = 15 / 0.414 = 36.232 cm (3 sig figs) So the negative point charge is located at x = 25+d = 61.23 cm Magnitude of the charge can now be found from either of the field strengths and distances. (10 N/C at 36.232 cm, or 5 N/C at 61.23). Both should give the same answer using Q = E * 4.pi.Eo.r^2 Q = 10 * 4.pi.Eo.(0.362)^2 = 1.46 * 10^-9 C

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