The electric field at the point x=5.00cm and y=0 points in the positive x direct

Question

The electric field at the point x=5.00cm and y=0 points in the positive x direction with a magnitude of 10.0N/C . At the point x=10.0cm and y=0 the electric field points in the positive x direction with a magnitude of 19.0N/C . Assume this electric field is produced by a single point charge.
A)Find the charge's location.
B)Positive or negative?
C)Find the magnitude of the charge

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 = 19/9 since E field is inversely proportional to d^2 Solve for d: 5+d = d*SQRT(19/9) = d*1.453 d = 5 / 0.453 = 11.0 cm (3 sig figs) So the negative point charge is located at x = 10+d = 21 cm Magnitude of the charge can now be found from either of the field strengths and distances. (19 N/C at 11 cm, or 9 N/C at 11+5 = 16 cm). Both should give the same answer using Q = E * 4.pi.Eo.r^2 Q = 19 * 4.pi.Eo.(0.11)^2 = 2.56 * 10^-11 C Q = 9 * 4.pi.Eo.(0.16)^2 = 2.56 * 10^-11 C [yes!]

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