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

Question

The electric field at the point x=5.00 cm and y=0 points in the positive x direction with a magnitude of 8.00 N/C. At the point x=10.0 cm and y=0 the electric field points in the positive x direction with a magnitude of 20.0 N/C. Assume this electric field is produced by a single point charge.

Part A
Find the charge's location. (in cm)

Part B
Find the magnitude of the charge. (in pC)

Explanation / Answer

Given that the magnitude of electric field at x = 5.00 cm is E1 = 8.00 N/C The magnitude of electric field at x = 10.00 cm is E2 = 20.0 N/C Since the magnitude of electric field at x=10.0 cm is larger than the x=5.00cm So the point charge is near to the x =10.0 cm Let the charge is negative From given data direction of field at x =5.00cm and x =10.0cm is along +x axis. It is possible only with the point charge placed at x > 10.0 cm -------------------------------------------------------------------------------------------- Let the separation between the charge point charge and x =10.0 cm is x The electric field at the point x = 10.0cm is E2 = kq/a^2 ------------- (1) The electric field at the point x = 5.0cm = 0.05m is E1 = kq/(a+0.05)^2 --------------(2) From equations (1) and (2) we get E2 / E1 = (a+5)^2 / a2 Taking square root on both sides ( E2 / E1)^(1/2) = (a+0.05) / a ( E2 / E1)^(1/2) = 1 + 0.05/ a ( 20.0N/C / 8.0N/C)^(1/2) = 1 + 0.05/ a ( 20.0N/C / 8.0N/C)^(1/2) = 1 + 0.05/ a 1.58 = 1+ 0.05/a 0.05/a = 0.58 a = 0.05/0.58 = 8.62 cm Then the position of the point charge is x = 10.0 cm + 8.62 cm = 18.62 cm ----------------------------------------------------------------------------------------- From equation (1) 20.0 N/C = (8.99*10^9 N.m^2/C^2)q / a^2 q = (20.0 N/C)a^2 / (8.99*10^9 N.m^2/C^2) q = (20.0 N/C)(0.0862m)^2 / (8.99*10^9 N.m^2/C^2) = 16.53*10^-12C = 16.53 pC

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