A single point charge is located at an unknown point on the x-axis. There are no

Question

A single point charge is located at an unknown point on the x-axis. There are no other charged objects nearby. You measure the electric field at x = +2.0 m to be S600 N/C, directed in the +x direction, while the field at x = +5.0 m has a magnitude of 1400 N/C. What is the sign and magnitude of the point charge? There are two possible solutions. First, we'll do the solution in which the charge on the point charge has the larger magnitude. For this solution, the sign and magnitude of the charge is and the charge is located at x = m. Now, we'll do the solution in which the charge on the point charge has the smaller magnitude. For this solution, the sign and magnitude of the charge is and the charge is located at x = m.

Explanation / Answer

Suppose the charge is at x = x0 position.
As the electric field at x = 2m is in +x direction so there are two possibilities:
a) if x0<2m, then the charge will be +ve.
b) if x0>2m then the charge will be negative.
Now calculating the magnitude,
at x = 2,E = 5600
E = kq/r^2
5600 = 9*10^9*q/(x-2)^2 ------(1)
at x = 5 E = 1400, then
1400 = 9*10^9*q/(x-5)^2 ----(2)
dividing eq. (1) by (2),
4 = (x0-5)^2/(x0-2)^2
1)considering possibility a (x0<2),
2 = (5-x0)/(2-x0)
x0 = -1 m then q = 5.6 uC.
2) considering possibility b (x0>2),
2 = (5-x0)/(x0-2)
x0 = 3m then q = -0.622 uC.

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