In: Physics
A QL = -22 ?C charge is placed on the x-axis at x = - 0.4 m. A QR = 22 ?C charge is placed at x = +0.4 m. (for all answers below assume that the unit vector x^ points toward positive x, and y^ points towards positive y.).
a.What is the x-component of the electric field at x = 0 m and y = 0.4m? (N/C)
b. What is the y-component of the electric field at x = 0 m and y = 0.4m? (N)
c What is the x-component of the electric field at x = 0 m and y = -0.4m? (N/C)
d. What is the y-component of the electric field at x = 0 m and y = -0.4m? (N/C)
B and D are 0 but I can't figure out A and C
The way you do these is first determine the direction of the E fields at 0,0.4
The field at 0,0.4 due to the positive charge points at 135 degrees
The field at 0,0.4 due to the negative charge points at 225 degrees
This is because the E field is OUT of positive charges and IN to negative charges.
The distance squared from each charge to 0,0.4 is 0.4^2+0.4^2 = 0.32m
Since both charges have the same magnitude and are the same distance from 0,0.4, their fields have the same intensity. The magnitude of the E fields at 0,0.4 is
k*22e-6/0.32 = 786,411 N/C
The net E field at 0,0.4 is the sum of the two vector E fields from the two charges.
Sum the horizontal components of the fields at 0,0.4:
786,411cos135 + 786,411cos225 = -1,112,153
Sum the vertical components:
786,411sin135 + 786,411sin225 = 0
This makes sense. The vertical components are equal in magnitude but opposite in direction while the horizontal components are equal in magnitude and both pointing in the negative x direction.
Therefore the net E field at 0,0.4 is
1,112,153 N/C at 180 degrees A)
0 B)
1,112,153 N/C at 180 degrees C)
0 D)
C) and D) have the same answers as A) and B) because of the symmetry. 0,-0.4 is the same distance from both charges and field directions are the same. The vertical components cancel and the horizontal components add.