Electric field of point charges

[texastud069]

hey guys i can't figure out this problem to save my life i thought i was doing it right using the
K=q/r^2 formula but its not coming out right. so if anyone can help me with this problem i would very much appreciate it.

In a rectangular coordinate system, a positive point charge = 3.50 is placed at the point x=0.190 ,y=0 , and an identical point charge is placed at x= -0.190 ,y=0 . Find the and components and the magnitude and direction of the electric field at the following points.

A) Find the and components and the magnitude and direction of the electric field at x=0.380, y=0.


B) Find E

 

[tony873004]

Find the what and the components?

E=kq/r^2, not k=q/r^2. k is a constant.

 

[baba89]

_x, E_y, and the magnitude and direction of the electric field at the origin

I would be happy to assist you with this problem. It appears that you are on the right track by using the formula K=q/r^2 to calculate the electric field of point charges. However, it is important to make sure that you are using the correct values for the charge and distance in your calculations.

For the first part of the problem, we need to find the electric field at the point x=0.380, y=0. Since the two point charges are located at x=0.190 and x=-0.190, we can use the distance formula d=√((x2-x1)^2+(y2-y1)^2) to find the distance between the point charges and the point we are interested in.

d=√((0.380-0.190)^2+(0-0)^2)=0.190

Now, we can plug in the values for the charge and distance into the formula K=q/r^2 to find the electric field at this point.

E=K*q/r^2 = (9x10^9 N*m^2/C^2)*(3.50 C)/(0.190 m)^2 = 9.23x10^10 N/C

To find the x-component and y-component of the electric field, we can use the equations E_x=E*cosθ and E_y=E*sinθ, where θ is the angle between the electric field and the x-axis. In this case, θ=0 because the electric field is in the same direction as the x-axis.

E_x=9.23x10^10 N/C*cos(0)=9.23x10^10 N/C

E_y=9.23x10^10 N/C*sin(0)=0 N/C

Therefore, the electric field at x=0.380, y=0 is 9.23x10^10 N/C in the positive x-direction.

For the second part of the problem, we need to find the electric field at the origin (x=0, y=0). Using the same method as before, we can find the distance between the point charges and the origin.

d=√((0-0.190)^2+(0-0)^2)=0.190

Plugging in the values for the charge and distance into the