In a rectangular coordinate system, a positive point charge q = 8.00 nC is place

Question

In a rectangular coordinate system, a positive point charge q = 8.00 nC is placed at the point x= 0.210 m , y=0, and an identical point charge is placed at x= -0.210 m , y=0. Find the x and y components and the magnitude and direction of the electric field at the following points.

part F) Find the x and y components of the electric field at x= 0.210 m , y= -0.450 m .

part G) Find the magnitude of the electric field at x= 0.210 m , y= -0.450 m .

part H) Find the directiion of the electric field at x= 0.210 m , y= -0.450 m .

part I) Find the x and y components of the electric field at x=0, y= 0.250 m .

part j) Find the magnitude of the electric field at x= 0, y= 0.250 m .

I followed 3 different solutions and kept getting the answer wrong. If someone could post a detailed solution with explanations for everything that would be appreciated. thanks

Explanation / Answer

Given in rectangular coordinate ssytem
point +ve charge q = 8 nC at (x = 0.21 m , y = 0)
another +q charge at (x = -0.21 m , y = 0)

F. electric field at point x = 0.210 m, y = -0.450 m
   Ex be the field in x direction, Ey in y direction
   then
   Ex = k*q*2*0.21/(0.45^2 + (2*0.21)^2)^3/2
   Ex = 129.368 V/m ( +ve x axis )

   Ey = -kq/0.45^2 - k*q*0.45/(0.45^2 + (2*0.21)^2)^3/2 = -493.374 V/m ( -ve y axis)

G> Magnitude of electric field = sqroot(Ex^2 + Ey^2)
   |E| = 510.05 V/m

H) direction of field = arctan(Ey/Ex) = -75.3 deg CCW from +x axis
I) at x = 0, y = 0.25 m, the y components of the electric field will add, the x components will cancel each other out ( due to symmetry)
   hence Ey = 2*k*q*0.25/(0.25^2 + 0.21^2)^3/2 = 1032.049 V/m

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