Find the electric field at the location of qa in the figure

Find the electric field at the location of q_a in the figure below, given that q_b = q_c = q_d = +1.80 nC, q = 1.00 nC, and the square is 18.5 cm on a side. (The +x axis is directed to the right.)

Solution

According to the given problem,

Lets consider the field at q_a due to the q_b is E₁,

The field from q_c to the charge q_a is E₂, The field from q_d to the charge q_a is E₃

And the field from q at the center to the charge q_a is E_4.

The directions of the fields at the upper left corner are

E₁ is pointing at 180°
E₂ is pointing at 90°
E₃ is pointing at 135°
E₄ is pointing at 315°

And calculate the distance from the charge

r₁² = r₂² = 0.185² = 0.0342 m²,

r₃² = 0.185²+0.185² = 0.06845m²

r₄² = 0.0925²+0.0925² = 0.0171m²

According to the columbs law,

E = kq/r²

E₁ = E₂ = k *1.80*10^-9/0.0342 = 473.15 N/C

E₃ = k*1.80*10^-9/0.06845 = 236.40 N/C

E₄ = k*1.0*10^-9/0.0171 = 525.73 N/C

Note: we didn\'t use keep negative sign for E₄ because the sign represent the field direction.

Now sum the 4 vector fields at the upper left corner:

Then,Horizontal components

E₁*cos180+E₂*cos90+E₃*cos135+E₄*cos315

-473.15 + 0 - 167.16 +371.75

E_x = -268.56 N/C

Finally Vertical components

E₁*sin180+E₂*sin90+E₃*sin135+E₄*sin315

= 0 + 473.15 + 167.16 - 371.75

E_y = 322.56 N/C

The magnetude if electric field is

E = Ex²+Ey²

E = 420 N/C

The direction of the electrci field at

arctan(322.56/-268.56) = -50.22°