Question

Using the symmetry of the arrangement, calculate the magnitude of the electric field in N/C at the center of the square given that qa = qb = −1.00 μC and qc = qd = + 4.93 μCq. Assume that the square is 5 m on a side.

Answer 1

Given that in the four corner of the square of side length 5 m four charge are placed as q_a,q_b,q_c and q_d respectively.lets assume the following image as a 2 Dimentional representation

(here in the below image the electric field in the y axis are to be E_ay,E_by,E_cy,E_dy respectively

In the above questien it is given that the value of charge are q_a = q_b = -1 and q_c = q_d = +4.93

assuming the centre as a unit point charge we can draw the electrric field lines as positive charge refers outward electric field and negetive charge refers inward electric field.now see these calculation where

• k    = 9 10⁹ N/m² (for vaccume medium aproximately)
• r distance between cente and corner of square i.e, half of the diagonal = m = 3.535 m

Calculating the value of electric field along X axis:

Now calculating the value of electric field along Y axis:

so along the x axis the electric field is zero and along the y axis the electric field has a value mention in above calculation.so the magnitude of the electric field is 6.039 10³ N/C along positive y axis.