In the figure the ofur particles ofrm a square of edge a 53

In the figure the ofur particles ofrm a square of edge a = 5.30 cm and have changes q_1 = 6.02nC, q_2 = -17.7 nC, q_3 = 17.7 nC, and q_4 = -6.02 nC. What is the magnitude of the net electric field produced by the particles at the square\'s center?

Solution

Side a= 5.3 cm, q1 = 6.02 nC, q2= -17.7nC, q3 =17.7nC, q4= -6.02nC

We use super position rule find the net electric field due to all the charges at the center of the square.

Q1 and q3 or on the ends of a diagonal and the net filed is along the diagonal

Net field at the center due to q1 and q3

                     E1 = 6.02e-9/4₀ r² + 17.7e-9/4₀ r²       , r = 2*5.3e-2 m

                           = 23.72e-9/4*8.852e-12*2*(5.3e-2)²

                             = 3.8e+4 N/C

By symmetry the net filed at the center of the square due to q2 and q4 is along the diagonal joining them

                    E2 = 3.8e+4 N/C

E1 and E2 are along the respective diagonals and are ^_r

The resultant field

    E = sqrt(E1² + E2² ) = sqrt(3.8² + 3.8²) e+4

            = 5.37e+4 N/C