Question

Calculate the electric field at one corner of a square 1.40 m on a side if the other three corners are occupied by 2.05 ? 10^-6 C charges.

Answer 1

Distance between diagonally opposite corner = r*sq rt 2

charge on each of three corners = q = 2.05x 10*-6 C

Electric field due to adjacent corner= E1= kq/r^2

Electric field due to adjacent corner=E1 =9*10^9*2.05*10^-6/1.40^2

Electric field due to adjacent corner =E1=9.41*10^-2N/C

Electric field E2due to second adjacent corner will be equal in magnitude but perpendicular to E1=9.41*10^-2N/C

Electric field due to diametrically opposite corner =E3=kq/2=E1/2=4.70*10^-2 N/C along diagonal

Resultant electric field is vector sum of E1,E2 and E3

Resultant electric field =sq rt2*E1+E1/2=E1(1.414+0.5)

Resultant electric field = E1(1.914)

Resultant electric field =[9.41*10^-2]*(1.914)N/C

Resultant electric field at one corner=18.016*10^-2N/C in the direction of diagonal away from the square