Question

calculate the magnitude of the electric field at one corner of a square 1.22 m on a side if the other three corners arr occupied by 3.75*10^-6 C charges.

Answer 1

Electric field is defined as the force experienced by a unit positive charge. Suppose a square ABCD is given where the charges are at B, C and D and electric field is to be worked out at A.

Electric field due to the charge B and D would be equal in magnitude
= kq/r^2 = (9*10^9) * (3.75 *10^-6) / (1.22)^2 N/C = 2.26* 10^4

The resultant of two such charges will be along CA
= 2cos45° * [2.26 x 10^4] N/C = 34370.34 N/C

Electric field due to charge at C will be along CA
= kq/(√2r)^2 = (9x10^9) * (3.75 x 10^-6) / (1.22√2)^2 = 1133 N

Total field = 34370.34++ 1133=35503 N/C in the direction CA.