Question

A) Calculate the magnitude of the electric field at one corner of a square 2.12 m on a side if the other three corners are occupied by 3.75×10−6 C charges.

B) What is the direction of the electric field at the corner?

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

E = kq/r^2

= (9x10^9) * (3.75 x 10^-6) / (2.12)^2 N/C

= 7.5 x 10^3
The resultant of two such charges will be along CA
= 2cos45° * [7.5 x 10^3] N/C = 10606.60 N/C

Electric field due to charge at C will be along CA
E = kq/(√2r)^2

= (9x10^9) * (3.75 x 10^-6) / (2.12√2)^2

= 3754.67 N/C

Total field = 10606.60 + 3754.67 = 14361.27 N/C in the direction CA.

Answer 2

=\(7.509 x 103).(7.509x103)2 +3.754x10 3 10.619 x10 +3.754x10 14.373×10° NUC Direction Along the line between the comer and centre of square about outward of the centre