Question

A) Calculate the electric field (magnitude and direction) at the upper right corner of a square 1.22 m on a side if the other three corners are occupied by 2.45

Answer 1

Let the four sides be A,B,C,D. A be the leftmost corner, B where electric field is to be found. C below B and D is the point left of C.


So Electric field at B due to C will be directed upward. Similarly Electric field at B due to A will be directed rightwards.

The magnitude of both the fields will be same and the resultant of the both will be directed towards the diagonal DB. The magnitude of the resultant will be sqrt[2] * [ k * q / (r*r)] where r = 0.22 m


Similarly field at B due to D will be in same direction and will be of magnitude [k * q/(r*r)] directed towards diagonal DB.


So net field will be the addition of both i.e.
sqrt[2] * [ k * q / (r*r)] + (k * q) / (diagonal of square * diagonal of square) where k = 9 *(10^9)

Calculating all, we will get E = 33.5669 * 10^3 N/C
Angle will be 45 degree