When a point charge of q is placed on one corner of a square, an electric field strength of 2 N/C is observed at the center of the square. Suppose three identical charges of q are placed on the remaining three corners of the square (so that each corner has q). What is the magnitude of the net electric field at the center of the square

0 N/C

Explanation:

Since the electric field due toeach charge q is at the same distance from the center of the square and has a magnitude E = 2 N/C, we resolve the components of each electric field to vertical and horizontal.

So, for each charge, the magnitude of the horizontal component is E' = Ecos45 and the magnitude of the vertical component is E" = Esin45.

For the charges on the bottom half of the square, the direction of the vertical component is upwards and the horizontal components of their electric fields cancel out since they are in opposite directions, the resultant electric field due to the bottom two charges is 2Esin45.

For the charges at the upper half of the square, the direction of the vertical component is downwards and the horizontal components of their electric fields cancel out since they are in opposite directions, the resultant electric field due to the top two charges is -2Esin45.

So, the resultant electric field at the center of the square is thus 2Esin45 + (-2Esin45) = 2Esin45 - 2Esin45 = 0 N/C  

i think the answer is a, hope this : )