Hi In the given problem, there is a combination of four charged particles. First of all, this is a positive charge Plus 5.0 Q. Then here this is another identical charge plus 5.0 Q. Then on the line joining them there is a midpoint of these two charges. And from this charge from this mid wind there is another charge that is also positive but Having a magnitude of 3.0 Q. And finally the fourth charge particles, this is negative minus 12 Q. And all these distances are given us B. Each. All these expenses are now as we know, electrical always goes away from the positive charge. So electrical, you do this Plus 5.0 Q will be downward and electrical due to this lower plus 5.0 Q will be upward and these two electric fields will be equal and opposite so we will cancel each other at and this observation point is marked as speak so at point B. Electric fields due to two Plus 5.0 Q charges are equal in magnitude as the distance the same. And the charges also saying so the magnet will be seen, so they are equal in magnitude but opposite in direction. So, net electric field at B. Due to these two charges will be zero. Now the remaining two charges Because of this plus 3.0 Q charge electric field, it is a way from the positive charge And you do this -12q. It is towards the negative charge. Like this. This will be so these are the two electric fields. Let it be even, Let it be E two. So finally, net electric field at this point B will be equal to as these two are opposite in direction, so they will be subtracted. So, first of all, due to this minus 12, you this is K. Q. The magnitude of charge. No need to use a sign as the electrical is a vector country. So this is K 12 Q by distance, which is too deep. The distance of -12 cue from this point, we will become deeply steeped, needs study and minus K trig. You divided by the stance. Again the square. So here it will be electric field at P E P is equal to Okay into 12. You buy for the square. Okay, Into 12. You by four. The square mine escape into three Q by T square. Now this is four. Threes are 12. So finally this is K three cube I. D square minus K three Q by the square. So this comes out to be zero. So there is no net electric field at observation point B. Thank you.