Two point charges of the same magnitude and opposite sign are fixed at points A and B. A third point charge is to be balanced at P by the electrostatic force due to these 2 charges. The Point P :
a)lies on the perpendicular bisector of line AB
b) is at the mid point of line AB
c) lies to left of A
d) none



  • Brainly User
If charge is placed to the left of A,then net force will be zero due to both the charges.
therefore c
orrect option is c.
1 5 1
It says D. @sweety2
but its a standard question r8!!!!
Test charge is always positive 
the magnitude of test charge is not given hence, we can place left & right of A on balancing magnitude of charge and distance of charge hence option (D) is correct.