Question

1 pt A charge of 2.09 × 10-9 C is placed at the origin, and a charge of 3.60 x 10-9 C is placed at -1.98 m. Find the position at which a third charge of 3.14 x 10-9 C can be placed so that the net electrostatic force on it is zero. (in m) AO 4.57 x 10-1 DO 7.32 x 10-1 EO 8.56 x 10-1 FO 1.00 GO 1.17 7. BO 5.35 x 10-1 CO 6.25 x 10-1 HO 1.37

Answer 1

The electrostatic force on a charge q1 due to a charge q2 at a distance r = Kq1q2/(rxr)

Therefore we assume the charge q3 to be at a distance x from the origin. Then we equate the electrostatic forces due to each charge q1 and q2 ( since theyretgonna act in the opposite direction, for net EF to be 0, they must be equal)

Therefore by solving the quadratic equation we find out the correct value of distance.

hence option E is correct.