The field will be zero where the distance to the distances are in the ratio sqrt(25/4)= 5/2, with the closest distance to the larger charge.
That will make the Coulombb forces equal and opposite. Since the sum of the two distances is 50 m, the location is 2/7 of the way from the end to the middle, at the 14.29 cm mark. That will be 35.71 cm from the 4.0 mC charge.
The ratio of the two distances is 2.5, as required.
A 25mC charge is placed at the end of a meter stick and a 4.omC charge is placed at the 50cm mark.At what position on the line joining the two charges is the electric field zero?
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