Answer:
50.4 N
Explanation:
Q1 = Q
Q2 = 4 Q
Distance = d
The force is given by
[tex]F = \frac{KQ_{1}Q_{2}}{d^{2}}[/tex]
[tex]1.60 = \frac{4KQ^{2}}{d^{2}}[/tex] .... (1)
Now,
Q3 = 2 Q
Q4 = 7 Q
distance = d/3
[tex]F' = \frac{9KQ_{3}Q_{4}}{d^{2}}[/tex]
[tex]F' = \frac{126KQ^{2}}{d^{2}}[/tex] .... (2)
Divide equation (2) by equation (1), we get
F' / 1.60 = 126 / 4
F' = 50.4 N
Thus, the force is 50.4 N.
