Four point charges have equal magnitudes. Three are positive, and one is negative, as the drawing shows. They are fixed in place
on the same straight line, and adjacent charges are equally separated by a distance d. Consider the net electrostatic force acting on each charge. Calculate the ratio of the largest to the smallest net force magnitude.
1 answer:
Answer:
Ans= 9
See attached picture for clearer solution.
Explanation:
The net electrostatic force acting on charge A = 2/ 2 + 2 /(2) 2 − 2 /(3) 2 = 2 / 2 (1 + 1/4 – 1/9 ) = 41/36 2/2 .
The net electrostatic force acting on charge B = 2/2 + 2/(2)2 − 2/2 = 1/4 2/d2 .
The net electrostatic force acting on charge C = 2/2 + 2/(2)2 + 2/2 = 2/2 (1 + 1 4 + 1) = 9/4 2/2 .
The net electrostatic force acting on charge D = 2/2+ 2 /(2)2 + 2/(3)2 = 2 /2 (1 + 1/4 + 1/9 ) = 49/36 2/ 2 .
The ratio of the largest to the smallest net force = 9/4*2/2 / 1/4 2/2 . = 9
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Answer:
Explanation:
The torque applied by a force can be calculated as
where
F is the magnitude of the force
d is the length of the arm
is the angle between the direction of the force and the arm
In this problem, we have
F = 15 N
d = 2.0 m
Substituting into the equation, we find