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
You might be interested in
Answer:
The acceleration and time are 588 m/s² and 0.071 sec respectively.
Explanation:
Given that,
Speed = 42.0 m/s
Distance = 1.50 m
(a). We need to calculate the acceleration
Using equation of motion
Put the value in the equation
(b). We need to calculate the time
Using equation of motion
Hence, The acceleration and time are 588 m/s² and 0.071 sec respectively.