Two positive charges repel each other with a force of 4x10^-4 when placed at a distance of 1 meter. If the the distance between them is increased by 2meter the force of repulsion will be_____ a) 21x10^-4N b) 8x10^-4N c) 2x10^-4N d) 4x10^-4N
Answers
hey hi...
sorry dude.but my answer is 10^-4. but this answer is not in your option...
plz make sure your option is correct.....
The force of repulsion when the distance is increased by 2 meters is
0.4 * 10^-4 N.
Given:
Force = 4x10^-4 N when x = 1 meter.
To Find:
The force of repulsion when the distance is increased by 2 meters.
Solution:
According to Coulomb’s law, the force of repulsion between two charged bodies is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
F = kq1q2 / r²
4 * 10^-4 = kq1q2 /1 -------------(1)
when r is increased by 2 meter, r = 1+2 = 3 meters.
F = kq1q2 / 3²
= kq1q2/1 * (1/9) -------------------(Using Equation 1)
= (4 * 10^-4 ) /9
= 0.4 * 10^-4 N
The force of repulsion when the distance is increased by 2 meters is
0.4 * 10^-4 N.
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