Question

An electrostatic force of attraction between 2 points A and B is 1000N. If the charge on A is increased bu 25% and that on B is reduced by 25% and the initial distance between them is decreased by 25%. Then find the new force of attraction between them​

Answer 1

Answer:

The new electrostatic force will be 1250 N.

Explanation:

Electrostatic force between two charged particles is directly proportional to the product of the charges and is inversely proportional to the square of the distance between them.

Here , q1 = first charge

q2 = second charge

r = distance between q1 and q2

k = 9× 10^9 (it's value remains unchanged)

So, Force F ={k×(charge 1)(charge 2)}/r^2}.

k is just a constant .

now, 1000= {k× (q1)(q2)}/r^2 ------ (equation 1)

It is given that, q1 is increased by 25%, so q1 becomes {(5q1)/4} and q2 is decreased by 25%, so q2 becomes {(3q2)/4}. Also , it is said that distance between them is decreased by 25%, so new distance becomes {(3r)/4}.

Hence, new force acting between them is

F' =[【k×{(5q1)/4}×{(3q2)/4}】/ (r/2)(r/2) ] .

On simplifying, we see,

F' = (5/4)×F. Since F = 1000N, so F'= (5×1000)÷4 N

or, F' is equal to 1250 N.