You are given two positively charged particles, of equal magnitude, separated by a distance, "d". What will happen to the force field between the two particles when "d" is doubled?

Force becomes 1/4th

Explanation:

Force between two positively charged particles is given by the equation

F = k×q1×q2/r² So, force is inversely proportional to square of r.

Here, k is a constant, q1 and q2 are the charges which are equal and r is the distance between them.

So if the distance is doubled, r becomes 2r and r² becomes (2r)² = 4r²Force, F then becomes 1/4th of the initial force.

The relationship of both of them

W= fd
where f is an applied force and d is the distance over which it is applied. when a parachuting skydiver is at constant velocity, he is experiencing "terminal velocity", the point at which his acceleration down is canceled out by air resistance. that means that the resisting force of the air is equal to his weight, but in the opposite direction. with this, we can solve for the work done by the force.
w = fd
w = (92.0)(9.81 m/s^2) (325)
w= 2.93 x 10^5 j
the value for work in this case is positive because it is pushing up on the skydiver, who has a downward force of gravity.
therefore the answer is 2.93 x 10^5 j