How does the magnitude of electric force compare with the charge between a pair of charged particles when they are brought to half their original distance of separation?

Answer 1

The force varies according to the square of the distance apart. If you move the particles twice as far apart, the force between them decreases by a factor of 2^2, that is to say, it decreases by a factor of four. [If you move them 3 times as far apart, the force decreases by 3^2, i.e. by 9 times.] This inverse square law also works for gravity and magnetism. Enjoy.

Answer 2

When one charge is doubled, you are going to still have a the same electric field: (2 opposites = attract) (2 similar = repel). But the electrical force will be greater since one has a strong electrical field than before.

Answer 3

The electrical force F between two charges Q and q separated by a distance r is given by:

F = Qq/4piEor2

where Eo is the permittivity of free space.

The electrical force follows an inverse square law, that is, the force is inversely proportional the square of their distance of separation.

Thus, if the distance is increased 3 times, the force should decrease by 9 times.

Answer 4

The magnitude of the force reduces to 1/4 of the original force.

This is because the force is proportional to 1/r2 where r is the distnace apart.

Answer 5

The Magnitude of the force varies INVERSELY to the Square of the Distance between the two Objects. So if you Halve the distance the Force increase by a factor of FOUR times.

Answer 6

The force is reduced by a factor of 4, f= k/r2, thus 2r gives f=k/(2r)2 = k/4r2.

Answer 7

The force changes inversely with the square of the distance between them.

Answer 8

The force is reduced to a quarter of its previous value.

Answer 9

it increases by a factor of four