The electrostatic force of attraction between two charged objects separated by a distance of 1.0 cm is given by f. if the distance between the objects were increased to 5.0 cm, what would be the electrostatic force of attraction between them?

New force between them will become times

Explanation:

Let the charge on both the object are and distance between them is is given 1 cm

So r = 1 cm = 0.01 m

Electric force between them is given F

According to Coulomb's between two charges is given by

According to question -----------eqn 1

Now distance is increased by 5 cm so new distance = 5+1 = 6 cm = 0.06 m

So new force ------------------eqn 2

Comparing eqn 1 and eqn 2

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weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards