Two point charges are separated by a distance r and repel each other with forces of magnitude F. If their separation is reduced to 0.281 times the original value, what is the magnitude of the forces of repulsion

12.66 times the original value.

Explanation:

Using the Coulomb's law that states that "the force of attraction or repulsion exerted between two point charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between these charges."

As stated in the law,

The force of repulsion (F) between two point charges is inversely proportional to the distance (r) between these two charges. This means that as the distance increases, the force will decrease and vice versa. i.e;

F ∝

=> F = k /

=> F = k

=>    --------------------------(i)

Where;

and are initial and final magnitudes of the force of repulsion/attraction

and are the initial and final distances of separation.

According to the question,

r has been reduced to 0.281 of the original value. This means that;

Substituting  into equation (i) gives;

=> x = x (0.281)²

=> x = x x

Dividing both sides by to eliminate it, we have;

=> = x

=> = x 0.07896  

Making the subject of the formula

=> = 12.66 x   -------------------- (ii)

From equation (ii),

It is evident that the magnitude of the forces of repulsion has increased to 12.66 times the original value.

50 inches should be the resultant vector

multiply 2kg by latent heat of vap