Question

A charge Q exerts a 12 N force on another charge q. If the distance between the charges is doubled, what is the magnitude of the force exerted on Q by q?​

Answer 1

Answer:

here if you will double the distance it will minimise force by 4 times because (2r) ka whole square will beome 4 ✖ r ka square then

12N / 4 will be resultant force

Answer 2

The require answer is 3N.

1.Electrostatic force of attraction between charges Q and q separated by r

$\frac{kQq}{r^{2} }$

2..Electrostatic force is equal and opposite,

Force exerted on Q=Force exerted on q.

Force on q by Q = $\frac{kQq}{r^{2} }$= 12N

if distance is between charges is doubled, r =2r.

Force on q by Q=Force on Q by q

$\frac{kQq}{(2r)^{2} }$

= $\frac{kQq}{4r^{2} }$

=$\frac{12}{4}$

=3N

Hence the require answer is 3N.

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