Question

A positive and a negative charge are initially 4cm apart when they are moved closer together so that they are now only 1cm apart so force between them is 4 times smaller than before, 4 times larger than before , 8 times larger than before , 16 times larger?​

Answer 1

Answer:

16 times larger

Explanation:

electrostatic force is inversely proportional to the (distance)^2

Answer 2

Answer:

The final force between charges is $16$ times greater than the original value.

Explanation:

• There is a force of attraction between unlike charges.
• There is a force of repulsion between like charges.
• The magnitude of the force depends on the medium between charges.

The formula for the force of attraction between two charges is:

$F=\frac{K*Q*q}{r^{2} }$

where $K$ is electrostatic constant.

           $Q,q$ are the two charges.

           $r$ is the distance.

Step 1:

It is given that initially, charges are $4$ cm apart.

It is clear from the formula that:

$F$ ∝ $\frac{1}{r^{2} }$

This means that when distance is reduced the force will increase by the square of the distance.

Step 2:

The final distance between the two charges is $1$ cm, the distance is reduced by  $4$ times.

So, Force is increased by  $4^{2}$ times which is equal to  $16$  times.

Therefore, the final force is $16$ times larger than before.