Question

The force of repulsion that two like charges exert on each other 5N. what will be if the distance between the charge is decreased to a quarter of its original value?​

Answer 1

The new force becomes One Ninth (1/9) of the original force.

The force between two point charges (let's say $\mathrm{Q} 1 and \mathrm{Q} 2$ ) is given by the following formula:

Force $=k \times Q 1 \times Q 2$divided by ( $r$ squared)

Here r is the distance.

If we multiply r by three then after squaring it will become $9 \times r$ squared.

Let's rewrite the formula and call it new Force:

New Force $=\mathrm{K} \times \mathrm{Q} 1 \times \mathrm{Q} 2$ divided by $(9 \times \mathrm{r}$ squared )

Now just separate the 9 :

New Force $=1 / 9(\mathrm{~K} \times \mathrm{Q} 1 \times \mathrm{Q} 2$divided by $(\mathrm{r}$ Squared ))

New Force $=1 / 9$ (Force)

So turns out that the new force becomes One Ninth (1/9) of the original force.

A force is an effect that can alter an object's motion according to physics. An object with mass can change its velocity, or accelerate, as a result of a force. An obvious way to define force is as a push or a pull. A force is a vector quantity since it has both magnitude and direction.

Learn more about force brainly.com/question/13191643

#SPJ9