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?​

Respuesta :

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

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

Force [tex]$=k \times Q 1 \times Q 2$[/tex]divided by ( [tex]$r$[/tex] squared)

Here r is the distance.

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

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

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

Now just separate the 9 :

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

New Force [tex]$=1 / 9$[/tex] (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 https://brainly.com/question/13191643

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