Question

Coulomb's Law Two point charges experience an attractive force of 10.8 N when they are separated by 2.4 m. VWhat force in newtons do they experience when their separation is 0.7 m?

Answer 1

The force between the charges when the separation decreases to 0.7 meters equals 126.955 Newtons

Step-by-step explanation:

We know that for two point charges of magnitude
`q_(1),q_(2)` the magnitude of force between them is given by

`F=(k_(e)q_(1)\cdot q_(2))/(r^(2))`

where

`k_(e)` is constant

`r` is the separation between the charges

Initially when the charges are separated by 2.4 meters the force can be calculated as

`F_(1)=(k_(e)\cdot q_(1)q_(2))/(2.4^(2))\\\\10.8=(k_(e)\cdot q_(1)q_(2))/(2.4^(2))\\\\\therefore k_(e)\cdot q_(1)q_(2)=10.8* 2.4^(2)=62.208`

Now when the separation is reduced to 0.7 meters the force is similarly calculated as

`F_(2)=(k_(e)\cdot q_(1)q_(2))/(0.7^(2))`

Applying value of the constant we get

`F_(1)=(62.208)/(0.7^(2))`

Thus
`F_(2)=126.955Newtons`