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 newton s do they experience when their separation is 0.7 m?

1 answer:

Answer:

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

Explanation:

We know that for two point charges of magnitude the magnitude of force between them is given by

where

is constant

is the separation between the charges

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

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

Applying value of the constant we get

Thus

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