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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