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1 answer:
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If the distance between two charges is halved, the electrical force between them increases by a factor 4.
In fact, the magnitude of the electric force between two charges is given by:
where
k is the Coulomb's constant
q1 and q2 are the two charges
r is the separation between the two charges
We see that the magnitude of the force F is inversely proportional to the square of the distance r. Therefore, if the radius is halved:
the magnitude of the force changes as follows:
so, the force increases by a factor 4.
In fact, the magnitude of the electric force between two charges is given by:
where
k is the Coulomb's constant
q1 and q2 are the two charges
r is the separation between the two charges
We see that the magnitude of the force F is inversely proportional to the square of the distance r. Therefore, if the radius is halved:
the magnitude of the force changes as follows:
so, the force increases by a factor 4.