An isolated point charge produces an electric field with magnitude E at a point 2 m away. At a point 1 m from the charge the magnitude of the electric field is

Respuesta :

Answer:

E(1m) = 4*E(2m)

Explanation:

By definition, an electric field is the electric force per unit charge, produced by a given charge distribution.

For a point charge, it is the electric force produced by the charge, over a positive test charge located at a distance d from the charge.

So, the E at a point 2 m away from the charge q, can be expressed as follows:

[tex]E = \frac{k*q}{(2m)^{2}} = \frac{k*q}{4} N/C[/tex]

At a point 1 m from the charge, the value of E is given by the following equation:

[tex]E = \frac{k*q}{(1m)^{2}} = \frac{k*q}{1} N/C[/tex]

As it can be easily seen, the magnitude of the electric field at 1 m from the charge creating it, is 4 times larger than the one at 2 m.

This is due to the electrostatic force obeys an inverse-square law, consequence of our universe be three-dimensional.

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