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A charge of 31.0 μC is to be split into two parts that are then separated by 24.0 mm, what is the maximum possible magnitude of the electrostatic force between those two parts?

Respuesta :

Answer:

1.72 x 10³ N.

Explanation:

When a charge is split equally and placed at a certain distance , maximum electrostatic force is possible.

So the charges will be each equal to

31/2 = 15.5 x 10⁻⁶ C

F = K Q q / r²

= [tex]\frac{9\times10^9\times(10.5)^2\times10^{-12}}{(24\times10^{-3})^2}[/tex]

= 1.72 x 10³ N.

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