A positively charged particle is in the center of a parallel plate capacitor that has charge +/- Q on it's plates. Suppose the distance between the plates is doubled, with the charged particle remaining in the center. Does the force on this particle increase, decrease, or remain the same?

the force remains constant if the charge does not change

Explanation:

In a capacitor the capacitance is given by

           C = ε₀ A / d

Where ε₀ is the permissiveness of emptiness, A is about the plates and d the distance between them.

The charge on the capacitor is given by the ratio

            Q = C ΔV

Let's apply these expressions to our problem, if the load remains constant

            C = Q / ΔV = ε₀ A / d

            ΔV / d = Q / ε₀ A

If the distance increases the capacitance should decrease, therefore if the charge is a constant the voltaje difference must increase

Now we can analyze the force on the test charge in the center of the capacitor

               ΔV = E d

               E= ΔV/d

               F = q E

              F = q ΔV / d

 Let's replace

          F = q Q /ε₀ A

From this expression we see that the force is constant since the voltage increase is compensated by increasing the distance, therefore the correct answer is that the force remains constant if the charge does not change

as we know our atmosphere or air have different levels. every level has other gasses, and every gas has another permeability. when the light comes from lower permeability to higher permeability, it changes its direction. so that means that when the light comes from the moon to the earth, it changes it's direction many more times. it basically means that the shape of the light part of the moon is changed on the earth.

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