Force between Two Charges with a Metal Sheet in Between: Decrease or Increase?

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Introducing a metal sheet between two charges affects the force between them, typically resulting in a decrease in that force. The metal sheet can become polarized, which influences the interaction between the charges. While the charges themselves are unchanged and their distance remains the same, the presence of the metal alters the electric field. The consensus is that the force decreases, although the reasoning behind this is not clearly explained in textbooks. Further discussion is encouraged to clarify the underlying principles.
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Homework Statement


When a metal sheet is introduced in between two charges placed in air what will happen to the force between them? Will it decrease, or, increase, or, become zero? What will be the explanation? kindly reply.


Homework Equations





The Attempt at a Solution

 
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Why has no one replied to my question? Had I asked any thing so trivial in nature or so tough to answer? I eagerly wait for the expert opinions on the subject from the forum-friends.
 
I'm definitely no expert, but my guess is that there is no effect on the force. By introducing the metal sheet, you neither add nor take away any of the charges. Nor do you increase/decrease their distance from each other.

The metal sheet if its thick may become polarized in which case the force you observe in may decrease, but it has not been taken away since it now is present in the metal itself.

Hope that helps.
 
Thank you for your reply. But the answer given for this question in books is "the force decreases". Only the explanation for "why it decreases" is not given. Hope this will initiate further discussion on this topic.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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