Can a charged sheet attract uncharged objects?

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A charged conducting sheet can attract uncharged objects through the process of electrostatic induction. When a positively charged sheet is near uncharged metal rods, it induces a negative charge on the side of the rods closest to the sheet, while the far side becomes positively charged. This creates an attractive force between the sheet and the rods, allowing the charged sheet to attract them. Additionally, the induced charges on one rod can further influence the other rod, leading to mutual attraction. The discussion emphasizes the role of electrostatic induction in explaining these interactions.
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Homework Statement


A large conducting sheet M is given a uniform charge density. Two uncharged small metal rods A and B are placed near the sheet on one side. Then
a)M attracts A
b)M attracts B
c)A attracts B
d)B attracts A

Homework Equations





The Attempt at a Solution


An electric field will be set up by the sheet but A and B are neutral. So how can it attract them?
 
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Good Morning Utkarshakash,

Have a look at "electrostatic induction".

J.
 
jfgobin said:
Good Morning Utkarshakash,

Have a look at "electrostatic induction".

J.

OK so if I assume the sheet to be +vely charged it will induce a -ve charge on the side of block A near to the sheet and +ve charge on the other end. Block A will then induce -ve charge on the side of B near to A and +ve charge to the other side. Am I right?
 
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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