How Do Electric Fields Differ Between Aluminum and Glass Plates?

AI Thread Summary
The electric field generated by a charged aluminum plate is calculated using the formula E = σ/2ε0, where σ is the surface charge density. In contrast, the glass plate, being an insulator, does not allow for the same distribution of charge and requires a different approach. The electric field for the glass plate is E = σ/ε0, reflecting its inability to conduct electricity like aluminum. This distinction leads to a stronger electric field for the glass plate compared to the aluminum plate when both have the same charge. Understanding these differences is crucial for analyzing electric fields in various materials.
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Homework Statement


A very large flat aluminum plate of area A has a total charge Q over its surface. The same charge is placed over the upper surface of a glass plate. Compare the electric fields.


Homework Equations





The Attempt at a Solution


I know how to get the electric field of the aluminum plate.
\sigma = Q/A
2EA = \sigmaA/e0
E = \sigma/2e0

For the glass plate, I'm not so sure... Initially I would think E = \sigma/e0, but it isn't...
 
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Realize that the aluminum plate is a conductor.
 
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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