A homemade capacitor question, Vf/2

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SUMMARY

The discussion centers on calculating the electric field between the plates of a homemade capacitor constructed from two 9-inch pie pans separated by 4 cm and connected to a 9-V battery. The key point of contention is whether to use the average voltage of 4.5 V or the full voltage of 9 V when estimating the electric field halfway between the plates. The solution manual asserts that the electric field is uniform, thus the full voltage should be applied. The discussion highlights the distinction between average voltage used in energy calculations and the uniform electric field assumption in this scenario.

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Homework Statement


A homemade capacitor is assembled by placing two 9-in.
pie pans 4 cm apart and connecting them to the opposite terminals
of a 9-V battery. Estimate (c) the electric field halfway between the
plates.

Homework Equations


V=E x d

The Attempt at a Solution


As it says "halfway" I assumed the voltage should be taken as "average voltage", 4.5 Volt. So my E value is different than solution manual. Solution manual says it's 9V as the field is uniform. If the field is uniform , what does it say "halfway" for?? We use average voltage Vf/2 when calculating energy stored in a capacitor why don't we use it here??
 
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The field may not be uniform (especially near the edges). They just said halfway to hint they meant somewhere where it probably was uniform.

The units of electric field are Volts per meter. eg 9V/0.04m or 4.5V/0.02m or ...

skepticwulf said:
We use average voltage Vf/2 when calculating energy stored in a capacitor why don't we use it here??

When you extract the energy from a capacitor the voltage changes from Vf to zero. The average voltage is Vf/2.

PS: It's easier to visualise this if you think about the voltage on a capacitor while it's charged or discharged by a constant current source.
 
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