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Amerez
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Consider a parallel plate capacitor, with distance between plates [itex] = d_1 [/itex]. As we know the voltage between them [itex] V = Ed [/itex]. The electric field of two parallel plates is perpendicular to the surface and of the same intensity no matter where we are between the surfaces (accurate for small d's).
Now, We approached the two parallel plates to each other so as the distance is reduced to half. [itex] d_2 = \frac{1}{2} d_1[/itex]. So now, the voltage between the plates is reduced to half too. This means that to transfer a positive charge from the negative plate to the positive plate you require half of the work now.
There is the contradiction: How is the work required is sliced in half to transfer a charge from a plate where the fields acting on it stayed the same to another plate where the fields on it stayed the same too??
Now, We approached the two parallel plates to each other so as the distance is reduced to half. [itex] d_2 = \frac{1}{2} d_1[/itex]. So now, the voltage between the plates is reduced to half too. This means that to transfer a positive charge from the negative plate to the positive plate you require half of the work now.
There is the contradiction: How is the work required is sliced in half to transfer a charge from a plate where the fields acting on it stayed the same to another plate where the fields on it stayed the same too??
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