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Assume the following problem:

Two parallel-plate capacitors with different distances between the plates are connected in parallel to a voltage source. A point positive charge is moved from a point 1 that is exactly in the middle between the plates of a capacitor $C_1$ to a point 2 (of a capacitor $C_2$) that lies at a distance from the negative plate of $C_2$ equal to half the distance between the plates of $C_1$. Is any work done in the process by the electric field?

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My Approach: Since the motion of the charge is perpendicular to the electric field of the capacitors, the work done in such a case would be zero. Hence, no work is done in the process by the electric field

Textbook Approach: The textbook says that the potential at point 1 is lower than the potential at point 2, hence $W = Q(V_2-V_1) > 0$ or positive work is done in moving the charge from 1 to 2

Where am I going wrong here? I don't see any mistake in my application of concepts. Is the electric field of a capacitor not always parallel and would I also have to account for edge effects or something similar?

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Is the electric field of a capacitor not always parallel and would I also have to account for edge effects or something similar?

Yes, this is the key. The field lines of a capacitor are not perpendicular to the plate everywhere. They are approximately perpendicular far from the edges, but as you get closer to the edge that approximation progressively fails. Outside the capacitor the field begins to look more like a dipole field than a uniform field.

Regardless of the shape of the field, the conservation of energy works. So it is more useful to focus on the potential rather than the E field.

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The electric fields only exist between the plates and they are different in the two capacitors. Hence it is incorrect to assume that charges moving perpendicular requires no work to be done.

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