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Imagine two large metal plates having charge density $+\sigma$ and $-\sigma$ placed close to each other with a distance $d$ between them. The question is that why are the two plates are at different potentials. If a charge $+q$ is brought by an external force from infinity to the $+ve$ plate from the opposite side of the $-ve$ plate, the force experienced by the charge should be zero as the the $+ve$ and the $-ve$ plates exert an equal and opposite force of $q\sigma/2\epsilon_0$ on the charge. So the work done is zero for either of the plates. The work done or potential should not change when the charge is brought from any side of the plate as both are state functions. But Electric Field does exist in between the plates. And it should because one plate has $+ve$ charges while the other has $-ve$. So there must be a potential difference in between them.

I know that I have made two contradictory statements above but I can't understand what's wrong in my understanding. Please explain.

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The reason you are facing this apparent contradiction is because you are bringing your test charge $+q$ to the plates from two different directions. In one case you start at $-\infty $ and in the other, at $+\infty $ You have to start from one fixed direction relative to which you assign all other potentials in space. Conventionally, this is chosen to be $+\infty $. The electric field outside the capacitor is zero, and inside it is $\frac{\sigma}{\epsilon_0}$. You take your test charge from $+\infty $ to the negatively charged plate, without feeling any force. The potential here is 0. Then you continue to move it in the same direction, encountering an electric force $\frac{q\sigma}{\epsilon_0}$ for a distance $d$. Work at the negative plate is $0$, so this is assigned a potential $0$. Then an additional work was done taking the charge from $-$ to $+$ plate, so that the potential (work per unit charge) at the $+$ plate is in an excess of $\frac{d\sigma}{\epsilon_0}$ relative to the $-$ plate and is assigned a potential of $\frac{d\sigma}{\epsilon_0}$ . Hence the potential difference between the plates of $\frac{d\sigma}{\epsilon_0}$ . Remember, potentials are assigned relative to a fixed point, not arbitrarily.

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  • $\begingroup$ I understand it now. Potential is measured with one "fixed" point, i.e. +infinity assuming it anywhere far. Work done for the force to move the charge from the "fixed" point to any one plate say +ve is zero, then it will be $-qd$$\sigma$/$\epsilon_0$ for the other. Thanks for answering my query. $\endgroup$ – Apoorv Potnis Aug 17 '16 at 17:54

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