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In many books of general physics, they prove the equations for capacitance of a cap of parallel plates as if the electric field where constant everywhere but what would happen if we take the real electric field that is non uniform at the edge of the parallel plates, would the capacitance be greater of smaller?

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First think of initially putting charges $\pm Q$ uniformly on the two plates and then letting the system arrive at the final configuration in isolation. If fringe effects were ignored, the initial uniform charge distribution on the plates will remain unchanged. In this case, the potential difference between the two plates will be $V_\infty=Q/C_\infty$, where $C_\infty$ is the capacitance ignoring the fringe effect. Now let us put fringe effects into the argument. The charges will re-distribute such that the initial uniform charge density will become non-uniform. The total energy stored in the capacitor will necessarily decrease. This implies $$ \frac{Q^2}C < \frac{Q^2}{C_\infty}\implies \quad \boxed{C > C_\infty}\ , $$ where $C$ is the (true) capacitance that takes fringe effects into account. Here is a more quantitive computation. (I have implicitly assumed that $d< \sqrt{A}$, where $d$ is the separation of the plates and $A$ the area.)

It is possible to exactly solve for the exact potential difference using the method of conformal mapping when the geometry of the capacitor is that of a long strip.

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  • $\begingroup$ Why would the energy decrese, and why cant you use the formula $U=\frac{C\Delta V^2}{2}$? $\endgroup$ Oct 16, 2014 at 21:29
  • $\begingroup$ The energy will decrease because a system not in equilibrium (here the system is the configuration of charges) will go to one of lower energy when it attains equilibrium. $\endgroup$
    – suresh
    Oct 17, 2014 at 2:04

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