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There's a caveat, which is often ignored, to the "easy" equation for parallel plate capacitors $C = \epsilon A / d$, namely that $d$ must be much smaller than the dimensions of the parallel plate.

Is there an equation that works for large $d$? I tried finding one and could not. (These two papers talk about fringing fields for disc-shape plates but don't seem to have a valid equation for $ d \to \infty$: http://www.santarosa.edu/~yataiiya/UNDER_GRAD_RESEARCH/Fringe%20Field%20of%20Parallel%20Plate%20Capacitor.pdf and http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.167.3361&rep=rep1&type=pdf)

My hand-waving intuition is that as $ d \to \infty$, $C$ should decrease to a constant value (which is the case for two spheres separated by a very large distance, where $C = 4\pi\epsilon_0/\left(1/R_1 + 1/R_2\right)$ ), because at large distances from each plate, the electric field goes as $1/R$, so the voltage line integral from one plate to the other will be a fixed constant proportional to charge $Q$.

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Your intuition is pretty solid. You can calculate the capacitance of an isolated sphere of radius $R$ relative to infinity from the fact that the potential at the surface of a sphere carrying charge $Q$ is $V=Q/(4\pi\epsilon_0R)$ and the definition of capacitance $Q=CV$. It's $$C=4\pi\epsilon_0R.$$ Capacitance is always $\epsilon$ times a length scale that corresponds to the size of the conductor and some unitless factor that depends on the geometry of the plates. In this particular case, as $d\gg a$, where $a$ is the widest dimension of the (assumed identical) capacitor plates, the capacitance will go like $$C=g\epsilon_0 a,$$ with $g$ the geometric factor that tends to some constant as $d/a\rightarrow\infty$ and to $\infty$ as $d/a\rightarrow 0$.

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As distance between plates increases, capacitance will decrease and , if it become infinitive the capacitance will become zero

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    $\begingroup$ The question focuses on a case where the distance is large, yet you seem to be referring to the usual equation $C=\frac{\epsilon \times A}{d}$. Please read this article (from the other answer). Also, it's usually recommended that you include more details in your answer. I'm quite sure this is incorrect, and there's no source/reasoning in this answer. $\endgroup$ – user191954 Jun 30 '18 at 13:06

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