= Solution
Put the positive plate at $z=d/2$ and the negative plate at $z=-d/2$. Reflection through $z=0$, combined with reversal of the two plate charges, makes the tangential fields cancel at every point of the mid-plane. Thus \b[$E_x=E_y=0$] at the corner as well as inside the <capacitor>.
For the <uniform-charge estimate at a capacitor corner>, assemble four identical square pairs around their common corner. Their normal fields there are equal by rotation <symmetry> and add to the central field $\sigma/\epsilon_0$ of the larger plate pair. With the bulk parallel-plate relation $\sigma\simeq\epsilon_0V/d$, the conventional corner estimate is
$$
\boxed{\mathbf E_{\rm corner}\simeq\left(0,0,-\frac{V}{4d}\right)\quad\text{(uniform charge sheets).}}
$$
This quarter-field result uses prescribed uniform surface charge, not <symmetry> alone for finite equipotential conductors. Real conducting plates redistribute charge near their edges; $d\ll a$ justifies the bulk relation away from the edges but does not remove the local corner fringe. For that exact conductor problem only the zero tangential components follow from the stated <reflection symmetry>; its normal coefficient requires a fringing-field boundary-value solution. The standard parallel-plate formula explicitly neglects edge effects.<University Physics Volume 2 by Samuel J. Ling and others>[Section 8.1, parallel-plate field and the approximation of negligible edge effects]{ref}
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