Would the capacitance be higher or lower than $\epsilon_0A/d$ if fringing effect was not neglected? If we were not to neglect the fringing of the electric field lines in a parallel-plate capacitor, would we calculate a higher or lower capacitance? I thought in energy related terms, now after considering increease in total energy and the formula for energy approximately will be $Q^2/2C$ so $C$ should be lower but i dont get if this is a exact (correct) reason ?( This was seen to be present in HRK chap 30 , ques 11)
 A: Fools rush in where angels fear to tread...
Let the positive plate (carrying charge $Q$) be surrounded by an imaginary Gaussian box, one face of which is a plane between the plates, parallel to them and extending well out beyond their edges. All the electric flux ($Q/\epsilon_0$) coming out from the box will come out through this plane, if we assume (realistically) that all the flux from the positive plate crosses the gap to the negative plate.
With no edge effects, all the flux would pass through an area of the plane equal to the plate area, $A$, so the field strength over that area of surface would be $Q/\epsilon_0A$. But with the 'bulging' of the field due to edge effects, the same flux, $Q/\epsilon_0$, will be spread over a wider area, and so the mean field strength will be less than $Q/\epsilon_0A$.
The pd, $V$, between the plates is $\int_s \mathbf E. d\mathbf s$ over any path running from one plate to the other. Therefore we can choose the easiest path – from the centre of one plate, straight across the gap to the centre of the other. Even over this central path, the field strength will be reduced below $Q/\epsilon_0$ by the bulging of the lines, as argued earlier. Therefore, for a given $Q$, $V$ will be reduced. This implies that the capacitance, $Q/V$, will be greater than $\epsilon_0 A/d$.
A: The actual capacitance is higher due to the edge effect because the field lines extend beyond the plates. See the following for a calculation https://www.calctown.com/calculators/edge-parallel
Hope this helps
