# What is the reason for the edge effect in capacitors?

The electric field lines bend at the edges of the capacitors like this: What is the reason for this? Any quick explanation as to why they bend?

To move a charge from one plate to the other plate work needs to be done as there is a potential difference between the plates.

As the electric field is conservative the amount work done is independent of the path taken.

If there was no electric field in the volume outside the parallel plate arrangement then no work would be needed to move a charge from one plate to the other via a path exclusively outside which means that there there must be an electric field outside the parallel plate arrangement.

The work done in moving a charge $q$ from one plate to the other is $\displaystyle \int_{\text{plate 1}}^{\text{plate 2}} \vec E \cdot \,d\vec r$ where $\vec E$ is the electric field and $d\vec r$ is an increment of the path between taken between the two plates.

If the path is chosen such that it is always along an electric field line the work done is $\displaystyle \int_{\text{plate 1}}^{\text{plate 2}} E \,d r$.
So to maintain a constant amount of work done between the two plates if the path length is longer (which it is true for your top most electric field line) the smaller must be the magnitude of the electric field.

Noting that the drawn electric field lines are there for illustrative purposes and that a smaller value of the electric field means that the (flux) density of field lines is less the field lines outside the parallel plate arrangement must be spaced further apart which requires them to bow outwards.

There is a further factor at play in that the edges of the plates have a greater curvature and so the charge density in that region will be larger.
This means that the electric field near the edges of the plates is actually larger than the electric field between the plates which in terms of work done by moving a charge along an electric field line means that the electric field "remote" from the plates must be weaker (greater spacing of electric field lines)to maintain the constancy of the work done in moving charge between the two plates.

There are two images in the article Solving the Generalized Poisson Equation with the Finite-Difference Method which illustrates the true complexity of the parallel plate arrangement which you usually get around by assuming that the linear dimensions of the plates are much greater than their separation. The electric field lines are imaginary lines, where on each point of the line the intensity of the field is tangent to that point. If in all points of a given area the intensity has the same magnitute and direction, then the field lines are parallel, equidistant to each other.

In a static charge distribuition, electric field is conservative. As the electric field along the axis perpendicular to the plates decreases as we go far from center, the field parallel to plate will try to cancel this effect to keep the field conservative. Hence the bending.

Possible correction: the electric field just above the curvature ($$E_1$$) is the same as the electric field between the parallel plates ($$E_2$$). $$E_1$$ from the same plate reduces like a sphere of charge ($$E=(kQ)/R$$), while $$E_2$$ above the same plate remains ideally constant. The combined $$E_1$$ from both plates is constant, but is weaker than $$E_2$$. ($$d_1$$) is the $$E_1$$ bowed field line. ($$d_2$$) is the $$E_2$$ unbowed distance between the plates. For the combined plates: ($$E_1 < E_2$$) and ($$d_1 > d_2$$). The voltage is same: $$V=(E_1)(d_1)=(E_2)(d_2)$$.