Why do sharp edges of a metallic conductor have more charges than flat edges? charges accumulate on sharp edges more than flat edges.


Like charges on the surface of a conductor repel each other. Let us consider two identical charges $q$ in points A and B on the surface of a conductor in an area where the curvature of the surface is $R$, so OA=OB=R (see the picture). Let us assume that the length of arc AB is $l$, and $l\ll R$. Then angle $\alpha=l/R$. The distance between the charges AB=$\delta=R\cdot2\sin\frac{\alpha}{2}$. The Coulomb force between the charges is directed along AB, and its magnitude is $\frac{q^2}{\delta^2}$. However, the normal component of the Coulomb force cannot move the charges, so we need to calculate the tangential component of the Coulomb force. The angle between AB and the red dashed tangent to the curve is $\frac{\alpha}{2}$, so the tangential projection of the Coulomb force is $$\frac{q^2}{\delta^2}\cos\frac{\alpha}{2}=\frac{q^2}{R^2\cdot 4\sin^2\frac{\alpha}{2}}\cos\frac{\alpha}{2}=\frac{q^2}{\frac{l^2}{4 x^2}\cdot 4\sin^2 x}\cos x=\frac{q^2}{l^2}\frac{x^2\cos x}{\sin^2 x},$$ where $x=\frac{l}{2 R}\ll 1$.
Let us use the following approximation: $$\frac{x^2\cos x}{\sin^2 x}\approx \frac{x^2(1-\frac{x^2}{2})}{(x-\frac{x^3}{6})^2}\approx\frac{x^2(1-\frac{x^2}{2})}{x^2-\frac{x^4}{3}}=\frac{1-\frac{x^2}{2}}{1-\frac{x^2}{3}}\approx(1-\frac{x^2}{2})(1+\frac{x^2}{3})\approx$$ $$\approx 1-\frac{x^2}{6}. $$Thus, if the arc length $l$ is constant and the radius of curvature decreases, $x$ increases, and the tangential component of the Coulomb force between the charges decreases. Therefore, one can expect that the charge density will be greater in the areas where the curvature is greater.

