E-field greater at sharp edges?

[kended]

Hello everyone,

Could someone explain or direct me to some detailed explanations as to why is the electric field at a conductor greatest where the curvature is sharpest?

I just can't seem to grasp this concept. I understand sharp edges do have this property so is this that there is more charge density at those points? And if so, what would prevent another point close by (on a rounded surface) to have the same charge density given that both points are at the same KV potential?

In another words, how is a sharp edge (high curvature) affecting the local charge distribution at an atomic level?

Thanks for enlightening me :)

Regards

 

[quanticism]

The charge density is indeed higher at sharper regions.

I think this site is quite good at explaining why.
http://www.physicsclassroom.com/class/estatics/u8l4d.cfm

Just scroll down to the "Electric Fields and Surface Curvature" section and it'll explain how the sharper regions results in a smaller component of the electron-electron electrostatic repulsive force being directed parallel to the conductor surface.

 

[wreckemtech]

Where there is a flat surface, the charge distribution is even because the exposed surface area of the object per unit volume of space surrounding it is constant. Where you have a sharp edge, or a pointed protrusion on an object, there is more surface area exposed per unit volume of space immediately surrounding it - and so a higher charge density in that region of 3D space.

The link posted by the other poster is very good, and should explain the rest.

 

[jartsa]

Let's consider this charged object:

O-------o

Two metal balls and a thin metal rod between them.

Let's say the small ball has a radius that is half of the big one's radius. So its capacitance is half of the other one's capacitance. So its charge is half of the big ball's charge.I'm lazy, so I leave this question as an exersice for the reader:
Why is the electric field near the small ball's surface twice as large as the electric field near the large ball's surface?

 

[pmacias]

,

This is a great question and one that has puzzled many scientists for a long time. The reason why the electric field is greater at sharp edges is due to a phenomenon called "electric field enhancement." This is when the electric field is concentrated in a small area, such as at a sharp edge, causing it to be stronger than in other areas.

To understand this better, we need to look at the atomic level. Atoms are made up of positively charged protons and negatively charged electrons. When an electric field is applied, the electrons in the atom are pulled towards the positive end of the field and the protons are pushed towards the negative end. This creates a separation of charge within the atom, with the negative charge being closer to the positive end of the field.

Now, when we have a sharp edge, the electric field is more concentrated in that area, causing a greater separation of charge within the atoms. This results in a higher charge density at the sharp edge, making the electric field stronger.

In terms of your question about why another point on a rounded surface wouldn't have the same charge density, it's important to remember that the electric field is not uniform and can vary depending on the shape and curvature of the surface. So, while two points may be at the same potential, their charge distribution will be different due to the shape of the surface.

I hope this helps to clarify the concept of electric field enhancement at sharp edges. It's a complex phenomenon that is still being studied and understood by scientists. Keep exploring and asking questions – that's what being a scientist is all about!