We are only considering the 2 horizontally placed charges in the sharp edge. But there are charges present in the vertical direction too.. And if you consider forces between any 2 vertically placed charges in the sharp edge.. it will be just like the circle, a huge component of the force will be acting along the direction of the surface.
Yes! You are absolutely right. Any charges placed vertically will experience tangential forces separating them. The charges on the surface of the conductor will redistribute until electrostatic equilibrium is reached.
@@melvinvaughn So does this mean that the charges which are already present in the vertical direction to the incoming charges, will repel the incoming charges away due to the tangential forces, pushing them along the walls of the surface?
First after you inject charges in the object, the charges beigin to repel each other. Let's start from the more spread portion. Charges repel each other with greater tangential force and reach the pointed portion. Once they reach there, the electric force acting between this charge in the pointed portion and the charge in the spread portion is less (because they are more distanced now) than the force between the adjacent charges in the pointed portion (less distance between them). Then the perpendicular component wins.
@@melvinvaughnyes then how come are the charges at equilibrium at sharp corners so closely apart even when the parallel forces are the same as in the case of round sphere ?
Thank you for the great explanation! What order of magnitude greater charge density should we be thinking of at a tip with a radius of for example 50 nanometers, compared to the charge density in the rest of the bulk material?
Yes but you assumed the charges to be accumulated in the sharp points already, and you are trying to prove it, but that is a self-proof, as you have already assumed it to be correct. How do the charges reach that configuration is my question.
@ErenMC_ suppose that initially the conductor is neutral. Now you put some charge on the flatter part. So the charges will feel repulsion, and as explained in the video, most of the force will be tangential. Only the tangential component will result in movement. Tangential component being large, it will move away. In the sharper region, though the force is large, the tangential component is veryyy small So there will be less movement. Hence more charge comes to rest at sharper ends. This is what I understood....
We are only considering the 2 horizontally placed charges in the sharp edge. But there are charges present in the vertical direction too.. And if you consider forces between any 2 vertically placed charges in the sharp edge.. it will be just like the circle, a huge component of the force will be acting along the direction of the surface.
Yes! You are absolutely right. Any charges placed vertically will experience tangential forces separating them. The charges on the surface of the conductor will redistribute until electrostatic equilibrium is reached.
@@melvinvaughn So does this mean that the charges which are already present in the vertical direction to the incoming charges, will repel the incoming charges away due to the tangential forces, pushing them along the walls of the surface?
First after you inject charges in the object, the charges beigin to repel each other. Let's start from the more spread portion. Charges repel each other with greater tangential force and reach the pointed portion. Once they reach there, the electric force acting between this charge in the pointed portion and the charge in the spread portion is less (because they are more distanced now) than the force between the adjacent charges in the pointed portion (less distance between them). Then the perpendicular component wins.
@@melvinvaughnyes then how come are the charges at equilibrium at sharp corners so closely apart even when the parallel forces are the same as in the case of round sphere ?
Now I understand! Thank you very much!
Glad it helped!
Thank you for the great explanation! What order of magnitude greater charge density should we be thinking of at a tip with a radius of for example 50 nanometers, compared to the charge density in the rest of the bulk material?
It just seems like you've saved me,thanks god
Thanks 💗🌸🙏🏾🤩Got the intuition🤩🙏🏾🖼
Wonderful!
very helpful thanks buddy
Good
Good explaination. thanks
Yes but you assumed the charges to be accumulated in the sharp points already, and you are trying to prove it, but that is a self-proof, as you have already assumed it to be correct. How do the charges reach that configuration is my question.
He didn't assume it
It's experimentally proven. He's just explaining why.
@@donaldthegreat5809 so there is no theoretical proof?
@@ErenMC_ there are many. One of which is so elegantly described in this video.
@ErenMC_ suppose that initially the conductor is neutral. Now you put some charge on the flatter part. So the charges will feel repulsion, and as explained in the video, most of the force will be tangential. Only the tangential component will result in movement. Tangential component being large, it will move away. In the sharper region, though the force is large, the tangential component is veryyy small So there will be less movement.
Hence more charge comes to rest at sharper ends. This is what I understood....