Charge density higher at sharp points

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1. Jul 16, 2015

Premanand

Hi... I want to know why charge density is higher at sharp points in a conductor? I have gone through the analogy of two spheres connected by a wire... But is there any other explanation which is not specific to spheres...?

2. Jul 16, 2015

ShayanJ

Hi
Its simple to explain it roughly. The surface of a conductor is an equipotential surface regardless of the electric field the conductor is in. But that is achieved by the induced surface charge distribution you're talking about. Now if some part of the conductor's surface is sharp, then the equipotential surface has to be sharp too, which means it should change direction faster. But that change in direction is actually a spatial derivative of the potential which is actually the definition of electric field. So in those sharp parts of the conductor, the change is faster and so the electric field should be larger. But a larger electric field needs a larger concentration of charges.

3. Jul 16, 2015

Premanand

Thank you... One more question... I was taught that electric potential is the energy that a charge would posses... So it means if the potential is high , the charge will have enough energy to cross the barrier of dielectric... However, even if the potential is low, if the points are sharp, it can still breakdown because the field strength is higher at that point.... How can that be explained qualitatively ? Electric potential is like wound up spring, so it is OK to assume that higher the potential, higher the energy per charge. So I assume this is the energy occupied by the charge by virtue of its position. How can we explain field concentration ? what is it analogous to ? Is it like steepness of terrain?
Sorry if the question is too basic.

4. Jul 16, 2015

ShayanJ

The point is, its wrong to say that "electric potential is the energy that a charge would posses". Potential energy(either electric or gravitational), is a property of all the participants in the interactions, not just a single one or even a group of them.
Another important point, is that the potential itself has no physical meaning and only the potential difference is significant which, through the definition of electric field as the spatial derivative of potential, means its the electric field that is significant. So its irrelevant whether potential is high or low, it just matters that the potential difference is enough or not.

5. Jul 17, 2015

Premanand

The energy required to move the charge through an equipotential surface is zero. Which means one can move the charge to sharp points at same potential without the expense of energy ? So I concentrated the charge at sharp point without expense of energy by virtue of geometry and let it rip through the dielectric at the same potential :)

6. Jul 17, 2015

ShayanJ

True.
You can bring the charge to the sharp point of the conductor but you can't "let it rip through the dielectric at the same potential" simply because the points out of the conductor are at different potentials. Remember that the surface of the conductor is an equipotential surface but if you want to go out of the conductor, your motion should have a component orthogonal to the surface which means you're going out of the equipotential surface and so the potential energy is changing.

7. Jul 17, 2015

Premanand

Sorry about that... I did not mean the dielectric is at that the same potential as the conductor, I meant that the potential of the conductor is unchanged. For instance, If I started with two plates, one at 80 kV and the other grounded separated by air of 75 mm. Say 75 mm is huge enough so that there is no breakdown at 80 kV. The field as you say are pointed towards grounded plate and the potential will decrease towards the grounded plate. Now if I place pointed edges on the plates, the geometry allows me to have a break down at 80 kV which was not previously possible. So I meant, I did not increase the potential of the plate but I increased the concentration of charges by introducing sharp points and shot out the charges. Is it correct?

8. Jul 17, 2015

ShayanJ

Yeah, that's correct.

9. Jul 17, 2015

Premanand

So if the breakdown actually happens, then there will be flow of current provided the voltage is maintained by a power supply. However, current is not due to fast moving electrons (which are the charge carriers) but it is due to the slow moving electrons transferring energy at a quicker pace. What is that transferred energy exactly ? Is it the momentum of electrons transferred by collision or a photon or something similar ?

10. Jul 17, 2015

ShayanJ

That's wrong.
I'm don't understand what you mean by fast and slow electrons.

11. Jul 18, 2015