# Electric Field perpendicular to surface

1. Oct 16, 2004

### Cyrus

hello,

Sorry to ask this question again, but im having a little trouble with it still. I found a pic on a website that should be helpful.

According to my physics text, the electric field has to emerge perpendicular from the surface. I can see how this is true. Since the surface is at equipotential, it takes no work to move along the surface. Because it takes no work the change in potential at two points must be zero, and E*dl must be perpendicular at all points. But what does that mean for a charge placed outside the body? In my pic you can see a charged particle outside the body. How would you draw the electric field there? Is the electric field just an extension of the electric field perpendicular at exactly that point out in space? (in my picture i exteneded the blue arrow wher the point charge would project itself onto the surface, and exteded it to reach the point charge, (the dotted brown extension arrow)).

Last edited: Dec 26, 2005
2. Oct 16, 2004

### Cyrus

I also have a follow up question. Lets say that we have a sphere, but at exactly one point on the sphere the potential is different than everywhere else on the sphere. Then that means the charge at the sphere (the electrons), would move towards that one spot. Obviously, all the electrons cant occupy the space since that area of potential difference is only a point. So would part of the sphere be at one potential and part of the sphere be at a different potential? Is it meaningful to ask a question like this, or is it wrong to say a place of different potentail on a sphere?

3. Oct 16, 2004

### Clausius2

Respect to your first question, I think the electric field will be complex but the arrows will curve in such a manner that makes possible a perpendicular entrance of E if the charge is negative. Your question is not trivial because the electric field will be distorsioned owing to the presence of the charge. Maybe I misunderstood your question.

Respect to your second question, I think it has no sense thinking in such potential distribution. Keep in mind the potential would not be derivable at all in that point. If so, the electric field would be infinite at that point. It's impossible to obtain an infinite electric field, therefore your question has no sense.