- #1
justaboy
- 13
- 0
Homework Statement
Calculate potential.
Last edited:
Why does the integral have to be evaluated over these limits? Do you have a ring of charge with inner diameter r1 and outer diameter r2? For an infinite plane, r goes to infinity which means that your potential diverges. This is to be expected for an infinite charged plate.justaboy said:Which isn't correct, considering the integral result has to be evaluated for r1 < r < r2. What's wrong here?
Limits of integration have physical significance. To integrate over the entire plane, you need to set the lower limit to zero and the upper limit to infinity.justaboy said:Oh, I don't mean r1 and r2 as being two separate radii, I just mean they are the limits of integration,
And where is the above potential zero, i.e. what is your reference point? It seems that it is zero on the plane at z=0. The summation over point charges assumes that the potential is zero at infinity, not at the origin. So where do you want your potential to be zero, at the plate or at infinity? No matter which you choose, you will end up with a very large (infinite) term to consider. This problem is artificial, in that there is no such thing as an infinite charge plate. One way to get around this is to integrate over r from zero to some radius R to get the potential V(z). Then consider what happens in the limit z/R << 1, i.e. do a Taylor expansion and see what you get.and no two limits of integration could give me the correct answer for the potential if that last step is correct. Is it only by evaluating over the surface that the potential diverges? Because the potential is finite and easily calculable using other integrals, like [tex]V(z)=\int{E*dz} = -\frac{n*z}{2\epsilon}[/tex]
Electric potential above a plane refers to the electrical potential energy per unit charge at any point above a conducting plane. It is a measure of the work required to bring a unit charge from infinity to that point above the plane.
The electric potential above a plane can be calculated using the formula V = kσz, where V is the potential, k is the Coulomb's constant, σ is the surface charge density of the plane, and z is the distance from the plane.
The electric potential above a plane is affected by the surface charge density of the plane, the distance from the plane, and the dielectric constant of the medium surrounding the plane.
Yes, the electric potential above a plane can be negative. This means that the work required to bring a unit charge from infinity to that point above the plane is negative, which indicates that the electric field is directed towards the plane.
The electric field above a plane is the negative gradient of the electric potential, meaning that it is the rate of change of the electric potential with respect to distance. This can be expressed as E = -∇V.