- #1
mdmartin
- 3
- 0
Homework Statement
Introduction to Electrodynamics by Griffiths, 3rd ed. pg 141-2, Example 3.8:
An uncharged metal sphere of radius R is placed in an otherwise uniform electric field E=(0,0,E). [The field will push positive charge to the northern surface of the sphere, leaving a negative charge on the southern surface. This induced charge, in turn, distorts the field in the neighborhood of the sphere. Find the potential in the region outside the sphere.
Solution: V(r,θ)=-E*(r-R^2/r^2)*cos(θ)
The first term is due to the external field; the second term is due to the induced charge.
If you want to know the induced charge density it can be calculated by:
σ(θ)=-ε*(∂V/∂r)|r=R =ε*E*(1+2*R^3/r^3)*cos(θ)|r=R = 3*ε*E*cos(θ)
My question: If we are being asked the induced charge density why don't we just use the voltage due to the induced charge? The author (and in other places I've found) use the entire potential but I haven't seen an explanation as to why.
Homework Equations
σ(θ)=-ε*(∂V/∂r)
The Attempt at a Solution
I understand the math; I'm not understanding the physics. The problem statement says that the induced charge distorts the field in the neighborhood of the sphere. So what I think is happening is similar to a system dynamics problem where there is an initial overshoot, followed by bounce-back until there is a steady-state that is reached. Initially, the external field induces a charge which then changes the field near the sphere, which in turn modifies the induced charge, etc. until steady state is reached. So I think it may just be bad terminology because in the description of the solution to the potential there is a clear distinction between an effect due to the external field and one due to the (what I'm calling the newly created) induced charge whereas, in reality, the charge is entirely induced and, therefore, the entire expression for the voltage must be used. Does that make sense?