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Next: Dielectric Breakdown and the Up: Electrostatic Potential Previous: The Poisson Equation

Conductors and Charge Sharing

We recall from the preceding section on Gauss's Law that the electric field vanishes inside a conductor in electrostatic equilibrium. Wires carrying current are not in equilibrium and we'll learn about them shortly. Conductors in equilibrium are known by their boring nature - nothing moves, as there are no unbalanced forces or torques.

Well, if there is no field, and the field is the derivative of the potential:


\begin{displaymath}
\vec{E} = -\vec{\nabla} V
\end{displaymath} (17)

then the potential inside a conductor must be constant. This is very useful to know. We say that a conductor at ESE is equipotential. A charge can be moved around anywhere inside without doing any net work on it.

This has profound global consequences on the shape of the field outside the conductor. Remember, the potential at any part of the surface of the conductor is the work per unit charge bringing a charge in from infinity to any point on the surface. No matter which path followed. There may be places where the field is strong, there may be places where it is weak, but the path integral of the field in from infinity has the same value for all paths. Wow!

There are two important properties of charged conductors that can be deduced from this simple result. One is how charge is shared between two conductors that are electrically connected with a conductor and left until they arrive at mutual electrostatic equilibrium (at the same potential). The other is the way that surplus charge and field strength is arranged on various curved conducting surfaces at ESE as a function of the radius of curvature. Lots of useful physics follows from understanding this.

Suppose we have two conducting spheres of radius $R_1$ and $R_2$ and put a charge $Q$ on either one of them. We then separate them by a lot farther than we'd like to walk3 and connect them with a conducting wire long enough for the charge to be shared between them. How is the charge distributed?

Well, the two conductors must be equipotential. If we remove the wire after it has done its work and the two spheres are far enough apart that the potential of one can pretty much be neglected on the surface of the other, and if a charge $q$ was tranferred when they reached ESE:


\begin{displaymath}
V_1 = \frac{k(Q-q)}{R_1} = \frac{kq}{R_2} = V_2
\end{displaymath} (18)

from which we can see that


\begin{displaymath}
\frac{Q_1}{R_1} = \frac{Q_2}{R_2}
\end{displaymath} (19)

where $Q_1 = Q - q$ is the final charge on sphere 1 and $Q_2 = q$ is the final charge on sphere 2. The charge on a given connected conducting surface at ESE varies linearly with the radius of curvature of that surface. The larger the radius of curvature (the flatter) the more of the charge. The charge on a pie-plate mostly resides where the shape is flat, not on the curved surfaces. How interesting.

Even more interesting, consider the field strength near the surfaces 1 and 2.


\begin{displaymath}
E_{1,r} = \frac{ k Q_1 }{ R_1^2 }
\end{displaymath} (20)

and


\begin{displaymath}
E_{2,r} = \frac{ k Q_2 }{ R_2^2 }
\end{displaymath} (21)

But $Q_1 = \frac{Q_2 R_1}{R_2}$, so


\begin{displaymath}
E_{1,r} = \frac{ k Q_1 }{ R_1^2 } = \frac{ k Q_2 }{ R_1 R_2} =
E_{2,r}\frac{R_2}{R_1}.
\end{displaymath} (22)

Don't be confused. In english, this is just ``The field is stronger on the most sharply curved surface'' (which has the smallest radius of curvature). How interesting indeed.

Charge likes to concentrate on the least curved surfaces of a conductor, but the field strength associated with that charge is highest at points of the greatest curvature. This is because the charge drops off linearly in the curvature, but the field strength goes up inverse quadratically, basically. From this we deduce an important consequence:

Sharp points of a charged electrical conductor have extremely strong fields.

For example, the radius of curvature of a needle tip might be on the order of microns (millionths of a meter). If the needle has any surplus charge at all on it, The field strength there might be hundreds of times stronger than on the more gently rounded sides. The field can even be strong enough to rip apart air molecules.

Whoa! Way cool! How does this work?

I'm glad you asked.


next up previous
Next: Dielectric Breakdown and the Up: Electrostatic Potential Previous: The Poisson Equation
Robert G. Brown 2002-01-30