Mean Value Theorem (Classical Electrodynamics) | Quantum Science Philippines

Mean Value Theorem (Classical Electrodynamics)

Roel N. Baybayon

MSPhysics1-MSU-IIT

—————————————————————————————

Problem 1.10
Prove the mean value theorem: For charge-free space the value of the electrostatic potential at any point is equal to the average of the potential over the surface of any sphere centered on that point.

Proof:

To prove this problem, we are going to use the Green’s  Second Identity which is given by, $\int_V \left(\phi\nabla^2 \psi-\psi \nabla^2 \phi\right)d^3v=\oint_S \left(\phi\frac{\partial\psi}{\partial n}-\psi\frac{\partial \phi}{\partial n}\right)da$.

Choosing $\phi=\Phi$ (the scalar potential), $\psi=\frac{1}{R}$ and $x'$ be the integration variable, we have $\int_V \left[\Phi(x')\nabla^2 \left(\frac{1}{R}\right)-\frac{1}{R} \nabla^2 \Phi(x')\right]d^3x'=\oint_S \left[\Phi(x')\frac{\partial}{\partial n}\left(\frac{1}{R}\right)-\frac{1}{R}\frac{\partial \Phi(x')}{\partial n}\right]d^2x'$.       Eq.(1)

Let us solve Eq.(1) term by term. For the first integral, $\int_V \Phi(x')\nabla^2 \left(\frac{1}{R}\right)d^3x'=-4\pi\int_V \Phi(x')\delta(x-x')d^3x'$,   since $\nabla^2 \frac{1}{R}=-4\pi\delta(x-x')$ $\int_V \Phi(x')\nabla^2 \left(\frac{1}{R}\right)d^3x'=-4\pi\Phi(x)$,  since $\int \delta (x-x')d^3x'=1$ if  V contains $x=x'$.

For the second integral, $-\int_V \frac{1}{R} \nabla^2 \Phi(x') d^3x'=\int_V \frac{1}{R} \frac{\rho}{\epsilon_o} d^3x'$,  since $\nabla^2 \Phi(x')=-\frac{\rho}{\epsilon_o}$.

But $\rho=0$ because there is no charge in the volume that we are integrating(Charge-free) . So the second integral becomes $-\int_V \frac{1}{R} \nabla^2 \Phi(x') d^3x'=0$.

For the third integral, $\begin{array}{rcl}\oint_S \Phi(x')\frac{\partial}{\partial n}\left(\frac{1}{R}\right) d^2x' & = & \oint_S \Phi(x')\left(-\frac{1}{R^2}\right) da\\ & =& -\frac{1}{R^2}\oint_S \Phi(x') d^2x'\end{array}$.

For the fourth integral, $-\oint_S\frac{1}{R} \frac{\partial \Phi(x')}{\partial n} d^2x'=-\oint_S\frac{1}{R}\left(\nabla \Phi(x')\cdot \hat{n'}\right) d^2x'$.

But $\vec{E}=-\nabla \Phi(x')$, then $-\oint_S\frac{1}{R} \frac{\partial \Phi(x')}{\partial n} d^2x'=\oint_S\frac{1}{R}\left(\vec{E}\cdot \hat{n'}\right) d^2x'$.

Using Divergence Theorem, $\int_V (\nabla\cdot\vec{A})d^3x=\oint_S(\vec{A}\cdot \hat{n}) da$,

the fourth integral becomes $-\oint_S\frac{1}{R} \frac{\partial \Phi(x')}{\partial n} d^2x'=\int_V\frac{1}{R}(\nabla\cdot \vec{E}) d^2x'$.

But $\nabla\cdot\vec{E}=\frac{\rho}{\epsilon_o}$ and $\rho=0$ (again, this is true for a charge-free volume! ), then the fourth integral  would be equal to zero, that is, $-\oint_S\frac{1}{R} \frac{\partial \Phi(x')}{\partial n} d^2x'=0$.

Thus, Eq.(1) is simplified into $-4\pi\Phi(x)=-\frac{1}{R^2}\oint_S \Phi(x') d^2x'$.

Hence, the scalar potential is then equal to $\Phi(x)=\frac{1}{4\pi R^2}\oint_S \Phi(x') d^2x'$.         Eq.(2)

Now, we have proven the mean value theorem.  Eq.(2) says that the potential at any point is equal to the average of the potential over the surface of any sphere centered on that point.

——————————————

2 Responses to “Mean Value Theorem (Classical Electrodynamics)”

1. Shabeeba shams Says:

Thank you.. You have explained it very well

2. Roel N. Baybayon Says:

You are so much welcome.