Electrodynamics/Laplace's Equation

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In theory, Coulombs Law can solve any electrostatic problem where the charge distribution is known. However, many cases, we do not know the distribution, as in the case when we have conductors. In these cases, we have to use other methods, such as solving the Laplace equation.

Laplace's Equation[edit]

If we use the Laplacian operator on the electric potential function over a region of the space where the charge density is not zero, we get a special equation called Poisson's Equation.


[Poisson's Equation]

\nabla^2 \varphi =
{\partial^2 \varphi\over \partial x^2 } +
{\partial^2 \varphi\over \partial y^2 } +
{\partial^2 \varphi\over \partial z^2 } = -\frac{\rho}{\epsilon}


If the charge density happens to be zero all over the region, the Poison's Equation becomes Laplace's Equation.

[Laplace's Equation]

\nabla^2 \varphi =
{\partial^2 \varphi\over \partial x^2 } +
{\partial^2 \varphi\over \partial y^2 } +
{\partial^2 \varphi\over \partial z^2 } = 0

Notice that there could still be an electric field even if the charge density is 0. The field, of course, originates from charges, but the charges could be outside the region we are interested in.

These equation are a second-order partial differential equation. Solutions to Laplace's Equation give the correct form of the electric potential in free space, satisfying the boundary conditions of the system under analysis. From the electrical potential φ one can calculate the electric field E using the gradient operator.

Of course, V is not unique. For any electric field, we have to set an arbitrary location (usually infinity) as having potential 0. However, it turns out that if you set the values of V at the boundaries of an region, then V is unique inside the region. This might seem obvious, but could be used to solve problems.

Solutions to Poisson's equation (and thus Laplace equation) with specified boundary conditions can be obtained in a number of ways:

Method of images[edit]

Suppose there is an infinite conducting plane at z=0. Suppose that there is a point charge q at (0,0,a), where a>0. What is the induced charge on the plane? What is the voltage distribution?

This seems like a hard problem to do with Coulombs law, or even solving Laplace's equation directly. However, we know that V=0 at infinity. Since the conducting plane will be an equipotential surface, we also know that the whole plane has V=0 (since otherwise, charge would flow to and from infinity). We thus know the boundary conditions.

Let us solve a different problem: Suppose there is a point charge q at (0,0,a) and another point charge -q at (0,0,-a). What is V? We know that V=\frac{1}{4 \pi \epsilon_0}\left[ \frac{-q}{\sqrt{x^2+y^2+(z-a)^2}}+\frac{q}{\sqrt{x^2+y^2+(z+a)^2}} \right] Notice that this V is 0 at infinity, and is 0 at the plane z=0. Thus, because of uniqueness of solution of Laplace equation after we specify boundary conditions, this V is the solution of original problem! Note the crucial role of the uniqueness theorem. Without it, no one would believe us if we claim that this is the solution of the first problem.

In general, if we have any object above an infinite conducting plane, to find the voltage distribution, we can use the method of images. We first reflect the object about the plane, and reverse the charges. This will create a mirror image (hence the name) that, together with the original charge, will create V=0 at the plane and at infinity, and thus, the potential set up by the object and the conducting plane is the same as the potential set up by the object and its mirror image.

Separation of variables[edit]

The method of images only work for certain distributions. However, the separation of variables is far more general. The premise is this: Suppose that V(x,y,z)=X(x)Y(y)Z(z). X(x) is free from y and z, and so on. Of course, this is not necessarily true. Then, Laplace equation becomes

YZ \frac{\partial^2 X}{\partial x^2}+XZ \frac{\partial^2 Y}{\partial y^2}+XY \frac{\partial^2 Z}{\partial z^2}=0

or

\frac{1}{X}\frac{\partial^2 X}{\partial x^2}+\frac{1}{Y}\frac{\partial^2 Y}{\partial y^2}+\frac{1}{Z}\frac{\partial^2 Z}{\partial z^2}=0

Now, suppose we vary x without changing y. Then, the expression\frac{1}{Y}\frac{\partial^2 Y}{\partial y^2}+\frac{1}{Z}\frac{\partial^2 Z}{\partial z^2} did not change, since that expression does not depend on x. Thus, we know that \frac{1}{X}\frac{\partial^2 X}{\partial x^2} must be constant. This becomes an ordinary differential equation, which is much easier to solve. In fact, it is either an exponential function, if the constant is positive, or a sinusoidal function, if the constant is negative. The sign of the constant has to be determined by the boundary conditions. Since the Laplace Equation is linear, all superpositions of solutions will be one as well, and so the general solution will be an infinite sum (usually a Fourier Series) with an infinite many parameters, which are to be determined from the boundary conditions.

Greens function approach[edit]

Numerical methods[edit]

Relaxation method[edit]

Finite element method[edit]