We are the **equations of Poisson and Laplace **for solving the problems related the electrostatic. These equations help to solve mainly, the problem in concern with the space change.

#### Space Change

For example, the space change exists in the space between the cathode and anode of a vacuum tube electrostatic valve.

#### Description

We can derive the equations of Poisson and Laplace by using Gauss’s theorem. Let us consider two parallel plates. Also, we place these plates vertically for better visualization. We name these plates as P and Q. We consider again, the plate P is positively and the plate Q is negatively charged.

We will also consider that, the same electric charge exists in the space between these two charged plates.

We can treat the charge between the plates as a space charge.

The space charge density is ρ, coulomb/meter^{3}. There is no variation of ρ vertically, it varies along the X-axis.

That means ρ varies depending upon the perpendicular distance from the plate P.

### Establishing the Poisson and Laplace Equations

Consider a strip in the space of thickness Δx at a distance x from the plate P.

Now, say the value of the electric field intensity at the distance x is E.

Now, the question is what will be the value of the electric field intensity at a distance x+Δx.

In general cases, the value of electric field intensity between two equally and oppositely charged plates remains constant throughout the space.

But here due to the presence of space charge, the electric field intensity is not constant. Moreover, it increases, with increasing distance from the plate P.

This is because as a point proceeds forward more electric charge comes behind it. Say the electric field intensity at the distance x + Δx is E + ΔE. So the rate of increase of electric field intensity along the x-axis is

As this is the rate of change so we can write the below-shown relation

So we can write,

Since the charge density (ρ) is constant along y and z-direction, the field intensity is also constant throughout the surface of the imaginary volume. So, the surface integral of the field intensity on the right surface is

Where A is the area of the surface. Similarly, the surface integral of the field intensity on the left surface is

Here the minus sign signifies that the direction of the field is inward from the left surface. Now, the closed integral of the intensity for the entire surface of the volume is

This equals to

Now, as per Gauss’s theorem, the closed surface integral of a volume with the electric field intensity is the ration of the changed trapped by the volume and the permittivity of the medium. So, we can write

Since ρ is the charge density, the charge trapped by the volume Δx A is Δx A ρ and ε is the permittivity of the medium. From here we can write,

Here, we have replaced E by the expression of voltage gradient. Now we write,

This is the Poisson Equation which tells that derivative of the voltage gradient in an electric field is minus space charge density by the permittivity. This is the one-dimensional equation when the field only changes along the x-axis. But when the value of the electric field intensity changes along all x, y, and z-direction then the equation will be,

Again, we denote,

So, finally, we can write

## Laplace Equations

If there is no space charge in the field, ρ = 0, so we can write the above equation as

This is the Laplace equation.

- Permittivity (Absolute Permittivity and Relative Permittivity)
- Electric Field Intensity or Electric Field Strength
- Gauss’s Theorem or Gauss’s Law
- Electrical Potential and Potential Difference or Voltage
- Breakdown Voltage and Dielectric Strength
- Electric Flux and Faraday’s Tubes
- Equations of Poisson and Laplace
- Coulomb’s Law Statement and Explanation