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Formula: **Wave Equation for E-Field**

$$\nabla^2 \, \class{purple}{\boldsymbol{E}} ~=~ \mu_0 \, \varepsilon_0 \, \frac{\partial^2 \class{purple}{\boldsymbol{E}}}{\partial t^2}$$

## Electric field

`$$ \class{purple}{\boldsymbol E} $$`Unit

`$$ \frac{\mathrm{V}}{\mathrm{m}} = \frac{\mathrm{N}}{\mathrm{C}} = \frac{\mathrm{kg} \, \mathrm{m}}{\mathrm{A} \, \mathrm{s}^3} $$`

Electric field specifies the force that would act on an electric charge if it were placed in a location where the electric field exists.

Solving the vectorial wave equation with the respective boundary conditions yields the electric field. For example, a simple solution of the wave equation yields the E-field in the form of plane waves.

## Vacuum Permittivity

`$$ \varepsilon_0 $$`Unit

`$$ \frac{\mathrm{As}}{\mathrm{Vm}} $$`

The vacuum permittivity is a physical constant that appears in equations involving electromagnetic fields. It has the following experimentally determined value:

`$$ \varepsilon_0 ~\approx~ 8.854 \, 187 \, 8128 ~\cdot~ 10^{-12} \, \frac{\mathrm{As}}{\mathrm{Vm}} $$`## Magnetic field constant

`$$ \mu_0 $$`Unit

`$$ \frac{\mathrm{Vs}}{\mathrm{Am}} = \frac{ \mathrm{kg} \, \mathrm{m} }{ \mathrm{A}^2 \, \mathrm{s}^2 } $$`

It is a natural constant and occurs whenever electromagnetic fields are involved. It has the value \( \mu_0 = 4\pi \cdot 10^{-7} \, \frac{ \text{N} }{ \text{A}^2 } \).

## Nabla operator

`$$ \nabla $$`Unit

`$$ \frac{1}{\mathrm m} $$`

The operator \(\nabla^2\) is applied to the electric field to differentiate the components of the E-field with respect to the spatial coordinates.

Applying \(\nabla^2\) to the E-field yields a vector quantity. The first component of this vector quantity is:
`
\[ \frac{\partial^2 E_x}{\partial x^2} + \frac{\partial^2 E_x}{\partial y^2} + \frac{\partial^2 E_x}{\partial z^2} ~=~ \mu_0 \, \varepsilon_0 \, \frac{\partial^2 E_x}{\partial t^2} \]
`