Lienard-Wiechert Potentials Calculator
Lienard-Wiechert Potentials Calculator computes potentials of a moving charge, visualizing scalar potential.
Formulas Used in Lienard-Wiechert Potentials Calculator
The calculator uses the following formulas for a point charge moving with constant velocity:
Scalar Potential:
\\[ \phi(\mathbf{r}, t) = \frac{1}{4\pi \epsilon_0} \frac{q}{|\mathbf{r} – \mathbf{r}_s(t_r)| (1 – \mathbf{\hat{n}} \cdot \boldsymbol{\beta}(t_r))} \\]Vector Potential:
\\[ \mathbf{A}(\mathbf{r}, t) = \frac{\mu_0}{4\pi} \frac{q \mathbf{v}(t_r)}{|\mathbf{r} – \mathbf{r}_s(t_r)| (1 – \mathbf{\hat{n}} \cdot \boldsymbol{\beta}(t_r))} \\]Retarded Time Condition:
\\[ t_r = t – \frac{|\mathbf{r} – \mathbf{r}_s(t_r)|}{c} \\]Definitions:
\\[ \mathbf{\hat{n}} = \frac{\mathbf{r} – \mathbf{r}_s(t_r)}{|\mathbf{r} – \mathbf{r}_s(t_r)|}, \quad \boldsymbol{\beta} = \frac{\mathbf{v}}{c} \\]Where:
- \\( \phi \\): Scalar potential (V)
- \\( \mathbf{A} \\): Vector potential (T·m/A)
- \\( q \\): Charge (C)
- \\( \mathbf{v} \\): Velocity of the charge (m/s)
- \\( \mathbf{r} \\): Observation point (m)
- \\( \mathbf{r}_s(t_r) \\): Source position at retarded time (m)
- \\( t_r \\): Retarded time (s)
- \\( c \approx 2.99792458 \times 10^8 \, \text{m/s} \\): Speed of light
- \\( \epsilon_0 = 8.854187817 \times 10^{-12} \, \text{F/m} \\): Permittivity of free space
- \\( \mu_0 = 4\pi \times 10^{-7} \, \text{T·m/A} \\): Permeability of free space
Example Calculations
Example 1: Standard Case
Input: q = 1e-9 C, v_x = 1e7 m/s, r_y = 0.1 m, t = 0 s
Solve for retarded time: \\( t_r = 0 – \frac{\sqrt{(v_x t_r)^2 + 0.1^2}}{c} \\)
\\[ t_r \approx -3.3356 \times 10^{-10} \, \text{s} \\] \\[ R = \sqrt{(1 \times 10^7 \cdot (-3.3356 \times 10^{-10}))^2 + 0.1^2} \approx 0.1 \, \text{m} \\] \\[ \mathbf{\hat{n}} \cdot \boldsymbol{\beta} = \frac{-1 \times 10^7 \cdot (-3.3356 \times 10^{-10})}{0.1} \cdot \frac{1 \times 10^7}{2.99792458 \times 10^8} \approx 0.03336 \\] \\[ \phi = \frac{1}{4\pi \cdot 8.854187817 \times 10^{-12}} \cdot \frac{1 \times 10^{-9}}{0.1 \cdot (1 – 0.03336)} \approx 92.52 \, \text{V} \\] \\[ A_x = \frac{4\pi \times 10^{-7}}{4\pi} \cdot \frac{1 \times 10^{-9} \cdot 1 \times 10^7}{0.1 \cdot (1 – 0.03336)} \approx 1.034 \times 10^{-4} \, \text{T·m/A} \\]Result: φ ≈ 92.52 V, A_x ≈ 1.034e-4 T·m/A at t = 0 s