Electromagnetic field of ball

Electromagnetic field of ball

Electromagnetic field of ball, if we do not take into account the influence of external fields and the environment, is completely determined by the equation of motion of electric charges in the substance of the ball and by Maxwell's equations. Due to the symmetry of the ball, the components of the electromagnetic field are most simply expressed in terms of spherical coordinates $~r,\;\theta ,\;\phi .$ However, in spherical coordinates, the scalar and vector Laplace operators, the gradient of a scalar function, the divergence, and the curl of a three-dimensional vector do not have the same form as their corresponding expressions in Cartesian coordinates (see Del in cylindrical and spherical coordinates).

For the electric scalar potential $~\varphi$ and for the magnetic vector potential $~{\mathbf A}$ of a uniformly charged ball rotating about its axis, the following equations follow from Maxwell's equations in case of homogeneous and isotropic media inside and outside the ball:

$~\Delta \varphi -{\frac {\varepsilon \mu }{c^{2}}}{\frac {\partial ^{2}\varphi }{\partial t^{2}}}=-{\frac {\gamma \rho _{{0q}}}{\varepsilon \varepsilon _{0}}},\qquad (1)$

$~\Delta {\mathbf A}-{\frac {\varepsilon \mu }{c^{2}}}{\frac {\partial ^{2}{\mathbf A}}{\partial t^{2}}}=-{\frac {\mu \gamma \rho _{{0q}}}{\varepsilon _{0}c^{2}}}{\mathbf v},\qquad (2)$

where $~\Delta$ is the Laplace operator; $~\gamma$ is the Lorentz factor; $~\rho _{{0q}}$ is the invariant charge density of the ball; $~\varepsilon$ is the relative permittivity; $~\varepsilon _{0}$ is the electric constant; is the speed of light; $~\mu$ is the relative permeability; $~{\mathbf v}$ is the linear velocity of rotation of a charged element of matter taken in the volume of the ball.

In this case, the Lorentz gauge condition is written as follows:

$~\nabla \cdot {\mathbf A}+{\frac {\varepsilon \mu }{c^{2}}}{\frac {\partial \varphi }{\partial t}}=0.\qquad (3)$

The electric field strength $~{\mathbf E}$ and the magnetic field $~{\mathbf B}$ are expressed in terms of the scalar and vector potentials:

$~{\mathbf E}=-\nabla \varphi -{\frac {\partial {\mathbf A}}{\partial t}}.\qquad (4)$

$~{\mathbf B}=\nabla \times {\mathbf A}.\qquad (5)$

When the ball is rotating at a constant angular velocity, the field is stationary and does not depend on time. This implies that all time derivatives in (1-4) are equal to zero.

The relative permittivity $~\varepsilon$ in the ball's substance and in the surrounding medium may have different values. The same applies to the relative permeability $~\mu$ .

For the sake of simplicity, the results presented below correspond to the values $~\varepsilon =1$ , $~\mu =1$ for the case where phenomena such as polarizability, magnetization and electrical conductivity are not taken into account inside and outside the ball. It is also assumed that all electromagnetic quantities are independent of time.

Contents

1 Fixed ball
2 Rotating ball

2.1 Scalar potential and electric field
2.2 Vector potential and magnetic field

3 References
4 See also
5 External links

Fixed ball

For a fixed uniformly charged ball, in (1) and in (4), $~\gamma =1$ , and the electric potential and electric field strength inside the ball are equal:^[1] ^[2]

$~\varphi _{i}={\frac {\rho _{{0q}}\left(3a^{2}-r^{2}\right)}{6\varepsilon _{0}}},$

$~{\mathbf E}_{i}={\frac {\rho _{{0q}}r}{3\varepsilon _{0}}}{\mathbf e}_{r},$

where is the radius of the ball; $~{\mathbf e}_{r}$ is a unit vector directed along the radial coordinate .

At the center of the ball, where , the potential $~\varphi _{i}$ has a maximum value, and on the surface of the ball, where , the potential decreases by one and a half. The internal electric field $~{\mathbf E}_{i}$ is zero at the center of the ball and increases proportionally to the radial coordinate .

The corresponding external electric potential and electric field strength outside the ball are as follows:

$~\varphi _{o}={\frac {\rho _{{0q}}a^{3}}{3\varepsilon _{0}r}},$

$~{\mathbf E}_{o}={\frac {\rho _{{0q}}a^{3}}{3\varepsilon _{0}r^{2}}}{\mathbf e}_{r}.$

Due to the absence of motion of electric charges in the fixed ball, the vector potential and magnetic induction are equal to zero throughout the system.

Rotating ball

Scalar potential and electric field

When a ball rotates with a constant angular velocity $~\omega$ , the Lorentz factor of the charged particles becomes a function of the radial coordinate and the angle $~\theta$ :

$~\gamma ={\frac {1}{{\sqrt {1-v^{2}/c^{2}}}}}={\frac {1}{{\sqrt {1-{\frac {\omega ^{2}r^{2}\sin ^{2}\theta }{c^{2}}}}}}}.$

Taking this into account, the solution of equation (1) for the scalar potential, as well as equation (4) for the components of the electric field strength inside a rotating uniformly charged ball in spherical coordinates, is as follows: ^[3]

$~\varphi _{i}\approx {\frac {\rho _{{0q}}a^{2}}{2\varepsilon _{0}}}+{\frac {\rho _{{0q}}\omega ^{2}a^{4}}{12c^{2}\varepsilon _{0}}}+{\frac {\rho _{{0q}}\omega ^{4}a^{6}}{30c^{4}\varepsilon _{0}}}+{\frac {c^{2}\rho _{{0q}}}{\omega ^{2}\varepsilon _{0}}}\left[{\sqrt {1-{\frac {\omega ^{2}r^{2}\sin ^{2}\theta }{c^{2}}}}}-1+\ln 2-\ln \left(1+{\sqrt {1-{\frac {\omega ^{2}r^{2}\sin ^{2}\theta }{c^{2}}}}}\right)\right]-$

$~-\left({\frac {\rho _{{0q}}}{12\varepsilon _{0}}}+{\frac {\rho _{{0q}}\omega ^{2}a^{2}}{60c^{2}\varepsilon _{0}}}+{\frac {\rho _{{0q}}\omega ^{4}a^{4}}{140c^{4}\varepsilon _{0}}}\right)r^{2}\left(3\cos ^{2}\theta -1\right)+\left({\frac {\rho _{{0q}}\omega ^{2}}{1120c^{2}\varepsilon _{0}}}+{\frac {\rho _{{0q}}\omega ^{4}a^{2}}{1680c^{4}\varepsilon _{0}}}\right)r^{4}\left(35\cos ^{4}\theta -30\cos ^{2}\theta +3\right)-$

$~-{\frac {\rho _{{0q}}\omega ^{4}r^{6}}{22176c^{4}\varepsilon _{0}}}\left(231\cos ^{6}\theta -315\cos ^{4}\theta +105\cos ^{2}\theta -5\right).$

$~E_{{ir}}\approx \left({\frac {\rho _{{0q}}}{6\varepsilon _{0}}}+{\frac {\rho _{{0q}}\omega ^{2}a^{2}}{30c^{2}\varepsilon _{0}}}+{\frac {\rho _{{0q}}\omega ^{4}a^{4}}{70c^{4}\varepsilon _{0}}}\right)r\left(3\cos ^{2}\theta -1\right)+{\frac {\rho _{{0q}}r\sin ^{2}\theta }{\varepsilon _{0}\left(1+{\sqrt {1-{\frac {\omega ^{2}r^{2}\sin ^{2}\theta }{c^{2}}}}}\right)}}-$

$~-\left({\frac {\rho _{{0q}}\omega ^{2}}{280c^{2}\varepsilon _{0}}}+{\frac {\rho _{{0q}}\omega ^{4}a^{2}}{420c^{4}\varepsilon _{0}}}\right)r^{3}\left(35\cos ^{4}\theta -30\cos ^{2}\theta +3\right)+{\frac {\rho _{{0q}}\omega ^{4}}{3696c^{4}\varepsilon _{0}}}r^{5}\left(231\cos ^{6}\theta -315\cos ^{4}\theta +105\cos ^{2}\theta -5\right).$

$~E_{{i\theta }}\approx -\left({\frac {\rho _{{0q}}}{2\varepsilon _{0}}}+{\frac {\rho _{{0q}}\omega ^{2}a^{2}}{10c^{2}\varepsilon _{0}}}+{\frac {3\rho _{{0q}}\omega ^{4}a^{4}}{70c^{4}\varepsilon _{0}}}\right)r\sin \theta \cos \theta +{\frac {\rho _{{0q}}r\sin \theta \cos \theta }{\varepsilon _{0}\left(1+{\sqrt {1-{\frac {\omega ^{2}r^{2}\sin ^{2}\theta }{c^{2}}}}}\right)}}+$

$~+\left({\frac {\rho _{{0q}}\omega ^{2}}{56c^{2}\varepsilon _{0}}}+{\frac {\rho _{{0q}}\omega ^{4}a^{2}}{84c^{4}\varepsilon _{0}}}\right)r^{3}\sin \theta \cos \theta \left(7\cos ^{2}\theta -3\right)-{\frac {\rho _{{0q}}\omega ^{4}}{528c^{4}\varepsilon _{0}}}r^{5}\sin \theta \cos \theta \left(33\cos ^{4}\theta -30\cos ^{2}\theta +5\right).$

$~E_{{i\phi }}=0.$

Outside the rotating ball, the scalar potential and electric field strength are given by the following formulas:

$~\varphi _{o}\approx {\frac {1}{r}}\left({\frac {\rho _{{0q}}a^{3}}{3\varepsilon _{0}}}+{\frac {\rho _{{0q}}\omega ^{2}a^{5}}{15c^{2}\varepsilon _{0}}}+{\frac {\rho _{{0q}}\omega ^{4}a^{7}}{35c^{4}\varepsilon _{0}}}\right)-{\frac {1}{r^{3}}}\left({\frac {\rho _{{0q}}\omega ^{2}a^{7}}{210c^{2}\varepsilon _{0}}}+{\frac {\rho _{{0q}}\omega ^{4}a^{9}}{315c^{4}\varepsilon _{0}}}\right)\left(3\cos ^{2}\theta -1\right)+$

$~+{\frac {\rho _{{0q}}\omega ^{4}a^{{11}}}{9240c^{4}\varepsilon _{0}r^{5}}}\left(35\cos ^{4}\theta -30\cos ^{2}\theta +3\right).$

$~E_{{or}}\approx {\frac {1}{r^{2}}}\left({\frac {\rho _{{0q}}a^{3}}{3\varepsilon _{0}}}+{\frac {\rho _{{0q}}\omega ^{2}a^{5}}{15c^{2}\varepsilon _{0}}}+{\frac {\rho _{{0q}}\omega ^{4}a^{7}}{35c^{4}\varepsilon _{0}}}\right)-{\frac {1}{r^{4}}}\left({\frac {\rho _{{0q}}\omega ^{2}a^{7}}{70c^{2}\varepsilon _{0}}}+{\frac {\rho _{{0q}}\omega ^{4}a^{9}}{105c^{4}\varepsilon _{0}}}\right)\left(3\cos ^{2}\theta -1\right)+$

$~+{\frac {\rho _{{0q}}\omega ^{4}a^{{11}}}{1848c^{4}\varepsilon _{0}r^{6}}}\left(35\cos ^{4}\theta -30\cos ^{2}\theta +3\right).$

$~E_{{o\theta }}\approx -{\frac {1}{r^{4}}}\left({\frac {\rho _{{0q}}\omega ^{2}a^{7}}{35c^{2}\varepsilon _{0}}}+{\frac {2\rho _{{0q}}\omega ^{4}a^{9}}{105c^{4}\varepsilon _{0}}}\right)\sin \theta \cos \theta +{\frac {\rho _{{0q}}\omega ^{4}a^{{11}}}{462c^{4}\varepsilon _{0}r^{6}}}sin\theta \cos \theta \left(7\cos ^{2}\theta -3\right).$

$~E_{{o\phi }}=0.$

Vector potential and magnetic field

The components of the vector potential $~{\mathbf A}$ and magnetic induction $~{\mathbf B}$ inside a uniformly charged ball rotating about its axis are given by equations (2) and (5) as follows: ^[4]

$~{\mathbf A}_{{ir}}=0.\qquad {\mathbf A}_{{i\theta }}=0.$

$~{\mathbf A}_{{i\phi }}\approx {\frac {c^{2}\rho _{{0q}}}{3\varepsilon _{0}\omega ^{3}r\sin \theta }}-{\frac {c^{2}\rho _{{0q}}\left(1-{\frac {\omega ^{2}r^{2}\sin ^{2}\theta }{c^{2}}}\right)^{{3/2}}}{3\varepsilon _{0}\omega ^{3}r\sin \theta }}-{\frac {\rho _{{0q}}r\sin \theta }{2\varepsilon _{0}\omega }}\left(1-{\frac {\omega ^{2}a^{2}}{3c^{2}}}-{\frac {\omega ^{4}a^{4}}{15c^{4}}}-{\frac {\omega ^{6}a^{6}}{35c^{6}}}\right)-$

$~-{\frac {\rho _{{0q}}\omega r^{3}\sin \theta \left(5\cos ^{2}\theta -1\right)}{40c^{2}\varepsilon _{0}}}\left(1+{\frac {2\omega ^{2}a^{2}}{7c^{2}}}+{\frac {\omega ^{4}a^{4}}{7c^{4}}}\right)+{\frac {\rho _{{0q}}\omega ^{3}r^{5}\sin \theta \left(21\cos ^{4}\theta -14\cos ^{2}\theta +1\right)}{1008c^{4}\varepsilon _{0}}}\left(1+{\frac {9\omega ^{2}a^{2}}{11c^{2}}}\right)-$

$~-{\frac {\rho _{{0q}}\omega ^{5}r^{7}\sin \theta \left(429\cos ^{6}\theta -495\cos ^{4}\theta +135\cos ^{2}\theta -5\right)}{54912c^{6}\varepsilon _{0}}}.$

$~{\mathbf B}_{{ir}}\approx {\frac {\rho _{{0q}}\cos \theta {\sqrt {1-{\frac {\omega ^{2}r^{2}\sin ^{2}\theta }{c^{2}}}}}}{\varepsilon _{0}\omega }}-{\frac {\rho _{{0q}}\cos \theta }{\varepsilon _{0}\omega }}\left(1-{\frac {\omega ^{2}a^{2}}{3c^{2}}}-{\frac {\omega ^{4}a^{4}}{15c^{4}}}-{\frac {\omega ^{6}a^{6}}{35c^{6}}}\right)-$

$~-{\frac {\rho _{{0q}}\omega r^{2}\cos \theta \left(5\cos ^{2}\theta -3\right)}{10c^{2}\varepsilon _{0}}}\left(1+{\frac {2\omega ^{2}a^{2}}{7c^{2}}}+{\frac {\omega ^{4}a^{4}}{7c^{4}}}\right)+{\frac {\rho _{{0q}}\omega ^{3}r^{4}\cos \theta \left(63\cos ^{4}\theta -70\cos ^{2}\theta +15\right)}{504c^{4}\varepsilon _{0}}}\left(1+{\frac {9\omega ^{2}a^{2}}{11c^{2}}}\right)-$

$~-{\frac {\rho _{{0q}}\omega ^{5}r^{6}\cos \theta \left(429\cos ^{6}\theta -693\cos ^{4}\theta +315\cos ^{2}\theta -35\right)}{6864c^{6}\varepsilon _{0}}}.$

$~{\mathbf B}_{{i\theta }}\approx -{\frac {\rho _{{0q}}\sin \theta {\sqrt {1-{\frac {\omega ^{2}r^{2}\sin ^{2}\theta }{c^{2}}}}}}{\varepsilon _{0}\omega }}+{\frac {\rho _{{0q}}\sin \theta }{\varepsilon _{0}\omega }}\left(1-{\frac {\omega ^{2}a^{2}}{3c^{2}}}-{\frac {\omega ^{4}a^{4}}{15c^{4}}}-{\frac {\omega ^{6}a^{6}}{35c^{6}}}\right)+$

$~+{\frac {\rho _{{0q}}\omega r^{2}\sin \theta \left(5\cos ^{2}\theta -1\right)}{10c^{2}\varepsilon _{0}}}\left(1+{\frac {2\omega ^{2}a^{2}}{7c^{2}}}+{\frac {\omega ^{4}a^{4}}{7c^{4}}}\right)-{\frac {\rho _{{0q}}\omega ^{3}r^{4}\sin \theta \left(21\cos ^{4}\theta -14\cos ^{2}\theta +1\right)}{168c^{4}\varepsilon _{0}}}\left(1+{\frac {9\omega ^{2}a^{2}}{11c^{2}}}\right)+$

$~+{\frac {\rho _{{0q}}\omega ^{5}r^{6}\sin \theta \left(429\cos ^{6}\theta -495\cos ^{4}\theta +135\cos ^{2}\theta -5\right)}{6864c^{6}\varepsilon _{0}}}.$

$~{\mathbf B}_{{i\phi }}=0.$

The components of the external vector potential and magnetic induction of a rotating charged ball are determined by the following formulas:

$~{\mathbf A}_{{or}}=0.\qquad {\mathbf A}_{{o\theta }}=0.$

$~{\mathbf A}_{{o\phi }}\approx {\frac {\rho _{{0q}}\omega a^{5}}{15c^{2}\varepsilon _{0}}}{\frac {\sin \theta }{r^{2}}}\left(1+{\frac {2\omega ^{2}a^{2}}{7c^{2}}}+{\frac {\omega ^{4}a^{4}}{7c^{4}}}\right)-{\frac {\rho _{{0q}}\omega ^{3}a^{9}}{630c^{4}\varepsilon _{0}}}{\frac {\sin \theta \left(5\cos ^{2}\theta -1\right)}{r^{4}}}\left(1+{\frac {9\omega ^{2}a^{2}}{11c^{2}}}\right)+$

$~+{\frac {\rho _{{0q}}\omega ^{5}a^{{13}}}{8008c^{6}\varepsilon _{0}}}{\frac {\sin \theta \left(21cos^{4}\theta -14\cos ^{2}\theta +1\right)}{r^{6}}}.$

$~{\mathbf B}_{{or}}\approx {\frac {2\rho _{{0q}}\omega a^{5}}{15c^{2}\varepsilon _{0}}}{\frac {\cos \theta }{r^{3}}}\left(1+{\frac {2\omega ^{2}a^{2}}{7c^{2}}}+{\frac {\omega ^{4}a^{4}}{7c^{4}}}\right)-{\frac {2\rho _{{0q}}\omega ^{3}a^{9}}{315c^{4}\varepsilon _{0}}}{\frac {\cos \theta }{r^{5}}}\left(5cos^{2}\theta -3\right)\left(1+{\frac {9\omega ^{2}a^{2}}{11c^{2}}}\right)+$

$~+{\frac {\rho _{{0q}}\omega ^{5}a^{{13}}}{4004c^{6}\varepsilon _{0}}}{\frac {\cos \theta }{r^{7}}}\left(63cos^{4}\theta -70cos^{2}\theta +15\right).$

$~{\mathbf B}_{{o\theta }}\approx {\frac {\rho _{{0q}}\omega a^{5}}{15c^{2}\varepsilon _{0}}}{\frac {\sin \theta }{r^{3}}}\left(1+{\frac {2\omega ^{2}a^{2}}{7c^{2}}}+{\frac {\omega ^{4}a^{4}}{7c^{4}}}\right)-{\frac {\rho _{{0q}}\omega ^{3}a^{9}}{210c^{4}\varepsilon _{0}}}{\frac {\sin \theta }{r^{5}}}\left(5cos^{2}\theta -1\right)\left(1+{\frac {9\omega ^{2}a^{2}}{11c^{2}}}\right)+$

$~+{\frac {5\rho _{{0q}}\omega ^{5}a^{{13}}}{8008c^{6}\varepsilon _{0}}}{\frac {\sin \theta }{r^{7}}}\left(21cos^{4}\theta -14cos^{2}\theta +1\right).$

$~{\mathbf B}_{{o\phi }}=0.$

The divergence of the vector $~{\mathbf A}$ in spherical coordinates is written as follows:

$~\nabla \cdot {\mathbf A}={\frac {1}{r^{2}}}{\frac {\partial \left(r^{2}A_{r}\right)}{\partial r}}+{\frac {1}{r\sin \theta }}{\frac {\partial \left(A_{\theta }\sin \theta \right)}{\partial \theta }}+{\frac {1}{r\sin \theta }}{\frac {\partial A_{\phi }}{\partial \phi }}.$

Substituting the components of the internal vector potential $~{\mathbf A}_{i}$ and the components of the external vector potential $~{\mathbf A}_{o}$ for $~{\mathbf A}$ in expression for divergence gives the relations $~\nabla \cdot {\mathbf A}_{i}=0$ and $~\nabla \cdot {\mathbf A}_{o}=0$ , since the vector potential does not depend on the angle $~\phi$ . These relations correspond to the Lorentz gauge condition (3), since the scalar potential $~\varphi$ is independent of time.

References

Feynman R., Leighton R. and Sands M. The Feynman Lectures on Physics. Vol. 2 (1964). Addison-Wesley, Massachusetts, Palo Alto, London.
Sergey G. Fedosin. The Electromagnetic Field of a Rotating Relativistic Uniform System. Chapter 2 in the book: Horizons in World Physics. Volume 306. Edited by Albert Reimer, New York, Nova Science Publishers Inc, pp. 53-128 (2021), ISBN: 978-1-68507-077-9, 978-1-68507-088-5 (e-book). https://doi.org/10.52305/RSRF2992.
Fedosin S.G. Electric field of rotating uniformly charged ball. TechRxiv. November 11, 2025. https://doi.org/10.36227/techrxiv.176289249.96428033/v1.
Fedosin S.G. Analysis of solution of equations for magnetic field of rotating ball using polynomials. Discover Physics, Vol. 2, 5 (2026). https://doi.org/10.1007/s44418-026-00008-w. TechRxiv. October 22, 2025. https://doi.org/10.36227/techrxiv.176116289.93994332/v1.

See also

External links

Electromagnetic field of ball in Russian

Source: http://sergf.ru/efben.htm

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