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Manishearth
Manishearth
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Not much sense. Your "center of charge" is nothing but the dipole moment divided by the net total charge. "Normalised dipole moment, if you will".

If you take $q|\vec v|$ instead of $q\vec v$, you get something related to current (generally current times a factor). Current is conserved at a junction.

Regarding your equal-and-opposite situation, the closest I can come up with is this: If you have a body emitting charge in free space (no magnetic field), there will be an equal-and-opposite displacement current (and thus an equal and opposite "displacement charge force"). Not that displacement current is not a real current, it's more of a mathematical convenience.

Firstly, the parallel between charge and mass comes from gravitational mass. Basically, Coulomb's law and the law of gravity are similar: $$\frac{kq_1q_2\hat r}{r^2}\sim-\frac{Gm_1m_2\hat r}{r^2}$$

These similarities are quite useful, for example, the shell theorem and Gauss' law work for both systems. There even are things like gravitomagnetism, though that deals with the similarities in the relativistic formulation. The similarity is only limited by the absence of "negative" gravitational mass.

However, these similarities don't extend to the inertial property of mass. Newton's laws deal with the inertial properties. The fact that attraction due to gravity is proportional to the inertial nature of a body can be said to be "chance" in the classical formulism. (More modern models explore and explain this link, IIRC)

Not much sense. Your "center of charge" is nothing but the dipole moment divided by the net total charge. "Normalised dipole moment, if you will".

If you take $q|\vec v|$ instead of $q\vec v$, you get something related to current (generally current times a factor). Current is conserved at a junction.

Regarding your equal-and-opposite situation, the closest I can come up with is this: If you have a body emitting charge in free space (no magnetic field), there will be an equal-and-opposite displacement current (and thus an equal and opposite "displacement charge force"). Not that displacement current is not a real current, it's more of a mathematical convenience.

Firstly, the parallel between charge and mass comes from gravitational mass. Basically, Coulomb's law and the law of gravity are similar: $$\frac{kq_1q_2\hat r}{r^2}\sim-\frac{Gm_1m_2\hat r}{r^2}$$

These similarities are quite useful, for example, the shell theorem and Gauss' law work for both systems. There even are things like gravitomagnetism, though that deals with the similarities in the relativistic formulation. The similarity is only limited by the absence of "negative" gravitational mass.

However, these similarities don't extend to the inertial property of mass. Newton's laws deal with the inertial properties. The fact that attraction due to gravity is proportional to the inertial nature of a body can be said to be "chance" in the classical formulism. (More modern models explore this link, IIRC)

Not much sense. Your "center of charge" is nothing but the dipole moment divided by the net total charge. "Normalised dipole moment, if you will".

If you take $q|\vec v|$ instead of $q\vec v$, you get something related to current (generally current times a factor). Current is conserved at a junction.

Regarding your equal-and-opposite situation, the closest I can come up with is this: If you have a body emitting charge in free space (no magnetic field), there will be an equal-and-opposite displacement current (and thus an equal and opposite "displacement charge force"). Not that displacement current is not a real current, it's more of a mathematical convenience.

Firstly, the parallel between charge and mass comes from gravitational mass. Basically, Coulomb's law and the law of gravity are similar: $$\frac{kq_1q_2\hat r}{r^2}\sim-\frac{Gm_1m_2\hat r}{r^2}$$

These similarities are quite useful, for example, the shell theorem and Gauss' law work for both systems. There even are things like gravitomagnetism, though that deals with the similarities in the relativistic formulation. The similarity is only limited by the absence of "negative" gravitational mass.

However, these similarities don't extend to the inertial property of mass. Newton's laws deal with the inertial properties. The fact that attraction due to gravity is proportional to the inertial nature of a body can be said to be "chance" in the classical formulism. (More modern models explore and explain this link)

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Manishearth
Manishearth
  • 19.1k
  • 6
  • 62
  • 104

Not much sense. Your "center of charge" is nothing but the dipole moment divided by the net total charge. "Normalised dipole moment, if you will".

If you take $q|\vec v|$ instead of $q\vec v$, you get something related to current (generally current times a factor). Current is conserved at a junction.

Regarding your equal-and-opposite situation, the closest I can come up with is this: If you have a body emitting charge in free space (no magnetic field), there will be an equal-and-opposite displacement current (and thus an equal and opposite "displacement charge force"). Not that displacement current is not a real current, it's more of a mathematical convenience.

Firstly, the parallel between charge and mass comes from gravitational mass. Basically, Coulomb's law and the law of gravity are similar: $$\frac{kq_1q_2\hat r}{r^2}\sim-\frac{Gm_1m_2\hat r}{r^2}$$

These similarities are quite useful, for example, the shell theorem and Gauss' law work for both systems. There even are things like gravitomagnetism, though that deals with the similarities in the relativistic formulation. The similarity is only limited by the absence of "negative" gravitational mass.

However, these similarities don't extend to the inertial property of mass. Newton's laws deal with the inertial properties. The fact that attraction due to gravity is proportional to the inertial nature of a body can be said to be "chance" in the classical formulism. (More modern models explore this link, IIRC)