What will happen to the force between two charged particles, if only one particle is placed inside the medium ? Like one in air and other one is in water( Or anything.. )
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2 Answers
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If the two media are gas or liquid, you can still use Coulomb's law. But if one of the media is a solid, then you cannot use Coulombs law, because the lattice structure of the solid creates itself a force that keeps the molecules in structure (has a mechanical effect).
Please see here:
answered Apr 26, 2019 at 20:59
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Non-coducting Materials, dielectrics, react to an applied electric field in various ways. A common one is the constituent molecules are distorted at their equilibrium positions os that there is a separation of charge. Consequently you get dipoles. Dipoles often cancel out within the material but accumulates at the surface. This Bound Charge effects the total electric field.
To handle this, we use the Displacement Field.
$$\vec{D}=\epsilon_0\vec{E}+\vec{P}$$
Where $\vec{E}$ is the electric field and $\vec{P}$ is the Polarization vector, the density of the induced dipoles within the material.
$\nabla \cdot \vec{E}=\rho_f+\rho_b$ where $\rho_f$ is the free charge and $\rho_b$ is the Bound Charge, produced by the distortion caused by the applied Electric field.
So in these Linear Materials, $\vec{P}=\epsilon_0\chi_e\vec{E}$ where $\chi_e$ is the susceptibility. Being the density of bound charge $\nabla \cdot \vec{P}=-\rho_b$
So we have $\vec{D}=\epsilon_0(1+\chi_e)\vec{E}=\epsilon\vec{E}$
And
$\nabla \cdot\vec{D}=\nabla \cdot (\epsilon_0\vec{E}+\vec{P})=(\rho_f+\rho_b)-\rho_b$
So $\nabla \cdot \vec{D}=\rho_f$
So unlike the electric field, the Displacement Field's normal component is continuous through an interface in the absence of free charge.
This enables you to use Gauss' Law with the Displacement field, then scaling via the permittivity to get the actual electric field.
