What i mean to ask is that, i learnt that EM waves are just disturbance in the already ever existing EM field, so when there is a charge in space, does it enhance the field in its required direction or it creates new ones, i had a perception that EM field exists every where with no or very little magnitude and the em waves are just disturbance in it, but then how can new field be created ,isn't it just enhancing it. Also, does extra change in EM fields have any effect on a propagating em wave aka disturbance in field.
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Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Mar 22 '23 at 13:55
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He is asking if charges create an electric field or change an electric field that already exists. Does an electric field exist when no charges are present? – mmesser314 Mar 22 '23 at 14:36
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This may help - In what medium are non-mechanical waves a disturbance? The aether? – mmesser314 Mar 22 '23 at 14:38
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You are correct in your statement that the EM field already exists everywhere, but with very small magnitude if there are no charges and currents around. That is the best way to think about it.
If there is a charge somewhere, its presence will "enhance" the field in that region, as you said. The more precise version of this statement is that the total field is given by the field of the charge (which I put at the origin, for simplicity), which is the Coulomb field \begin{equation} \mathbf{E}_{\text{charge}}(\mathbf{x}) = \frac{1}{4 \pi \epsilon_0} \frac{q}{|x|^2}\hat{\mathbf{x}}, \end{equation} added to the field that was already present in that region without the charge, $\mathbf{E}_{\text{background}}(\mathbf{x})$. The total field is then \begin{equation} \mathbf{E}_{\text{total}}(\mathbf{x}) = \mathbf{E}_{\text{background}}(\mathbf{x})+ \mathbf{E}_{\text{charge}}(\mathbf{x}). \end{equation}
This is called the superposition principle, and is a consequence of the fact that the Maxwell equations are linear: the sum of two solutions (fields) is also a valid solution. If there are no charges or currents close to your charge, the background field will be very weak, so we may approximate $\mathbf{E}_{\text{total}}(\mathbf{x}) \approx \mathbf{E}_{\text{charge}}(\mathbf{x})$. It does not make a lot of sense to say the charge "creates" a new field, it simply disturbs, or changes, the value of the electric field close to it, as you stated.
And yes, the fields due to charges change the way waves propagate. The usual wave equations for the EM fields are usually derived in the vacuum
\begin{align} \frac{1}{c^2} \frac{\partial^2 \mathbf{E}}{\partial t^2} - {\nabla}^2 \mathbf{E} &= 0 \\ \frac{1}{c^2} \frac{\partial^2 \mathbf{B}}{\partial t^2} - {\nabla}^2 \mathbf{B} &= 0. \end{align}
If there are charges and currents, the equations change to \begin{align} \frac{1}{c^2} \frac{\partial^2 \mathbf{E}}{\partial t^2} - {\nabla}^2 \mathbf{E} &= - \left( \frac{1}{\epsilon_0} \mathbf{\nabla} \rho + \mu_0 \frac{\partial \mathbf{J}}{\partial t} \right) \\ \frac{1}{c^2} \frac{\partial^2 \mathbf{B}}{\partial t^2} - {\nabla}^2 \mathbf{B} &= \mu_0 \nabla \times \mathbf{J}, \end{align} where $\rho$ is the density of charge and $\mathbf{J}$ is the density of current. These new terms will alter the solutions to the equations considerably, and will depend on exactly what $\rho$ and $\mathbf{J}$ are for each situation. Of course, the disturbance that $\rho$ and $\mathbf{J}$ cause on the fields will be very small if you are not close to the charges and currents, so if your wave comes from far away it will propagate approximately as a vacuum EM wave for some time. Once the wave gets close to the disturbances, however, the vacuum equations no longer hold as a good approximation and the charges and currents start to influence the propagation considerably, according to the last two equations.
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