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Consider a +Q charged particle is travelling towards another test charge +Q. Now what would be the difference in electric field experienced by the test charge(avoid the gradual decrease in distance between them)? Would the field lines look compressed and effective field strength increased for the test charge?

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2 Answers 2

A test charge "feels" only an external field. If another charge is approaching, it will experience a stronger field due to $1/R^2$ field strength dependence.

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So other than this distance dependent field strength no other possibilities are there? –  Inquisitive Dec 28 '12 at 20:28
    
Well, you can consider retarded fields as more exact, if necessary (a retarded near field and radiation). –  Vladimir Kalitvianski Dec 28 '12 at 20:30
    
Not understood, please explain it little more? –  Inquisitive Dec 28 '12 at 20:32

If you are looking for an effect separate from the particle's position, at classical velocities there isn't one. The electric field is \begin{equation} \mathbf{E}=\mathbf{E}(\mathbf{r},t), \end{equation} that is, the electric field is only a function of the position $\mathbf{r}$ and the time $t$. At any given instant in time, the force a test charge 'feels' due to another charge depends only on its position $\mathbf{r}$, and not on its velocity.

This velocity independence breaks down when the charges' relative velocities approach the speed of light. If a reference frame has an electric field, a frame boosted with the respect to the reference appears to have some magnetic field. For a frame boosted by a velocity $\mathbf{v}=v_x \mathbf{\hat{x}}$ where the separation $\mathbf{r}$ between the charges is given by $\mathbf{r}=r\mathbf{\hat{x}}$ (in other words, the charges are moving directly toward each other), so that \begin{equation} \mathbf{\beta}=\beta_x=v_x/c, \end{equation} and \begin{equation} \gamma=\left[1-\left(\frac{v_x}{c}\right)^2\right]^{-1/2}, \end{equation} then for an electric field in the frame of the stationary charge $\mathbf{E}=E_x\mathbf{\hat{x}}+E_y\mathbf{\hat{y}}+E_z\mathbf{\hat{z}}$ with a background magnetic field ($\mathbf{B}$), the test charge will 'see' fields $\mathbf{E}'$ and $\mathbf{B}'$ given by \begin{equation} \mathbf{E}'=\gamma(\mathbf{E}+\beta_x \mathbf{\hat{x}}\times\mathbf{B}) - \frac{\gamma^2\beta_x^2}{\gamma+1}(\mathbf{\hat{x}}\cdot\mathbf{E})\mathbf{\hat{x}}\\ \mathbf{B}'=\gamma(\mathbf{B}-\beta_x \mathbf{\hat{x}}\times\mathbf{E}) - \frac{\gamma^2\beta_x^2}{\gamma+1}(\mathbf{\hat{x}}\cdot\mathbf{B})\mathbf{\hat{x}} \end{equation}

(Source, J.D. Jackson 1999, section 11.10.)

The end result is that electric fields in a rest frame look like magnetic fields from a moving frame.

Interestingly, if the test particle is moving directly toward the charge, the electric and magnetic field along its trajectory will always be the classical one and relativity will have no effect. It is only when the boost has a component perpendicular to the rest-frame fields that the boost-frame fields are different.

There is compression of the field lines at relativistic velocities, but again, only for field lines that are not parallel to the velocity. If you picture the field lines radiating out of a stationary charge, then a moving charge looks similar, but with the field lines perpendicular to the boost velocity more tightly bunched together.

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