Consider two positive charges $q_1$ and $q_2$ separated by $r$ distance. The are both moving with $v$ velocity (same direction). Considering the electric field are moving with $c$ velocity, what would be the Electric force on $q_2$ by $q_1$?
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Duplicate : Charges and relative motion. – Frobenius Feb 23 '22 at 12:25
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1Does this answer your question? Charges and relative motion – Frobenius Feb 23 '22 at 12:25
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The fields due to Point charge, that we get using Lienard-Wiechert potentials: $$ \text{E} = {q\over 4\pi\epsilon_0}{|\vec{r} - \vec{r'}| \over [(\vec{r} - \vec{r'})\cdot\vec{u}]^3} \cdot [(c^2-v^2)\vec{u} + (\vec{r}-\vec{r'})\times(\vec{u}\times \vec{a})] $$ $$ B = {1\over c} (\vec{r}-\vec{r'}) \times E $$
Now, for a point charge moving with constant velocity, the value of these fields along the direction of motion are:
$$ E = {q\over 4\pi \epsilon_0}{1-v^2/c^2\over R^2}\hat{R} $$ $$ B = {1\over c^2} (\vec{v}\times \vec{E}) = 0 $$ And now we can calculate the Force using Lorentz law: $$ F = {q_1 q_2\over 4\pi \epsilon_0}{1-v^2/c^2\over r^2} $$
Atul Kumar
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Please do not post complete answers to questions that look like homework questions. This is not a site for doing other people's homework. This is a site for education, and posting complete answers is the opposite of that. Please reword the answer to give guidance, not an answer. Or if you prefer, just delete the answer. – garyp Feb 23 '22 at 12:30
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Well how did you get $B= 0$?. Both charges are moving so there should be a magnetic field. Neither $v$ nor $E$ is zero here. – James Webb Feb 24 '22 at 01:55