A point charge `q` moves unifromly and rectilnearly with a relativistic with a relativistic velocity of light `(beta = v//c)`. Find the electric field strength `E` produced by the charge at the point whose radius vector relatives to the charge is equal to `r` and forms an angle `theta` with its velocity vector.

Explanation

Suppose the charge `q` moves in the positive direction of the x-axis of the frame `K`. Let us go over to the moving frame `K'`, at whose origin the charge is at rest. We take the `x` and `x'` axes of the two frames to be coincident, and the `y & y'` axes, to be parallel.
In the `K'` frame, `vec(E) = (1)/(4pi epsilon_(0)) (q vec(r))/(r^(3))`,
and this has the following components,
`E'_(x) = (1)/(4pi epsilon_(0)) (q x')/(r'^(3)), E'_(y) = (1)/(4pi epsilon_(0)) (q y')/(r'^(3))`
Now let us go back to the frame `K`. At the moment, when the origins of the two frames coincide, we take `t = 0`. Then,
`x = r cos theta = x' sqrt(1 - (v^(2))/(c^(2))), y = r sin theta = y'`
Also, `E'_(x) = E'_(x), E_(y) = E'_(y)//sqrt(1 - v^(2)//c^(2))`
Form these equations, `r'^(2) = (r^(2)(1 - beta^(2) sin^(2) theta))/(1 - beta^(2))`
`vec(E) = (q)/(4pi epsilon_(0)) (1)/(r^(3) (1 - beta^(2) sin^(2) theta)^(3//2)) [ (1 - beta^(2))^(3//2)(x' hat(i) + (y')/(sqrt(1 - beta^(2)))hat(j))]`
`= (q vec(r) (1 - beta^(2)))/(4pi epsilon_(0) r^(3) (1 - beta^(2) sin^(2) theta)^(3//2))`