Question

A charge Q is moving with a uniform velocity . find an expression for the vector potential at a distance r from it.

Answer 1

solution

According to question let us suppose a charge Q is moving with uniform velocity v in positive x-axis.the charge Q is at origin"O"at time t=0.

at time t

its position are x=vt, y=z=0 or (vt, 0,0)

then retarded time can be written as

t'=(t - r'/c )................equation.(A)

where r' is distance to the point P(x,y,z)from charge Q at retarded time(t').

at the earlier time the charge Q was at position

x=vt'

y=0,

z=0.

or position (vt', 0,0).

therefore retarded position vector for Point P can be expressed as below

$r'=\sqrt{(x-vt')^2+y^2+z^2}$ ....equation(B)

By equation (A) and (B) we can rewrite r' as furture

$c^2(t-t')^2=(x-vt')^2+y^2+z^2$

$(v^2-c^2)t^2-2(xv-c^2t)t'+x^2+y^2+z^2-c^2t^2=0$

this is the quadratic equation in t'(retarded time) it's roots can be calculated by using SHRIDHAR AACHARYA formula

value of t' putting in equation (A)

we got as below

r' = c(t - t') ........equation(C)

scaler potential can be written for a charge Q which is moving with uniform velocity v is given by

$\phi(x,y,z,t)=\frac{q}{4\pi\in_0(r'-v.r'/c)}$

$\phi(x,y,z,t)=\frac{q}{4\pi\in_0(\sqrt{(x-vt)^2+(1-v^2/c^2)(y^2+z^2)}}$

$\phi(x,y,z,t)=\frac{q}{4\pi\in_0\sqrt{(1-v^2/c^2)}(\sqrt{((x-vt)/(\sqrt(1-v^2/c^2))^2+(y^2+z^2)}}$

relation between vector potential and scaler potential is

$A=\frac{v\phi}{c^2}$

for a charge Q at a time t and position (vt,0,0)

then vector potential is given by as below

$A_x=\frac{q}{4\pi\in_0\sqrt{(1-v^2)}(\sqrt{(x-vt)^2/((1-v^2)+(y^2+z^2)}}$ ........ equation (D)

and A_y=A_z=0

Here vector potentials are zero in x and y directions because of this charge Q is moving in positive x direction (according our initial consideration).