Physics Help Forum Help on finding electric field of a moving point charge
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 May 19th 2011, 06:53 AM #1 Member   Join Date: Nov 2010 Posts: 40 Help regarding electric field of a moving point charge My questions are what the book said after working out the solution of E that I have no issue of finding. The original question is to find E and B at a field point pointed by r due to a moving point charge pointed by w(t_r) at retard time moving at a constant velocityv. . I also need help in finding B. For constant velocity, $\vec E_{(\vec r, t)} = \frac q {4\pi\epsilon_0}\frac {\eta}{(\vec{\eta}\cdot\vec u)^3} [(c^2-v^2)\vec u ]$ Where $\vec{\eta} = \vec r -\vec w(t_r), \; \vec u=c\hat{\eta}-\vec v\;,\; \vec v = \frac { d \vec w(t_r)}{dt_r}$ I have no problem finding the answer: $\vec E_{(\vec r, t)} = \frac q {4\pi\epsilon_0} \frac {1-\frac{v^2}{c^2}} {\left (1-\frac{v^2}{c^2} sin^2\theta \right )^{\frac 2 3}}\frac {\hat R}{R^2}, \;\hbox { where }\; \;\vec R = \vec r –t\vec v$ Last edited by yungman; Jun 6th 2011 at 12:21 AM.
 May 19th 2011, 06:59 AM #2 Member   Join Date: Nov 2010 Posts: 40 Below are the three statements directly quoted by the book and I have no idea what they mean: [BOOK] 1) Because of the $sin^2\theta$ in the denominator, the field of a fast-moving charge is flattened out like a pancake in the direction perpendicular to the motion. 2) In the forward and backward directions, E is reduced by a factor $\left (1 - {\frac {v^2}{c^2}\right )$ relative to the field of a charge at rest; 3) In the perpendicular direction it is enhanced by a factor $\frac {1} {\sqrt { \left (1 - \frac {v^2}{c^2}\right )}}$ [END BOOK] 1), R is vector from the point charge at PRESENT time, to the field point. This vector is not necessary perpendicular to the velocity vector no matter what velocity, this mean \theta a variable. Can anyone explain to me? Thanks Alan Last edited by yungman; May 19th 2011 at 07:54 AM.
 May 19th 2011, 08:58 AM #3 Member   Join Date: Nov 2010 Posts: 40 In finding B $\vec B =\frac 1 c \hat{\eta}\times \vec E = \frac 1 c \hat{\eta} \times \frac {q\left ( 1-\frac {v^2}{c^2}\right )}{4\pi\epsilon_0\left ( 1-\frac {v^2}{c^2} sin^2\theta \right )^{\frac 3 2}} \frac {\vec R}{R^3}$ In two different ways, I get different answer. 1) $\hat {\eta} \times \vec R = \frac {1}{\eta} \left [(\vec r -t_r \vec v)\times(c\frac{\vec{\eta}}{\eta}-t\vec v) \right ]=\frac {1}{\eta}\left [-\frac {ct_r}{\eta} \vec r\times \vec v -t(\vec r \times \vec v) -\frac{ct_r}{\eta}(\vec v\times \vec r)\right ]=-\frac {t} {\eta} (\vec r \times \vec v)$ $\Rightarrow\;\vec B = \frac 1 c \frac {q\left ( 1-\frac {v^2}{c^2} \right )}{4\pi\epsilon_0\eta \left ( 1-\frac {v^2}{c^2} sin^2\theta\right )^{\frac 3 2}R^3} (t\vec v \times \vec r)$ 2) $\hat{\eta}=\frac{\vec r-t_r\vec v}{\eta}=\frac {(\vec r-t\vec v)+(t-t_r)\vec v}{\eta}=\frac {\vec R}{\eta} + \frac {\vec v}{c} \Rightarrow \vec B = \frac 1 c \frac {q\left ( 1-\frac {v^2}{c^2} \right )}{4\pi\epsilon_0\eta \left ( 1-\frac {v^2}{c^2} sin^2\theta\right )^{\frac 3 2}R^3}(\vec v \times \vec r)$ Notice a difference of t between the two method? I cannot resolve this. Last edited by yungman; May 20th 2011 at 10:49 AM.
 Jun 3rd 2011, 10:06 AM #4 Member   Join Date: Nov 2010 Posts: 40 Hi Dan I know I should not bump post, but I have not been able to resolve this question. Please help if you can, even pointing to some site will be appreciated. Sorry
Jun 5th 2011, 12:19 PM   #5
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 Originally Posted by yungman Below are the three statements directly quoted by the book and I have no idea what they mean: [BOOK] 1) Because of the $sin^2\theta$ in the denominator, the field of a fast-moving charge is flattened out like a pancake in the direction perpendicular to the motion. 2) In the forward and backward directions, E is reduced by a factor $\left (1 - {\frac {v^2}{c^2}\right )$ relative to the field of a charge at rest; 3) In the perpendicular direction it is enhanced by a factor $\frac {1} {\sqrt { \left (1 - \frac {v^2}{c^2}\right )}}$ [END BOOK]
I think all three have essentially the same answer.

Consider the static case: E is radially outward from the charge.

Back to the moving case: v/c < 1, so 1 > 1 - (v/c)^2 > 0. So in the forward direction (theta = 0) the field is proportional to 1 - (v/c)^2 so it is reduced by that factor from the static case. ie it is smaller. In the perpendicular direction theta = pi / 2, and the field is proportional to the inverse square of 1 - (v/c)^2, which is greater than 1. This means the field is larger than in the static case. So the field is smaller in the front (and back) and larger out to the sides than it is for a radially symmetric field. The shape is somewhat reminiscent of a flattened disk, ie a "pancake."

 1), R is vector from the point charge at PRESENT time, to the field point. This vector is not necessary perpendicular to the velocity vector no matter what velocity, this mean \theta a variable. Can anyone explain to me? Thanks Alan
No idea about this one.

-Dan
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Jun 5th 2011, 12:45 PM   #6
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 Originally Posted by yungman In finding B $\vec B =\frac 1 c \hat{\eta}\times \vec E = \frac 1 c \hat{\eta} \times \frac {q\left ( 1-\frac {v^2}{c^2}\right )}{4\pi\epsilon_0\left ( 1-\frac {v^2}{c^2} sin^2\theta \right )^{\frac 3 2}} \frac {\vec R}{R^3}$ In two different ways, I get different answer. 1) $\hat {\eta} \times \vec R = \frac {1}{\eta} \left [(\vec r -t_r \vec v)\times(c\frac{\vec{\eta}}{\eta}-t\vec v) \right ]=\frac {1}{\eta}\left [-\frac {ct_r}{\eta} \vec r\times \vec v -t(\vec r \times \vec v) -\frac{ct_r}{\eta}(\vec v\times \vec r)\right ]=-\frac {t} {\eta} (\vec r \times \vec v)$ $\Rightarrow\;\vec B = \frac 1 c \frac {q\left ( 1-\frac {v^2}{c^2} \right )}{4\pi\epsilon_0\eta \left ( 1-\frac {v^2}{c^2} sin^2\theta\right )^{\frac 3 2}R^3} (t\vec v \times \vec r)$
I'm not sure what happened here, but I'd leave everything in terms of r and v:
$\begin{array}{lclcl} \hat{\eta} \times \vec{R} = \frac{1}{\eta}[ (\vec{r} - t_r \vec{v}) \times (\vec{r} - t \vec{v})] \\ . \\ = \frac{1}{\eta}[ \vec{r} \times \vec{r} - t \vec{r} \times \vec{v} - t_r \vec{v} \times \vec{r} + tt_r \vec{v} \times \vec{v} ] \\ . \\= \frac{1}{\eta}[-t \vec{r} \times \vec{v} - t_r \vec{v} \times \vec{r}] = - \frac{t - t_r}{\eta} \vec{r} \times \vec{v} = -\frac{1}{c} \vec{r} \times \vec{v} \end{array}$

which is the same as you get in your second method.

-Dan
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 Jun 5th 2011, 10:55 PM #7 Member   Join Date: Nov 2010 Posts: 40 Thanks Dan I need to take time to read this first. Alan

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