Mixing
How to find the critical point for this coulomb field
$begingroup$
Two equal positive charges are at distance $d$, $-d$ from the origin on the $y$ axis. What is the distance on the $x$ axis beyond which a small perturbation in $y$ will move a particle away from the $x$ axis?
My attempt is to consider the y components of electr ic field where $y<d$ $$frac{y-d}{(x^2+(y-d)^2)^{3/2}}+frac{d+y}{(x^2+(d+y)^2)^{3/2}}=0$$
and solve for $x$ as $y/d to 0$. But this gives me nonsense unsolvable equations.
The answer should be $x = sqrt{2}d$.
As illustrated above, We are finding the horizontal distance to where the lines start to curve away.
physics vector-fields perturbation-theory electromagnetism
asked Jan 16 at 14:00
Dis-integratingDis-integrating
1,036426
$endgroup$
add a comment |
$begingroup$
Two equal positive charges are at distance $d$, $-d$ from the origin on the $y$ axis. What is the distance on the $x$ axis beyond which a small perturbation in $y$ will move a particle away from the $x$ axis?
My attempt is to consider the y components of electr ic field where $y<d$ $$frac{y-d}{(x^2+(y-d)^2)^{3/2}}+frac{d+y}{(x^2+(d+y)^2)^{3/2}}=0$$
and solve for $x$ as $y/d to 0$. But this gives me nonsense unsolvable equations.
The answer should be $x = sqrt{2}d$.
As illustrated above, We are finding the horizontal distance to where the lines start to curve away.
physics vector-fields perturbation-theory electromagnetism
asked Jan 16 at 14:00
Dis-integratingDis-integrating
1,036426
$endgroup$
$begingroup$
Are you still interested in a solution? I see that you already got a response on the physics site.
$endgroup$
– Lee David Chung Lin
Jan 31 at 3:54
add a comment |
$begingroup$
Two equal positive charges are at distance $d$, $-d$ from the origin on the $y$ axis. What is the distance on the $x$ axis beyond which a small perturbation in $y$ will move a particle away from the $x$ axis?
My attempt is to consider the y components of electr ic field where $y<d$ $$frac{y-d}{(x^2+(y-d)^2)^{3/2}}+frac{d+y}{(x^2+(d+y)^2)^{3/2}}=0$$
and solve for $x$ as $y/d to 0$. But this gives me nonsense unsolvable equations.
The answer should be $x = sqrt{2}d$.
As illustrated above, We are finding the horizontal distance to where the lines start to curve away.
physics vector-fields perturbation-theory electromagnetism
asked Jan 16 at 14:00
Dis-integratingDis-integrating
1,036426
$endgroup$
Two equal positive charges are at distance $d$, $-d$ from the origin on the $y$ axis. What is the distance on the $x$ axis beyond which a small perturbation in $y$ will move a particle away from the $x$ axis?
My attempt is to consider the y components of electr ic field where $y<d$ $$frac{y-d}{(x^2+(y-d)^2)^{3/2}}+frac{d+y}{(x^2+(d+y)^2)^{3/2}}=0$$
and solve for $x$ as $y/d to 0$. But this gives me nonsense unsolvable equations.
The answer should be $x = sqrt{2}d$.
As illustrated above, We are finding the horizontal distance to where the lines start to curve away.
physics vector-fields perturbation-theory electromagnetism
physics vector-fields perturbation-theory electromagnetism
asked Jan 16 at 14:00
Dis-integratingDis-integrating
1,036426
asked Jan 16 at 14:00
Dis-integratingDis-integrating
1,036426
edited Feb 18 at 0:16
Dis-integrating
asked Jan 16 at 14:00
Dis-integratingDis-integrating
1,036426
asked Jan 16 at 14:00
Dis-integratingDis-integrating
1,036426
asked Jan 16 at 14:00
Dis-integratingDis-integrating
1,036426
1,036426
$begingroup$
Are you still interested in a solution? I see that you already got a response on the physics site.
$endgroup$
– Lee David Chung Lin
Jan 31 at 3:54
add a comment |
$begingroup$
Are you still interested in a solution? I see that you already got a response on the physics site.
$endgroup$
– Lee David Chung Lin
Jan 31 at 3:54
$begingroup$
Are you still interested in a solution? I see that you already got a response on the physics site.
$endgroup$
– Lee David Chung Lin
Jan 31 at 3:54
$begingroup$
Are you still interested in a solution? I see that you already got a response on the physics site.
$endgroup$
– Lee David Chung Lin
Jan 31 at 3:54
add a comment |
1 Answer
1
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oldest
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$begingroup$
Your goal is to show the critical point where the vertical electric field $E_y$ at given tiny $y>0$ goes from negative (towards the $x$-axis) to positive (away from $x$-axis$), where
begin{align}
E_y &equiv frac{d+y}{(x^2+(d+y)^2)^{3/2}} - frac{d-y}{(x^2+(y-d)^2)^{3/2}} \
&= frac{d}{ (x^2 + d^2)^{3/2} } left( frac{ 1 + y/d }{ displaystyle bigl( 1 + frac{ 2dy + y ^2}{ x^2 + d^2 } bigr)^{3/2} } -frac{ 1 - y/d }{ displaystyle bigl( 1 - frac{ 2dy - y ^2}{ x^2 + d^2 } bigr)^{3/2} } right)end{align}
Since we are only interested solving $E_y = 0$, the overall factor in front can be dropped for convenience.
$$begin{aligned}
E_y &propto bigl(1 + frac{y}d bigr) left( 1 + frac{ 2dy + y ^2}{ x^2 + d^2 } right)^{-3/2} - bigl(1 - frac{y}d bigr) left( 1 - frac{ 2dy - y ^2}{ x^2 + d^2 } right)^{-3/2} \
&= bigl(1 + frac{y}d bigr) left( 1 + frac{ 2d }{ x^2 + d^2 } bigl( 1 + frac{y}{2d}bigr)y right)^{-3/2} - bigl(1 - frac{y}d bigr) left(
1 - frac{ 2d }{ x^2 + d^2 } bigl( 1 - frac{y}{2d}bigr)y right)^{-3/2}
end{aligned}$$
As you intended to do, let's consider the series expansion for $displaystyle epsilon equiv frac{y}{2d} ll 1$. Use the shorthand $displaystyle K equiv frac{ 4d^2 }{ x^2 + d^2 }$ for this factor that doesn't involve $epsilon$.
$$begin{aligned}
E_y &propto bigl(1 + frac{y}d bigr) left( 1 + K bigl( 1 + frac{y}{2d}bigr)frac{y}{2d} right)^{-3/2} - bigl(1 - frac{y}d bigr) left(
1 - K bigl( 1 - frac{y}{2d}bigr) frac{y}{2d} right)^{-3/2}\
&=bigl(1 + 2epsilon bigr) left( 1 + K bigl( 1 + epsilon bigr) epsilon right)^{-3/2} - bigl(1 - 2epsilon bigr) left(
1 - K bigl( 1 - epsilon bigr) epsilon right)^{-3/2}\
&=begin{aligned}[t]
&bigl(1 + 2epsilon bigr) left( 1 - frac32 K (1+epsilon)epsilon + frac1{2!}frac32 frac52 K^2 (1+epsilon)^2epsilon^2 - frac1{3!}frac32 frac52 frac72 K^3 (1+epsilon)^3epsilon^3 + cdots right) \
&qquad - bigl(1 - 2epsilon bigr) left( 1 - frac32 bigl( -K (1-epsilon)epsilon bigr) + frac1{2!}frac32 frac52 bigl( -K (1-epsilon)epsilon bigr)^2 - frac1{3!}frac32 frac52 frac72 bigl( -K (1-epsilon)epsilon bigr)^3 + cdots right)
end{aligned}
end{aligned}$$
Here we have $(1 + 2epsilon) cdot A_1 - (1 - 2epsilon) cdot A_2 = A_1 - A_2 + 2epsilon cdot ( A_1 + A_2)$ where
$$begin{aligned}
A_1 &= 1 - frac32 K (1 + epsilon)epsilon + frac{15}8 K^2 (1 + epsilon)^2epsilon^2 - frac{35}{16} K^3 (1 + epsilon)^3epsilon^3 + cdots \
A_2 &= 1 + frac32 K (1-epsilon)epsilon + frac{15}8 K^2 (1-epsilon)^2epsilon^2 + frac{35}{16} K^3 (1-epsilon)^3epsilon^3 + cdots end{aligned}$$
so that
$$begin{aligned}
A_1 - A_2 & = -3 K epsilon + frac{15}8 K^2 (4epsilon)epsilon^2 - frac{35}{16} K^3 (2 + 6epsilon^2)epsilon^3 + cdots \
A_1 + A_2 & = 2 - 3 K epsilon^2 + frac{15}8 K^2 ( 2+2epsilon^2)epsilon^2 + frac{35}{16} K^3 (6 epsilon + 2epsilon^3)epsilon^3 + cdots
end{aligned}$$
$impliesbegin{aligned}[t]
E_y &propto -3 K epsilon + O( epsilon^3 ) + 2epsilon cdot bigl( 2 + O(epsilon^2) bigr) \
&= (4-3K)epsilon + O( epsilon^3 )
end{aligned}$
Therefore the critical point $E_y = 0$ is given by
begin{align}
4 - 3K &= 0 &&implies frac{ 4d^2 }{ x^2 + d^2} = frac43 && implies 2d^2 = x^2 &&
end{align}
which is the desired $x = sqrt{2} d$.
answered Feb 2 at 9:35
Lee David Chung LinLee David Chung Lin
4,34531241
$endgroup$
$begingroup$
I think your answer is right but aren't you using loads of redundant terms in your taylor approximations? It's hard to read..
$endgroup$
– Dis-integrating
Feb 18 at 0:17
$begingroup$
I was just trying to provide all the details. Indeed this is routine expansion and nothing particularly profound about it. Since you posted the question here after already got an instruction on the physics site, I presumed you had difficulties wading through the algebra. Whether a particular algebraic manipulation is "redundant" or "helpful" is really personal, and obviously I guessed wrong what you want. If you have already found elsewhere a version (or written up yourself) that is "easy to read" then I'm glad for you.
$endgroup$
– Lee David Chung Lin
yesterday
$begingroup$
Now, about the technical stuff. (1) The choices to denote $epsilon$ and $K$ as such is intended to make the expressions shorter. You can try other ways (with different multiplicative factors), and most likely you'll end up with more terms and longer equations. (2) The use of $A_1$ and $A_2$ is intended to highlight the parallel and complementarity of the two parts.
$endgroup$
– Lee David Chung Lin
yesterday
$begingroup$
The bottomline is, I'm sorry if my solution as seen (with those designs and intentions ) doesn't work for you. There's the edit functionality, and you can modify my post to suit your needs. Or, you are very welcomed to type up your own approach as a separate answer post and "accept" it, to show future readers a derivation that can be deemed better under some criteria. Editing to improve the content and self-answering are common practice that are encouraged on this site.
$endgroup$
– Lee David Chung Lin
yesterday
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
Your goal is to show the critical point where the vertical electric field $E_y$ at given tiny $y>0$ goes from negative (towards the $x$-axis) to positive (away from $x$-axis$), where
begin{align}
E_y &equiv frac{d+y}{(x^2+(d+y)^2)^{3/2}} - frac{d-y}{(x^2+(y-d)^2)^{3/2}} \
&= frac{d}{ (x^2 + d^2)^{3/2} } left( frac{ 1 + y/d }{ displaystyle bigl( 1 + frac{ 2dy + y ^2}{ x^2 + d^2 } bigr)^{3/2} } -frac{ 1 - y/d }{ displaystyle bigl( 1 - frac{ 2dy - y ^2}{ x^2 + d^2 } bigr)^{3/2} } right)end{align}
Since we are only interested solving $E_y = 0$, the overall factor in front can be dropped for convenience.
$$begin{aligned}
E_y &propto bigl(1 + frac{y}d bigr) left( 1 + frac{ 2dy + y ^2}{ x^2 + d^2 } right)^{-3/2} - bigl(1 - frac{y}d bigr) left( 1 - frac{ 2dy - y ^2}{ x^2 + d^2 } right)^{-3/2} \
&= bigl(1 + frac{y}d bigr) left( 1 + frac{ 2d }{ x^2 + d^2 } bigl( 1 + frac{y}{2d}bigr)y right)^{-3/2} - bigl(1 - frac{y}d bigr) left(
1 - frac{ 2d }{ x^2 + d^2 } bigl( 1 - frac{y}{2d}bigr)y right)^{-3/2}
end{aligned}$$
As you intended to do, let's consider the series expansion for $displaystyle epsilon equiv frac{y}{2d} ll 1$. Use the shorthand $displaystyle K equiv frac{ 4d^2 }{ x^2 + d^2 }$ for this factor that doesn't involve $epsilon$.
$$begin{aligned}
E_y &propto bigl(1 + frac{y}d bigr) left( 1 + K bigl( 1 + frac{y}{2d}bigr)frac{y}{2d} right)^{-3/2} - bigl(1 - frac{y}d bigr) left(
1 - K bigl( 1 - frac{y}{2d}bigr) frac{y}{2d} right)^{-3/2}\
&=bigl(1 + 2epsilon bigr) left( 1 + K bigl( 1 + epsilon bigr) epsilon right)^{-3/2} - bigl(1 - 2epsilon bigr) left(
1 - K bigl( 1 - epsilon bigr) epsilon right)^{-3/2}\
&=begin{aligned}[t]
&bigl(1 + 2epsilon bigr) left( 1 - frac32 K (1+epsilon)epsilon + frac1{2!}frac32 frac52 K^2 (1+epsilon)^2epsilon^2 - frac1{3!}frac32 frac52 frac72 K^3 (1+epsilon)^3epsilon^3 + cdots right) \
&qquad - bigl(1 - 2epsilon bigr) left( 1 - frac32 bigl( -K (1-epsilon)epsilon bigr) + frac1{2!}frac32 frac52 bigl( -K (1-epsilon)epsilon bigr)^2 - frac1{3!}frac32 frac52 frac72 bigl( -K (1-epsilon)epsilon bigr)^3 + cdots right)
end{aligned}
end{aligned}$$
Here we have $(1 + 2epsilon) cdot A_1 - (1 - 2epsilon) cdot A_2 = A_1 - A_2 + 2epsilon cdot ( A_1 + A_2)$ where
$$begin{aligned}
A_1 &= 1 - frac32 K (1 + epsilon)epsilon + frac{15}8 K^2 (1 + epsilon)^2epsilon^2 - frac{35}{16} K^3 (1 + epsilon)^3epsilon^3 + cdots \
A_2 &= 1 + frac32 K (1-epsilon)epsilon + frac{15}8 K^2 (1-epsilon)^2epsilon^2 + frac{35}{16} K^3 (1-epsilon)^3epsilon^3 + cdots end{aligned}$$
so that
$$begin{aligned}
A_1 - A_2 & = -3 K epsilon + frac{15}8 K^2 (4epsilon)epsilon^2 - frac{35}{16} K^3 (2 + 6epsilon^2)epsilon^3 + cdots \
A_1 + A_2 & = 2 - 3 K epsilon^2 + frac{15}8 K^2 ( 2+2epsilon^2)epsilon^2 + frac{35}{16} K^3 (6 epsilon + 2epsilon^3)epsilon^3 + cdots
end{aligned}$$
$impliesbegin{aligned}[t]
E_y &propto -3 K epsilon + O( epsilon^3 ) + 2epsilon cdot bigl( 2 + O(epsilon^2) bigr) \
&= (4-3K)epsilon + O( epsilon^3 )
end{aligned}$
Therefore the critical point $E_y = 0$ is given by
begin{align}
4 - 3K &= 0 &&implies frac{ 4d^2 }{ x^2 + d^2} = frac43 && implies 2d^2 = x^2 &&
end{align}
which is the desired $x = sqrt{2} d$.
answered Feb 2 at 9:35
Lee David Chung LinLee David Chung Lin
4,34531241
$endgroup$
$begingroup$
I think your answer is right but aren't you using loads of redundant terms in your taylor approximations? It's hard to read..
$endgroup$
– Dis-integrating
Feb 18 at 0:17
$begingroup$
I was just trying to provide all the details. Indeed this is routine expansion and nothing particularly profound about it. Since you posted the question here after already got an instruction on the physics site, I presumed you had difficulties wading through the algebra. Whether a particular algebraic manipulation is "redundant" or "helpful" is really personal, and obviously I guessed wrong what you want. If you have already found elsewhere a version (or written up yourself) that is "easy to read" then I'm glad for you.
$endgroup$
– Lee David Chung Lin
yesterday
$begingroup$
Now, about the technical stuff. (1) The choices to denote $epsilon$ and $K$ as such is intended to make the expressions shorter. You can try other ways (with different multiplicative factors), and most likely you'll end up with more terms and longer equations. (2) The use of $A_1$ and $A_2$ is intended to highlight the parallel and complementarity of the two parts.
$endgroup$
– Lee David Chung Lin
yesterday
$begingroup$
The bottomline is, I'm sorry if my solution as seen (with those designs and intentions ) doesn't work for you. There's the edit functionality, and you can modify my post to suit your needs. Or, you are very welcomed to type up your own approach as a separate answer post and "accept" it, to show future readers a derivation that can be deemed better under some criteria. Editing to improve the content and self-answering are common practice that are encouraged on this site.
$endgroup$
– Lee David Chung Lin
yesterday
add a comment |
$begingroup$
Your goal is to show the critical point where the vertical electric field $E_y$ at given tiny $y>0$ goes from negative (towards the $x$-axis) to positive (away from $x$-axis$), where
begin{align}
E_y &equiv frac{d+y}{(x^2+(d+y)^2)^{3/2}} - frac{d-y}{(x^2+(y-d)^2)^{3/2}} \
&= frac{d}{ (x^2 + d^2)^{3/2} } left( frac{ 1 + y/d }{ displaystyle bigl( 1 + frac{ 2dy + y ^2}{ x^2 + d^2 } bigr)^{3/2} } -frac{ 1 - y/d }{ displaystyle bigl( 1 - frac{ 2dy - y ^2}{ x^2 + d^2 } bigr)^{3/2} } right)end{align}
Since we are only interested solving $E_y = 0$, the overall factor in front can be dropped for convenience.
$$begin{aligned}
E_y &propto bigl(1 + frac{y}d bigr) left( 1 + frac{ 2dy + y ^2}{ x^2 + d^2 } right)^{-3/2} - bigl(1 - frac{y}d bigr) left( 1 - frac{ 2dy - y ^2}{ x^2 + d^2 } right)^{-3/2} \
&= bigl(1 + frac{y}d bigr) left( 1 + frac{ 2d }{ x^2 + d^2 } bigl( 1 + frac{y}{2d}bigr)y right)^{-3/2} - bigl(1 - frac{y}d bigr) left(
1 - frac{ 2d }{ x^2 + d^2 } bigl( 1 - frac{y}{2d}bigr)y right)^{-3/2}
end{aligned}$$
As you intended to do, let's consider the series expansion for $displaystyle epsilon equiv frac{y}{2d} ll 1$. Use the shorthand $displaystyle K equiv frac{ 4d^2 }{ x^2 + d^2 }$ for this factor that doesn't involve $epsilon$.
$$begin{aligned}
E_y &propto bigl(1 + frac{y}d bigr) left( 1 + K bigl( 1 + frac{y}{2d}bigr)frac{y}{2d} right)^{-3/2} - bigl(1 - frac{y}d bigr) left(
1 - K bigl( 1 - frac{y}{2d}bigr) frac{y}{2d} right)^{-3/2}\
&=bigl(1 + 2epsilon bigr) left( 1 + K bigl( 1 + epsilon bigr) epsilon right)^{-3/2} - bigl(1 - 2epsilon bigr) left(
1 - K bigl( 1 - epsilon bigr) epsilon right)^{-3/2}\
&=begin{aligned}[t]
&bigl(1 + 2epsilon bigr) left( 1 - frac32 K (1+epsilon)epsilon + frac1{2!}frac32 frac52 K^2 (1+epsilon)^2epsilon^2 - frac1{3!}frac32 frac52 frac72 K^3 (1+epsilon)^3epsilon^3 + cdots right) \
&qquad - bigl(1 - 2epsilon bigr) left( 1 - frac32 bigl( -K (1-epsilon)epsilon bigr) + frac1{2!}frac32 frac52 bigl( -K (1-epsilon)epsilon bigr)^2 - frac1{3!}frac32 frac52 frac72 bigl( -K (1-epsilon)epsilon bigr)^3 + cdots right)
end{aligned}
end{aligned}$$
Here we have $(1 + 2epsilon) cdot A_1 - (1 - 2epsilon) cdot A_2 = A_1 - A_2 + 2epsilon cdot ( A_1 + A_2)$ where
$$begin{aligned}
A_1 &= 1 - frac32 K (1 + epsilon)epsilon + frac{15}8 K^2 (1 + epsilon)^2epsilon^2 - frac{35}{16} K^3 (1 + epsilon)^3epsilon^3 + cdots \
A_2 &= 1 + frac32 K (1-epsilon)epsilon + frac{15}8 K^2 (1-epsilon)^2epsilon^2 + frac{35}{16} K^3 (1-epsilon)^3epsilon^3 + cdots end{aligned}$$
so that
$$begin{aligned}
A_1 - A_2 & = -3 K epsilon + frac{15}8 K^2 (4epsilon)epsilon^2 - frac{35}{16} K^3 (2 + 6epsilon^2)epsilon^3 + cdots \
A_1 + A_2 & = 2 - 3 K epsilon^2 + frac{15}8 K^2 ( 2+2epsilon^2)epsilon^2 + frac{35}{16} K^3 (6 epsilon + 2epsilon^3)epsilon^3 + cdots
end{aligned}$$
$impliesbegin{aligned}[t]
E_y &propto -3 K epsilon + O( epsilon^3 ) + 2epsilon cdot bigl( 2 + O(epsilon^2) bigr) \
&= (4-3K)epsilon + O( epsilon^3 )
end{aligned}$
Therefore the critical point $E_y = 0$ is given by
begin{align}
4 - 3K &= 0 &&implies frac{ 4d^2 }{ x^2 + d^2} = frac43 && implies 2d^2 = x^2 &&
end{align}
which is the desired $x = sqrt{2} d$.
answered Feb 2 at 9:35
Lee David Chung LinLee David Chung Lin
4,34531241
$endgroup$
$begingroup$
I think your answer is right but aren't you using loads of redundant terms in your taylor approximations? It's hard to read..
$endgroup$
– Dis-integrating
Feb 18 at 0:17
$begingroup$
I was just trying to provide all the details. Indeed this is routine expansion and nothing particularly profound about it. Since you posted the question here after already got an instruction on the physics site, I presumed you had difficulties wading through the algebra. Whether a particular algebraic manipulation is "redundant" or "helpful" is really personal, and obviously I guessed wrong what you want. If you have already found elsewhere a version (or written up yourself) that is "easy to read" then I'm glad for you.
$endgroup$
– Lee David Chung Lin
yesterday
$begingroup$
Now, about the technical stuff. (1) The choices to denote $epsilon$ and $K$ as such is intended to make the expressions shorter. You can try other ways (with different multiplicative factors), and most likely you'll end up with more terms and longer equations. (2) The use of $A_1$ and $A_2$ is intended to highlight the parallel and complementarity of the two parts.
$endgroup$
– Lee David Chung Lin
yesterday
$begingroup$
The bottomline is, I'm sorry if my solution as seen (with those designs and intentions ) doesn't work for you. There's the edit functionality, and you can modify my post to suit your needs. Or, you are very welcomed to type up your own approach as a separate answer post and "accept" it, to show future readers a derivation that can be deemed better under some criteria. Editing to improve the content and self-answering are common practice that are encouraged on this site.
$endgroup$
– Lee David Chung Lin
yesterday
add a comment |
$begingroup$
Your goal is to show the critical point where the vertical electric field $E_y$ at given tiny $y>0$ goes from negative (towards the $x$-axis) to positive (away from $x$-axis$), where
begin{align}
E_y &equiv frac{d+y}{(x^2+(d+y)^2)^{3/2}} - frac{d-y}{(x^2+(y-d)^2)^{3/2}} \
&= frac{d}{ (x^2 + d^2)^{3/2} } left( frac{ 1 + y/d }{ displaystyle bigl( 1 + frac{ 2dy + y ^2}{ x^2 + d^2 } bigr)^{3/2} } -frac{ 1 - y/d }{ displaystyle bigl( 1 - frac{ 2dy - y ^2}{ x^2 + d^2 } bigr)^{3/2} } right)end{align}
Since we are only interested solving $E_y = 0$, the overall factor in front can be dropped for convenience.
$$begin{aligned}
E_y &propto bigl(1 + frac{y}d bigr) left( 1 + frac{ 2dy + y ^2}{ x^2 + d^2 } right)^{-3/2} - bigl(1 - frac{y}d bigr) left( 1 - frac{ 2dy - y ^2}{ x^2 + d^2 } right)^{-3/2} \
&= bigl(1 + frac{y}d bigr) left( 1 + frac{ 2d }{ x^2 + d^2 } bigl( 1 + frac{y}{2d}bigr)y right)^{-3/2} - bigl(1 - frac{y}d bigr) left(
1 - frac{ 2d }{ x^2 + d^2 } bigl( 1 - frac{y}{2d}bigr)y right)^{-3/2}
end{aligned}$$
As you intended to do, let's consider the series expansion for $displaystyle epsilon equiv frac{y}{2d} ll 1$. Use the shorthand $displaystyle K equiv frac{ 4d^2 }{ x^2 + d^2 }$ for this factor that doesn't involve $epsilon$.
$$begin{aligned}
E_y &propto bigl(1 + frac{y}d bigr) left( 1 + K bigl( 1 + frac{y}{2d}bigr)frac{y}{2d} right)^{-3/2} - bigl(1 - frac{y}d bigr) left(
1 - K bigl( 1 - frac{y}{2d}bigr) frac{y}{2d} right)^{-3/2}\
&=bigl(1 + 2epsilon bigr) left( 1 + K bigl( 1 + epsilon bigr) epsilon right)^{-3/2} - bigl(1 - 2epsilon bigr) left(
1 - K bigl( 1 - epsilon bigr) epsilon right)^{-3/2}\
&=begin{aligned}[t]
&bigl(1 + 2epsilon bigr) left( 1 - frac32 K (1+epsilon)epsilon + frac1{2!}frac32 frac52 K^2 (1+epsilon)^2epsilon^2 - frac1{3!}frac32 frac52 frac72 K^3 (1+epsilon)^3epsilon^3 + cdots right) \
&qquad - bigl(1 - 2epsilon bigr) left( 1 - frac32 bigl( -K (1-epsilon)epsilon bigr) + frac1{2!}frac32 frac52 bigl( -K (1-epsilon)epsilon bigr)^2 - frac1{3!}frac32 frac52 frac72 bigl( -K (1-epsilon)epsilon bigr)^3 + cdots right)
end{aligned}
end{aligned}$$
Here we have $(1 + 2epsilon) cdot A_1 - (1 - 2epsilon) cdot A_2 = A_1 - A_2 + 2epsilon cdot ( A_1 + A_2)$ where
$$begin{aligned}
A_1 &= 1 - frac32 K (1 + epsilon)epsilon + frac{15}8 K^2 (1 + epsilon)^2epsilon^2 - frac{35}{16} K^3 (1 + epsilon)^3epsilon^3 + cdots \
A_2 &= 1 + frac32 K (1-epsilon)epsilon + frac{15}8 K^2 (1-epsilon)^2epsilon^2 + frac{35}{16} K^3 (1-epsilon)^3epsilon^3 + cdots end{aligned}$$
so that
$$begin{aligned}
A_1 - A_2 & = -3 K epsilon + frac{15}8 K^2 (4epsilon)epsilon^2 - frac{35}{16} K^3 (2 + 6epsilon^2)epsilon^3 + cdots \
A_1 + A_2 & = 2 - 3 K epsilon^2 + frac{15}8 K^2 ( 2+2epsilon^2)epsilon^2 + frac{35}{16} K^3 (6 epsilon + 2epsilon^3)epsilon^3 + cdots
end{aligned}$$
$impliesbegin{aligned}[t]
E_y &propto -3 K epsilon + O( epsilon^3 ) + 2epsilon cdot bigl( 2 + O(epsilon^2) bigr) \
&= (4-3K)epsilon + O( epsilon^3 )
end{aligned}$
Therefore the critical point $E_y = 0$ is given by
begin{align}
4 - 3K &= 0 &&implies frac{ 4d^2 }{ x^2 + d^2} = frac43 && implies 2d^2 = x^2 &&
end{align}
which is the desired $x = sqrt{2} d$.
answered Feb 2 at 9:35
Lee David Chung LinLee David Chung Lin
4,34531241
$endgroup$
Your goal is to show the critical point where the vertical electric field $E_y$ at given tiny $y>0$ goes from negative (towards the $x$-axis) to positive (away from $x$-axis$), where
begin{align}
E_y &equiv frac{d+y}{(x^2+(d+y)^2)^{3/2}} - frac{d-y}{(x^2+(y-d)^2)^{3/2}} \
&= frac{d}{ (x^2 + d^2)^{3/2} } left( frac{ 1 + y/d }{ displaystyle bigl( 1 + frac{ 2dy + y ^2}{ x^2 + d^2 } bigr)^{3/2} } -frac{ 1 - y/d }{ displaystyle bigl( 1 - frac{ 2dy - y ^2}{ x^2 + d^2 } bigr)^{3/2} } right)end{align}
Since we are only interested solving $E_y = 0$, the overall factor in front can be dropped for convenience.
$$begin{aligned}
E_y &propto bigl(1 + frac{y}d bigr) left( 1 + frac{ 2dy + y ^2}{ x^2 + d^2 } right)^{-3/2} - bigl(1 - frac{y}d bigr) left( 1 - frac{ 2dy - y ^2}{ x^2 + d^2 } right)^{-3/2} \
&= bigl(1 + frac{y}d bigr) left( 1 + frac{ 2d }{ x^2 + d^2 } bigl( 1 + frac{y}{2d}bigr)y right)^{-3/2} - bigl(1 - frac{y}d bigr) left(
1 - frac{ 2d }{ x^2 + d^2 } bigl( 1 - frac{y}{2d}bigr)y right)^{-3/2}
end{aligned}$$
As you intended to do, let's consider the series expansion for $displaystyle epsilon equiv frac{y}{2d} ll 1$. Use the shorthand $displaystyle K equiv frac{ 4d^2 }{ x^2 + d^2 }$ for this factor that doesn't involve $epsilon$.
$$begin{aligned}
E_y &propto bigl(1 + frac{y}d bigr) left( 1 + K bigl( 1 + frac{y}{2d}bigr)frac{y}{2d} right)^{-3/2} - bigl(1 - frac{y}d bigr) left(
1 - K bigl( 1 - frac{y}{2d}bigr) frac{y}{2d} right)^{-3/2}\
&=bigl(1 + 2epsilon bigr) left( 1 + K bigl( 1 + epsilon bigr) epsilon right)^{-3/2} - bigl(1 - 2epsilon bigr) left(
1 - K bigl( 1 - epsilon bigr) epsilon right)^{-3/2}\
&=begin{aligned}[t]
&bigl(1 + 2epsilon bigr) left( 1 - frac32 K (1+epsilon)epsilon + frac1{2!}frac32 frac52 K^2 (1+epsilon)^2epsilon^2 - frac1{3!}frac32 frac52 frac72 K^3 (1+epsilon)^3epsilon^3 + cdots right) \
&qquad - bigl(1 - 2epsilon bigr) left( 1 - frac32 bigl( -K (1-epsilon)epsilon bigr) + frac1{2!}frac32 frac52 bigl( -K (1-epsilon)epsilon bigr)^2 - frac1{3!}frac32 frac52 frac72 bigl( -K (1-epsilon)epsilon bigr)^3 + cdots right)
end{aligned}
end{aligned}$$
Here we have $(1 + 2epsilon) cdot A_1 - (1 - 2epsilon) cdot A_2 = A_1 - A_2 + 2epsilon cdot ( A_1 + A_2)$ where
$$begin{aligned}
A_1 &= 1 - frac32 K (1 + epsilon)epsilon + frac{15}8 K^2 (1 + epsilon)^2epsilon^2 - frac{35}{16} K^3 (1 + epsilon)^3epsilon^3 + cdots \
A_2 &= 1 + frac32 K (1-epsilon)epsilon + frac{15}8 K^2 (1-epsilon)^2epsilon^2 + frac{35}{16} K^3 (1-epsilon)^3epsilon^3 + cdots end{aligned}$$
so that
$$begin{aligned}
A_1 - A_2 & = -3 K epsilon + frac{15}8 K^2 (4epsilon)epsilon^2 - frac{35}{16} K^3 (2 + 6epsilon^2)epsilon^3 + cdots \
A_1 + A_2 & = 2 - 3 K epsilon^2 + frac{15}8 K^2 ( 2+2epsilon^2)epsilon^2 + frac{35}{16} K^3 (6 epsilon + 2epsilon^3)epsilon^3 + cdots
end{aligned}$$
$impliesbegin{aligned}[t]
E_y &propto -3 K epsilon + O( epsilon^3 ) + 2epsilon cdot bigl( 2 + O(epsilon^2) bigr) \
&= (4-3K)epsilon + O( epsilon^3 )
end{aligned}$
Therefore the critical point $E_y = 0$ is given by
begin{align}
4 - 3K &= 0 &&implies frac{ 4d^2 }{ x^2 + d^2} = frac43 && implies 2d^2 = x^2 &&
end{align}
which is the desired $x = sqrt{2} d$.
answered Feb 2 at 9:35
Lee David Chung LinLee David Chung Lin
4,34531241
edited Feb 2 at 9:46
answered Feb 2 at 9:35
Lee David Chung LinLee David Chung Lin
4,34531241
answered Feb 2 at 9:35
Lee David Chung LinLee David Chung Lin
4,34531241
answered Feb 2 at 9:35
Lee David Chung LinLee David Chung Lin
4,34531241
4,34531241
$begingroup$
I think your answer is right but aren't you using loads of redundant terms in your taylor approximations? It's hard to read..
$endgroup$
– Dis-integrating
Feb 18 at 0:17
$begingroup$
I was just trying to provide all the details. Indeed this is routine expansion and nothing particularly profound about it. Since you posted the question here after already got an instruction on the physics site, I presumed you had difficulties wading through the algebra. Whether a particular algebraic manipulation is "redundant" or "helpful" is really personal, and obviously I guessed wrong what you want. If you have already found elsewhere a version (or written up yourself) that is "easy to read" then I'm glad for you.
$endgroup$
– Lee David Chung Lin
yesterday
$begingroup$
Now, about the technical stuff. (1) The choices to denote $epsilon$ and $K$ as such is intended to make the expressions shorter. You can try other ways (with different multiplicative factors), and most likely you'll end up with more terms and longer equations. (2) The use of $A_1$ and $A_2$ is intended to highlight the parallel and complementarity of the two parts.
$endgroup$
– Lee David Chung Lin
yesterday
$begingroup$
The bottomline is, I'm sorry if my solution as seen (with those designs and intentions ) doesn't work for you. There's the edit functionality, and you can modify my post to suit your needs. Or, you are very welcomed to type up your own approach as a separate answer post and "accept" it, to show future readers a derivation that can be deemed better under some criteria. Editing to improve the content and self-answering are common practice that are encouraged on this site.
$endgroup$
– Lee David Chung Lin
yesterday
add a comment |
$begingroup$
I think your answer is right but aren't you using loads of redundant terms in your taylor approximations? It's hard to read..
$endgroup$
– Dis-integrating
Feb 18 at 0:17
$begingroup$
I was just trying to provide all the details. Indeed this is routine expansion and nothing particularly profound about it. Since you posted the question here after already got an instruction on the physics site, I presumed you had difficulties wading through the algebra. Whether a particular algebraic manipulation is "redundant" or "helpful" is really personal, and obviously I guessed wrong what you want. If you have already found elsewhere a version (or written up yourself) that is "easy to read" then I'm glad for you.
$endgroup$
– Lee David Chung Lin
yesterday
$begingroup$
Now, about the technical stuff. (1) The choices to denote $epsilon$ and $K$ as such is intended to make the expressions shorter. You can try other ways (with different multiplicative factors), and most likely you'll end up with more terms and longer equations. (2) The use of $A_1$ and $A_2$ is intended to highlight the parallel and complementarity of the two parts.
$endgroup$
– Lee David Chung Lin
yesterday
$begingroup$
The bottomline is, I'm sorry if my solution as seen (with those designs and intentions ) doesn't work for you. There's the edit functionality, and you can modify my post to suit your needs. Or, you are very welcomed to type up your own approach as a separate answer post and "accept" it, to show future readers a derivation that can be deemed better under some criteria. Editing to improve the content and self-answering are common practice that are encouraged on this site.
$endgroup$
– Lee David Chung Lin
yesterday
$begingroup$
I think your answer is right but aren't you using loads of redundant terms in your taylor approximations? It's hard to read..
$endgroup$
– Dis-integrating
Feb 18 at 0:17
$begingroup$
I think your answer is right but aren't you using loads of redundant terms in your taylor approximations? It's hard to read..
$endgroup$
– Dis-integrating
Feb 18 at 0:17
$begingroup$
I was just trying to provide all the details. Indeed this is routine expansion and nothing particularly profound about it. Since you posted the question here after already got an instruction on the physics site, I presumed you had difficulties wading through the algebra. Whether a particular algebraic manipulation is "redundant" or "helpful" is really personal, and obviously I guessed wrong what you want. If you have already found elsewhere a version (or written up yourself) that is "easy to read" then I'm glad for you.
$endgroup$
– Lee David Chung Lin
yesterday
$begingroup$
I was just trying to provide all the details. Indeed this is routine expansion and nothing particularly profound about it. Since you posted the question here after already got an instruction on the physics site, I presumed you had difficulties wading through the algebra. Whether a particular algebraic manipulation is "redundant" or "helpful" is really personal, and obviously I guessed wrong what you want. If you have already found elsewhere a version (or written up yourself) that is "easy to read" then I'm glad for you.
$endgroup$
– Lee David Chung Lin
yesterday
$begingroup$
Now, about the technical stuff. (1) The choices to denote $epsilon$ and $K$ as such is intended to make the expressions shorter. You can try other ways (with different multiplicative factors), and most likely you'll end up with more terms and longer equations. (2) The use of $A_1$ and $A_2$ is intended to highlight the parallel and complementarity of the two parts.
$endgroup$
– Lee David Chung Lin
yesterday
$begingroup$
Now, about the technical stuff. (1) The choices to denote $epsilon$ and $K$ as such is intended to make the expressions shorter. You can try other ways (with different multiplicative factors), and most likely you'll end up with more terms and longer equations. (2) The use of $A_1$ and $A_2$ is intended to highlight the parallel and complementarity of the two parts.
$endgroup$
– Lee David Chung Lin
yesterday
$begingroup$
The bottomline is, I'm sorry if my solution as seen (with those designs and intentions ) doesn't work for you. There's the edit functionality, and you can modify my post to suit your needs. Or, you are very welcomed to type up your own approach as a separate answer post and "accept" it, to show future readers a derivation that can be deemed better under some criteria. Editing to improve the content and self-answering are common practice that are encouraged on this site.
$endgroup$
– Lee David Chung Lin
yesterday
$begingroup$
The bottomline is, I'm sorry if my solution as seen (with those designs and intentions ) doesn't work for you. There's the edit functionality, and you can modify my post to suit your needs. Or, you are very welcomed to type up your own approach as a separate answer post and "accept" it, to show future readers a derivation that can be deemed better under some criteria. Editing to improve the content and self-answering are common practice that are encouraged on this site.
$endgroup$
– Lee David Chung Lin
yesterday
add a comment |
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$begingroup$
Are you still interested in a solution? I see that you already got a response on the physics site.
$endgroup$
– Lee David Chung Lin
Jan 31 at 3:54