Torque acting on dipole

[songoku]

Homework Statement

Please see below

Relevant Equations

$\vec \tau=\vec p \times \vec E$

$\vec E=\frac{kQ}{r^3} \vec r$

Electric field acting on the dipole due to $+Q$ is the sum of $E$ on $+q$ and $-q$
$$\vec E=\frac{kQ}{(a^2+d^2)^{\frac{3}{2}}} (-a\hat x+d \hat y)+\frac{kQ}{(a^2+d^2)^{\frac{3}{2}}} (a\hat x+d \hat y)$$
$$=\frac{2kQd}{(a^2+d^2)^{\frac{3}{2}}} \hat y$$

So,
$$\vec \tau=\vec p \times \vec E$$
$$=-P\hat x \times \frac{2kQd}{(a^2+d^2)^{\frac{3}{2}}} \hat y$$
$$=-\frac{2kQdP}{(a^2+d^2)^{\frac{3}{2}}} \hat z$$

But the answer key has no 2 in the numerator. What is my mistake?

Thanks

 

[anuttarasammyak]

Electric field acting on the dipole due to +Q is the sum of E on +q and −q

How about dealing forces on +q and -q individually to calculate torque by them ?

 

[haruspex]

I think you have a different problem. Check the signs in the right half.

$$\vec E=\frac{kQ}{(a^2+d^2)^{\frac{3}{2}}} (-a\hat x+d \hat y)+\frac{kQ}{(a^2+d^2)^{\frac{3}{2}}} (a\hat x+d \hat y)$$

Which way should the net force be, looking at the diagram?
And in your final answer, should the torque increase as d increases and should it decrease as a increases?

 

[Orodruin]

The problem is counting the electric field twice …

You either use the electric field at the dipole and apply $\vec \tau = \vec p \times \vec E$ or you compute the forces on the charges individually and compute the torque from those.

 

[songoku]

How about dealing forces on +q and -q individually to calculate torque by them ?

By using this method, I got the vertical force on $+q$ to be $F=\frac{kQqd}{(a^2+d^2)^{\frac{3}{2}}} \hat y$ and the torque to be:
$$\vec \tau=2 (-a \hat x) \times \frac{kQqd}{(a^2+d^2)^{\frac{3}{2}}} \hat y$$
$$=-\frac{2akQqd}{(a^2+d^2)^{\frac{3}{2}}} \hat z$$
$$=-\frac{kQdP}{(a^2+d^2)^{\frac{3}{2}}}\hat z$$

The answer is correct but I still don't understand why my method in #1 is wrong

I think you have a different problem. Check the signs in the right half.

Sorry I still don't understand my mistake

Which way should the net force be, looking at the diagram?

Net force on the dipole is to the left

And in your final answer, should the torque increase as d increases and should it decrease as a increases?

From the equation, I think as $d$ increases (keeping $a$ constant), the torque will increase until max then decreases

Keeping $d$ constant, the torque will decrease then increase

The problem is counting the electric field twice …

Isn't the electric field used in $\vec \tau=\vec p \times \vec E$ is the sum of electric field on $+q$ and $-q$?

Thanks

 

[anuttarasammyak]

The answer is correct but I still don't understand why my method in #1 is wrong

Not a total force but forces on different parts of a body matter in considering torque.

PS. the total force seems to have minus direction-x not y.

 

[songoku]

Not a total force but forces on different parts of a body matter in considering torque.

PS. the total force seems to have minus direction-x not y.

In post #1, I use electric field, not force

 

[haruspex]

The answer is correct but I still don't understand why my method in #1 is wrong

As @Orodruin posted, you are double-counting the field. There is no principle that says you can take the field at one place and the field at another and add them. You can add two independent fields acting at the same point, but the two fields you added are the same field.

Sorry I still don't understand my mistake

Because it was my mistake. I was thinking of the force, not the field.

 

[Orodruin]

Isn't the electric field used in $\vec \tau=\vec p \times \vec E$ is the sum of electric field on $+q$ and $-q$?

Absolutely not! It is the electric field at the dipole. The approximation mafe is that it is a point dipole.

 

[anuttarasammyak]

In post #1, I use electric field, not force

Not only field E but also forces qE , - qE have different values at the different location of charges. Here E is the field by Q not by q, -q where force does not contribute to torque and cancel through the bar between.

 

[Orodruin]

Here E is the field by Q not by q, -q.

No, it isn’t. The field $\vec E$ in $\vec\tau = \vec p \times \vec E$ is the field at the idealized dipole and absolutely nothing else.

 

[anuttarasammyak]

No, it isn’t. The field E→ in τ→=p→×E→ is the field at the idealized dipole and absolutely nothing else.

I agree with you. @songoku calculated E by Q only. It works because field by -q on q does not contribute to torque and translational motion of the dipole by cancelling pair.

 

[Orodruin]

I agree with you. @songoku calculated E by Q only.

No, they did not. They added the fields at q and at -q.

It works because field by -q on q does not contribute to torque and translational motion of the dipole by cancelling pair.

No, it doesn’t work. You cannot add fields at different points.

 

[anuttarasammyak]

[edit] forces on charges from the bar should be also accounted for full investigation.

 

[Orodruin]

View attachment 355402

That’s adding forces, not fields. And it is not using $\vec \tau = \vec p \times \vec E$ as the OP does.

 

[Steve4Physics]

@songoku, it’s probably worth noting that $\vec \tau=\vec p \times \vec E$ applies when the field is uniform over the dipole (so the field has the same magnitude and direction at each end of the dipole).

This does not apply in the present problem, though it would be a sensible approximation if $d \gg a$; in this case you would use the field’s value at the centre of the dipole.

@anuttarasammyak's suggestion in Post #2 (consider the individual forces acting on +q and -q) looks likethe simplest approach.

Minor edit.

 

[Orodruin]

@songoku, it’s probably worth noting that $\vec \tau=\vec p \times \vec E$ applies when the field is uniform over the dipole (so the field has the same magnitude and direction at each end of the dipole).

I would phrase this differently: It applies to an ideal dipole, ie, in the situation when $a\to 0$ with constant dipole moment.

Calling the charge configuration shown a dipole is also somewhat misleading. Yes, the leading contribution to the far field is the dipole moment, but it is not a pure dipole - it also has higher moments, just no monopole moment.

 

[kuruman]

The statement of the problem can be criticized on the basis that it does not specify the point about which the torque is to be calculated. The reference point is irrelevant if the expression $\vec \tau = \vec p \times \vec E$ is used. However, this expression, as has already been pointed out, is valid for a "point" dipole, i.e. with the understanding that the electric field is uniform over the "length" of the dipole. This is not the case here because $d\approx 2\times(2a).$

The forces acting on the system are $$\mathbf F_1=\frac{kQq(-a\mathbf {\hat x}+d \mathbf {\hat y})}{(d^2+a^2)^{3/2}}~;~~\mathbf F_2=-\frac{kQq(a\mathbf {\hat x}+d \mathbf {\hat y})}{(d^2+a^2)^{3/2}}\implies \mathbf F_{\text{net}}= -\frac{2kQq a}{(d^2+a^2)^{3/2}}\mathbf {\hat x}.$$This net force produces a torque that changes the "orbital" angular momentum about any point that does not lie on the line joining $+q$ and $-q$. Judging from the known answer, this torque is not what the problem is asking us to find.

The problem is asking about the torque that changes the "spin" angular momentum about the midpoint between the charges. Then $$\mathbf N_{\text{spin}}=(-a\mathbf {\hat x})\times\mathbf F_1+(a\mathbf {\hat x})\times\mathbf F_2=-\frac{2kQq ad}{(d^2+a^2)^{3/2}}\mathbf {\hat z}.$$ As always, one should not mention "torque" or "angular momentum" without specifying the origin with respect to which these position-dependent vectors are defined.

 

[Orodruin]

The reference point is irrelevant if the expression τ→=p→×E→ is used.

This is correct only if the net force on the dipole is zero, which is not generally the case even for an ideal dipole ($\vec F = (\vec p \cdot \nabla)\vec E$). However, the unstated assumption is usually wrt the dipole position.

 

[songoku]

@songoku, it’s probably worth noting that $\vec \tau=\vec p \times \vec E$ applies when the field is uniform over the dipole (so the field has the same magnitude and direction at each end of the dipole).

This does not apply in the present problem, though it would be a sensible approximation if $d \gg a$; in this case you would use the field’s value at the centre of the dipole.

I don't think the electric field is uniform over the dipole but it has same magnitude and direction at each end of the dipole for calculating the torque (I am only considering the vertical component)

By considering the explanation from @Orodruin and @haruspex, I got the answer using $\vec \tau=\vec p \times \vec E$. Is it only coincidental?

Thanks

 

[jbriggs444]

By considering the explanation from @Orodruin and @haruspex, I got the answer using $\vec \tau=\vec p \times \vec E$. Is it only coincidental?

Ummm, you did not evaluate $\vec{E}$ at the position of the dipole. Instead, you evaluated [the relevant component of] $\vec{E}$ at the ends of the dipole.

This is obviously the right thing to do. But it is not the same as computing $\vec p \times \vec E$.

 

[kuruman]

By considering the explanation from @Orodruin and @haruspex, I got the answer using τ→=p→×E→. Is it only coincidental?

Consider this general derivation of the torque on the dipole about the mid-point between the charges.
Let $2d$ be the separation between the charges.
Let charge $(-q)$ be at location $\mathbf a$ where the electric field is $\mathbf E$
Let charge $(+q)$ be at location $-\mathbf a$ where the electric field is $\mathbf E + \delta \mathbf E ~,~(\delta \mathbf E \neq 0).$
The dipole moment is $\mathbf p=-2q\mathbf a.$
The net torque about the midpoint is
$$\begin{align} & \mathbf N_{\text{net}}=(-\mathbf a)\times(+q)(\mathbf E +\delta \mathbf E)+\mathbf a\times(-q)\mathbf E =-2q\mathbf a\times \mathbf E-q\mathbf a\times \delta \mathbf E \nonumber \\
& \mathbf N_{\text{net}}=\mathbf p\times\mathbf E+\frac{1}{2}\mathbf p\times \delta \mathbf E. \nonumber
\end{align}$$Here, $\delta \mathbf E$ is in the negative x-direction which makes it parallel to the dipole moment. Therefore, the second term in the above expression vanishes and you are left with $\mathbf p\times\mathbf E.$

 

[Steve4Physics]

I don't think the electric field is uniform over the dipole but it has same magnitude and direction at each end of the dipole for calculating the torque (I am only considering the vertical component)

The field directions at each end of the dipole are different (by about 30-40 degrees on the Post #1 diagram).

But in this question the x-conponents of the field do not give rise to torque. By using the y-components only, you have allowed for the field directions.

By considering the explanation from @Orodruin and @haruspex, I got the answer using $\vec \tau=\vec p \times \vec E$. Is it only coincidental?

You haven't shown your working so we can't tell.
Edit - sorry, please ignore.

 

[Orodruin]

But in this question the x-conponents of the field do not give rise to torque.

… relative to the dipole’s center it should be noted. They will provide a net torque around any point offset from that center in the y-direction.

 

[Orodruin]

Since it has been mentioned a couple of times in this thread, the torque - even on an ideal dipole - will generally depend on the position relative to which the torque is considered (just as any force-torque couple) as the net force on the dipole is non-zero. The idealised forms of the force and torque (relative to the position of the ideal dipole) are given by
$$
\vec F = (\vec p \cdot \nabla) \vec E, \qquad \vec \tau = \vec p \times \vec E.
$$
Assuming the dipole is located at $\vec x$, the torque relative to a different position $\vec y$ is given by
$$
\vec \tau_{\vec y} = \vec \tau + (\vec x - \vec y) \times \vec F
= \vec p \times \vec E + (\vec x - \vec y) \times [(\vec p \cdot \nabla)\vec E]
= \vec p \times \vec E + (\vec p \cdot \nabla)[(\vec x - \vec y) \times \vec E] - [(\vec p \cdot \nabla)\vec x]\times \vec E
= \vec p \times \vec E + (\vec p \cdot \nabla)[(\vec x - \vec y) \times \vec E] - \vec p \times \vec E
= (\vec p \cdot \nabla)[(\vec x - \vec y) \times \vec E]
$$

 

[kuruman]

Assuming the dipole is located at $\vec x$, the torque relative to a different position $\vec y$ is given by
$$
\vec \tau_{\vec y} = \vec \tau + (\vec x - \vec y) \times \vec F
= \vec p \times \vec E + (\vec x - \vec y) \times [(\vec p \cdot \nabla)\vec E]
= \vec p \times \vec E + (\vec p \cdot \nabla)[(\vec x - \vec y) \times \vec E] - [(\vec p \cdot \nabla)\vec x]\times \vec E
= \vec p \times \vec E + (\vec p \cdot \nabla)[(\vec x - \vec y) \times \vec E] - \vec p \times \vec E
= (\vec p \cdot \nabla)[(\vec x - \vec y) \times \vec E]
$$

I am confused by this expression. If I omit the intermediate steps to fit it in the screen, I get
$$\vec \tau_{\vec y} = \vec \tau + (\vec x - \vec y) \times \vec F=\dots =(\vec p \cdot \nabla)[(\vec x - \vec y) \times \vec E].$$ Shouldn't $~ \vec \tau_{\vec y}~$ reduce to $~\vec \tau_{\vec x}=\vec {\tau} =\vec p\times E~$ when $\vec y=\vec x~$? It seems that it does so in the beginning of the expression but not in the end result.

 

[Orodruin]

I am confused by this expression. If I omit the intermediate steps to fit it in the screen, I get
$$\vec \tau_{\vec y} = \vec \tau + (\vec x - \vec y) \times \vec F=\dots =(\vec p \cdot \nabla)[(\vec x - \vec y) \times \vec E].$$ Shouldn't $~ \vec \tau_{\vec y}~$ reduce to $~\vec \tau_{\vec x}=\vec {\tau} =\vec p\times E~$ when $\vec y=\vec x~$? It seems that it does so in the beginning of the expression but not in the end result.

The assumption in the derivation is that $\nabla$ should be seen as acting only on $\vec x$. Directly letting $\vec y = \vec x$ in the expression violates this. You can still use the expression, but you must let $\vec y = \vec x$ only after evaluating the derivative.

Edit: A better notation might be $\vec\tau (\vec x, \vec y) = (\vec p \cdot \nabla_{\vec x})[(\vec x - \vec y) \times \vec E(\vec x)]$. It is then a bit clearer that the function $\vec\tau (\vec x, \vec y)$ is defined as the expression after evaluating the derivative and taking $\vec\tau (\vec x, \vec x)$ is what you obtain after inserting $\vec y = \vec x$ into that expression.

 

[kuruman]

Ah, yes. A subscripted del operator makes it clearer. Thanks.

 

[songoku]

Thank you very much for all the help and explanation anuttarasammyak, haruspex, Orodruin, Steve4Physics, kuruman, jbriggs444