External forces required to move an electric dipole quasi-statically

[vcsharp2003]

Homework Statement

What forces would be required to move an electric dipole that is inclined at 60° from $\infty$ to a position inside a uniform electric field area so that the dipole remains inclined to the horizontal at 60° throughout its movement?

Relevant Equations

I tried to visualise this scenario using diagrams below.

If the dipole is to be in equilibrium at all positions as it's moved so that it's always inclined at 60° to the horizontal, then the torque due to electric field needs to be balanced by torque due to external forces $F_{ext}$ as shown in above diagrams. But such external forces will not make the dipole move forward horizontally since it will only prevent any rotation of the dipole without causing any translational motion.

What other external force would be needed to be applied to the dipole to make it move horizontally towards left quasi-statically? A force at the center of dipole towards the left cannot be applied since the two charges of a dipole are not connected by any physical rod.

Ultimately, my goal is to prove that the work done by external forces in moving a dipole from infinity to the final position in uniform electric field is not $0$ and also that this work is negative.

 

[BvU]

This seems to me a very artificial problem statement:
A uniform E field that is not uniform
A dipole that isn't a dipole -- or is it ?
(two separate charges but yes/not connected?)

I think some clarification and focusing is needed..

$\$

 

[vcsharp2003]

This seems to me a very artificial problem statement:
A uniform E field that is not uniform
A dipole that isn't a dipole -- or is it ?
(two separate charges but yes/not connected?)

I think some clarification and focusing is needed..

$\$

I see. Probably, one can assume that the dipole charges are connected since the dipole in a water molecule is like a connected pair of equal and opposite charges.

I did mention uniform electric in my question. But the diagrams posted are hand drawn, and therefore the electric field lines may not look exactly equally spaced

 

[hutchphd]

Homework Statement: What forces would be required to move an electric dipole that is inclined at 60° from $\infty$ to a position inside a uniform electric field area so that the dipole remains inclined to the horizontal at 60° throughout its movement?
Relevant Equations: I tried to visualise this scenario using diagrams below.

Ultimately, my goal is to prove that the work done by external forces in moving a dipole from infinity to the final position in uniform electric field is not 0 and also that this work is negative.

Is this result supposed to be independent of the path from infinity? What if you interchange the signs of the charges: is the result still always negative? There are many reasons this leads to impossibilities.

 

[vcsharp2003]

Is this result supposed to be independent of the path from infinity?

It should be since electric force is a conservative force.

What if you interchange the signs of the charges: is the result still always negative?

Probably yes. If the final position is as shown in third diagram, then the work done by external forces would be the same if at $\infty$ the bottom charge was -ve or +ve. This should be because electric field is a conservative field.

 

[Steve4Physics]

Homework Statement: What forces would be required to move an electric dipole that is inclined at 60° from $\infty$ to a position inside a uniform electric field area so that the dipole remains inclined to the horizontal at 60° throughout its movement?
.
Ultimately, my goal is to prove that the work done by external forces in moving a dipole from infinity to the final position in uniform electric field is not $0$ and also that this work is negative.

This might be too simple-minded but from a classical point of view:

Assume that the uniform electric field extends through all of space.

Rotation needs to be prevented by application of a constant couple (a pair of equal magnitude, antiparallel forces with different lines of action). The couple does no work as there is no rotation.

Assume constant velocity - in order to avoid having to consider energy-loss associated with accelerating charges. No external net force is needed (Newton's 1st law).

There is no change in translational and rotational kinetic energies, no change in potential energy and no radiation losses. So the net work done is zero.

 

[vcsharp2003]

This might be too simple-minded but from a classical point of view:

Assume that the uniform electric field extends through all of space.

Rotation needs to be prevented by application of a constant couple (a pair of equal magnitude, antiparallel forces with different lines of action). The couple does no work as there is no rotation.

Assume constant velocity - in order to avoid having to consider energy-loss associated with accelerating charges. No external net force is needed (Newton's 1st law).

There is no change in translational and rotational kinetic energies, no change in potential energy and no radiation losses.

But isn't the potential energy of a system of charges in a certain position obtained by determining the work done by external forces in moving the system from infinity to the current position?

If the work done by external forces is 0 then potential energy of dipole in final position would be zero, but we know the potential energy is $- \vec {p} .\vec {E}$ which is not 0 for final position.

 

[BvU]

from $\infty$ to a position inside a uniform electric field area so that the dipole remains inclined to the horizontal at 60° throughout its movement?

The electric field you seem to have in mind is impossible: field lines do not stop at arbitrary points in space.

A uniform E field that is not uniform

$\$

 

[Steve4Physics]

But isn't the potential energy of a system of charges in a certain position obtained by determining the work done by external forces in moving the system from infinity to the current position?

If you have a conservative force, the potential energy associated with that particular force is obtained as you describe (with an appropriate minus sign and assuming no change in kinetic erergy).

But in the current scenario, there is is a zero net external electric force. So the work done is zero and the change in electric potential energy is zero.

If the work done by external forces is 0 then potential energy of dipole in final position would be zero, but we know the potential energy is $- \vec p \dot \vec E$ which is not 0 for final position.

At infinity, the dipole is still in the uniform electric electric and oriented at 60º (unless you haven't described the scenario accurately). The initial and final dipole energies are equal. It is the change we are interested in - and the change is zero.

 

[kuruman]

If the work done by external forces is 0 then potential energy of dipole in final position would be zero, but we know the potential energy is $- \vec {p} .\vec {E}$ which is not 0 for final position.

You are forgetting that the initial potential energy is also $U_i=-\mathbf{p}\cdot \mathbf{E}$. There is no change in potential energy, therefore no net work done on the dipole.

You can also look at it this way. As @Steve4Physics remarked, for the dipole to maintain its orientation, you need an additional pair forces that cancel exactly the forces on the charges exerted by the external field. This additional pair could be a second electric field that cancels the original field or a fair of fingers holding the charges. Regardless of how the additional forces are generated, the effect is the same. There is no unbalanced force on the dipole in which case Newton's first law takes over.

 

[kuruman]

Ultimately, my goal is to prove that the work done by external forces in moving a dipole from infinity to the final position in uniform electric field is not $0$ and also that this work is negative.

I assume that by "work" you mean the work done on the dipole by the conservative electrical forces, not by the agent moving the dipole. Your starting position is an assembled dipole at infinity, you ending position is a dipole in a uniform field at 60° relative to the field. That's all that matters because the work done by the electrical forces is independent of the path. Note that the uniform electric field does not have to extend to infinity but can be created between the plates of a parallel plate capacitor.

Can you devise a path to show that the total work done by the electrical forces on the charges is $W=\mathbf{p}\cdot \mathbf{E}$ where $\mathbf{E}$ is the electric field between the plates?

 

[vcsharp2003]

The electric field you seem to have in mind is impossible: field lines do not stop at arbitrary points in space.
$\$

Ok. I get what you're saying. I meant something like the field between two oppositely charged parallel plates, in which case the uniform field would be confined to a limited space.

 

[vcsharp2003]

I assume that by "work" you mean the work done on the dipole by the conservative electrical forces, not by the agent moving the dipole. Your starting position is an assembled dipole at infinity, you ending position is a dipole in a uniform field at 60° relative to the field. That's all that matters because the work done by the electrical forces is independent of the path. Note that the uniform electric field does not have to extend to infinity but can be created between the plates of a parallel plate capacitor.

Can you devise a path to show that the total work done by the electrical forces on the charges is $W=\mathbf{p}\cdot \mathbf{E}$ where $\mathbf{E}$ is the electric field between the plates?

Is below the correct answer to your question?

I assume an electric field between a parallel plate capacitor as shown in diagram below. My goal is to find the work done by electrostatic forces on the dipole as it's moved from outside the field to into the field.

The dashed vertical lines through the electric field denote the equipotential lines that will be perpendicular to the horizontal electric field for potentials $V_1$ and $V_2$. Now potential at a point is defined as the negative of the work done by the electrostatic force in moving a charge from infinity to that point.

So, $W_1= - (-q)V_1$ and $W_2= - (+q)V_2$, where $W_1$ and $W_2$ are the work done by electrostatic force on charges $q_1$ and $q_2$ respectively as these charges are moved from infinity to their final positions.

So, $W_{net} = W_1 + W_2 = qV_1- qV_2= q(V_1 - V_2)$.

Since $E = - \dfrac {dV}{dx}$ so $E = \dfrac {- (V_2-V_1)}{d \times cos{\theta}}$.

Thus we get $W_{net} = qEd cos {\theta} = \vec p . \vec E$

 

[kuruman]

That works fine. A more elegant way to do this is to insert the dipole between the plates with its moment perpendicular to the electric field. Then the charges are moving along an equipotential and the work done by the electrical forces is zero. Once in, you rotate the dipole and calculate the work done by the net electric torque $W=\int \tau~ d\theta.$

 

[vcsharp2003]

That works fine. A more elegant way to do this is to insert the dipole between the plates with its moment perpendicular to the electric field. Then the charges are moving along an equipotential and the work done by the electrical forces is zero. Once in, you rotate the dipole and calculate the work done by the net electric torque $W=\int \tau~ d\theta.$

Yes, that is more easier than what I did.

In my derivation in post#13, each charge of +q and -q was moved separately from $\infty$ to their final positions, whereas in your approach the whole dipole is moved as one unit which seems a better approach.