Why Does the Potential Energy Calculation for a Dipole Yield a Negative Value?

[guyvsdcsniper]

Homework Statement

A water molecule perpendicular to an electric field has 5.00×10−22 J more potential energy than a water molecule aligned with the field. The dipole moment of a water molecule is 6.2×10−30Cm.

What is the strength of the electric field?

Relevant Equations

U=-pEcos(theta)

So the change in potential energy is ∆U = Uf-Ui. Final minus initial. If i solve the above problem like this I end up with a negative value. The way the person in the attached work solved the problem, is they used ∆U = Ui-Uf. How are the switching Ui and Uf? What is it I am missing?

 

[haruspex]

What exactly do you think it means when it says "aligned"? Can that be in the reverse direction?
Which choice leads to a lower PE than when at right angles?
What angle does that correspond to?

 

[vcsharp2003]

What exactly do you think it means when it says "aligned"? Can that be in the reverse direction?
Which choice leads to a lower PE than when at right angles?
What angle does that correspond to?

PE is a scalar quantity and so it can be +ve or -ve or 0. If $PE_1=0$ and $PE_2= -5$ then $PE_1 > PE_2$. So, if $PE = 0$ it doesn't mean it's a minimum. $PE= 0$ could be a maximum.

• Aligned means electric field is in same or opposite direction to dipole moment.
• Yes, aligned could mean anyone of following two cases: dipole moment and electric field in same direction OR dipole moment and electric field in opposite direction.
• PE is always lowest in equilibrium position i.e. when no net torque acts on the dipole, which will be when dipole is aligned with electric field according to above bullet point. PE will be maximum at extremes i.e. when dipole moment is perpendicular to electric field.
• Angle of 0 or 180 degrees between dipole moment and electric field results in lowest PE. Angle of 90 degrees results in highest PE.

 

[haruspex]

PE is always lowest in equilibrium position i.e. when no net torque acts on the dipole, which will be when dipole is aligned with electric field according to above bullet point. PE will be maximum at extremes i.e. when dipole moment is perpendicular to electric field.

No, equilibrium does not necessarily correspond to minimum PE. A pencil standing on its point could theoretically be in equilibrium, though highly unstable.
Likewise, though torque is maximised for the dipole in the perpendicular position, PE is not.

 

[vcsharp2003]

No, equilibrium does not necessarily correspond to minimum PE. A pencil standing on its point could theoretically be in equilibrium, though highly unstable.
Likewise, though torque is maximised for the dipole in the perpendicular position, PE is not.

I was thinking of simple harmonic motion analogy where PE is min at equilibrium position i.e. at the middle of extreme positions. So where would PE be a maximum for dipole placed in a uniform electric field? It appears that when $\theta = 180^o$ then $PE = - | \vec P | | \vec E | \cos {180^0} = | \vec P | | \vec E |$ would be maximum since PE at this position takes a +ve value.

 

[haruspex]

I was thinking of simple harmonic motion analogy where PE is min at equilibrium position i.e. at the middle of extreme positions. So where would PE be a maximum for dipole placed in a uniform electric field?

Suppose the external field is oriented left to right, i.e. it would push an isolated positive charge to the right. So it would push the positive side of a dipole to the right and the negative side to the left.
Consider the centre of the dipole as fixed, since there is no net force from the external field. The max PE is when each charge has furthest to go, i.e. the positive is on the left, negative on the right. That means the dipole's field is not only parallel to the applied field but acts in the same direction.
The minimum PE is the reverse position, with the two fields in opposition.

 

[vcsharp2003]

Suppose the external field is oriented left to right, i.e. it would push an isolated positive charge to the right. So it would push the positive side of a dipole to the right and the negative side to the left.
Consider the centre of the dipole as fixed, since there is no net force from the external field. The max PE is when each charge has furthest to go, i.e. the positive is on the left, negative on the right. That means the dipole's field is not only parallel to the applied field but acts in the same direction.
The minimum PE is the reverse position, with the two fields in opposition.

I get it using the mathematical relationship of $U = - \vec P \cdot \vec E$, but still trying to understand how you've explained.

 

[vcsharp2003]

Suppose the external field is oriented left to right, i.e. it would push an isolated positive charge to the right. So it would push the positive side of a dipole to the right and the negative side to the left.
Consider the centre of the dipole as fixed, since there is no net force from the external field. The max PE is when each charge has furthest to go, i.e. the positive is on the left, negative on the right. That means the dipole's field is not only parallel to the applied field but acts in the same direction.
The minimum PE is the reverse position, with the two fields in opposition.

I think I get your explanation. What you're saying is to look at the PE of system consisting of the + and - charges in the two positions shown in diagram below. We could consider that initially the two dipole charges are at center of dipole and then we separate them so each is at a distance $a$ from the center of dipole. Clearly +ve work is being by external force on each +ve and -ve charge as they are separated to their dipole positions in top dipole which means PE is increasing. In bottom dipole, -ve work is being by external force which means PE is decreasing.

 

[haruspex]

I think I get your explanation. What you're saying is to look at the PE of system consisting of the + and - charges in the two positions shown in diagram below. We could consider that initially the two dipole charges are at center of dipole and then we separate them so each is at a distance $a$ from the center of dipole. Clearly +ve work is being by external force on each +ve and -ve charge as they are separated to their dipole positions in top dipole which means PE is increasing. In bottom dipole, -ve work is being by external force which means PE is decreasing.

View attachment 289767

Yes.