Potential of a dipole in E field

AI Thread Summary
The discussion focuses on deriving the potential energy of a dipole in an electric field, specifically aiming to show that U = -p.E. The user attempts to start from the potential of two charges but encounters difficulties in transitioning from U = q(VB - VA) to the desired equation. The challenge arises from the differing distances in the potential terms and the absence of an acosθ term in their approximation. It is noted that the potential energy of the dipole's interaction with the external field varies with angle, while the interaction potential remains constant. The conversation highlights the importance of considering the external field's contribution to the potential energy.
raggle
Messages
8
Reaction score
0

Homework Statement



I'm going over some EM notes and I found a derivation for the potential energy of a dipole in an E field which first found the torque on the dipole and then integrated.
I had a go at a derivation that goes the other way, starting from the potential of the 2 charges. I've ran into a bit of a problem midway through and I can't see how to get past it.

Homework Equations



Want to show:
U = -p.E

The Attempt at a Solution


U = q(VB-VA)

now I don't know how to get from here to what I want to show. Both of the potentials look like \frac{q}{4πε<sub>0</sub>r} , where the r is different in either potential.
The only thing I can think of is to use an approximation to r, but that doesn't give an acosθ term in the numerator.
 
Physics news on Phys.org
raggle said:

Homework Statement



I'm going over some EM notes and I found a derivation for the potential energy of a dipole in an E field which first found the torque on the dipole and then integrated.
I had a go at a derivation that goes the other way, starting from the potential of the 2 charges. I've ran into a bit of a problem midway through and I can't see how to get past it.

Homework Equations



Want to show:
U = -p.E

The Attempt at a Solution


U = q(VB-VA)

now I don't know how to get from here to what I want to show. Both of the potentials look like \frac{q}{4πε_0 r} , where the r is different in either potential.
The only thing I can think of is to use an approximation to r, but that doesn't give an acosθ term in the numerator.

The charges of the dipole are in an external electric field. By changing the angle, that potential energy gained from the field changes, the potential energy of their interaction does not. You ignored the contribution of the external field to the potential energy.

As for the potential energy of the interaction between two point charges, it is \frac{q_1 q_2}{4 \pi \epsilon_0 r} where r is the distance between the charges.

ehild
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top