Force on an electric dipole

[wam_mi]

Homework Statement

A dipole, consisting of a charge -q at position vector r and a charge +q at position vector r+s, is sitting in an external electric field E(ext).

(a) Write down the total force on the dipole F(dipole) due to the external field

(b) Show that when s is small,
F(dipole) approx. = (dot product of p and del operator)*E(ext)
where p is the electric dipole moment, p = q s

(c) In what limit does the result become exact?

Homework Equations

p = q s
F = torque = cross product (p and E(ext))

The Attempt at a Solution

(a) F(dipole) = cross product (p and E(ext)) = q* cross product (s and E(ext))

stuck on (b) and (c)


Please help!

 

[anathema]

Hello, fellow scientist. I would be happy to assist you with this problem.

(b) To show that F(dipole) approx. = (dot product of p and del operator)*E(ext) when s is small, we can expand the cross product using the vector triple product rule. This gives us:

F(dipole) = q*(s x E(ext)) = q*(s x (E(ext) x s))

Since s is small, we can approximate E(ext) x s as E(ext) x 0 = 0. Therefore, we have:

F(dipole) = q*(s x 0) = 0

Now, using the definition of the electric dipole moment, p = q*s, we can rewrite this as:

F(dipole) = (p x E(ext)) = (p x (E(ext) x s))

Using the vector triple product rule again, we get:

F(dipole) = (p x (E(ext) x s)) = (E(ext) x (p x s))

Since s is small, we can approximate (p x s) as p x 0 = 0. Therefore, we have:

F(dipole) = (E(ext) x 0) = 0

This means that when s is small, F(dipole) approx. = 0. However, we know that this is not the case because the dipole experiences a torque in an external electric field. Therefore, we need to consider higher order terms in our approximation.

Using the vector triple product rule again, we can expand (p x (E(ext) x s)) to get:

F(dipole) = (p x (E(ext) x s)) = (p x (E(ext) x 0)) + (p x (E(ext) x (s - 0)))

Since E(ext) x 0 = 0, we are left with:

F(dipole) = (p x (E(ext) x (s - 0)))

Using the definition of the dot product, we can rewrite this as:

F(dipole) = (p x (E(ext) x (s - 0))) = (p . (E(ext) x (s - 0)))

Finally, using the vector triple product rule again, we can expand (E(ext) x (s - 0)) to get:

F(d

 

[thomate1]

(b) When s is small, we can approximate the cross product of s and E(ext) as s x E(ext) = sE(ext) where s is the magnitude of the displacement vector s. Therefore, F(dipole) = q* sE(ext) = (dot product of p and del operator)*E(ext) since the del operator acting on E(ext) gives the gradient of the electric field, which is equivalent to multiplying E(ext) by the magnitude of the displacement vector s.

(c) The result becomes exact when s approaches zero, meaning that the two charges in the electric dipole are very close to each other. This is because in this limit, the dipole moment p approaches zero and the cross product of p and E(ext) becomes negligible. Therefore, the force on the dipole is solely determined by the gradient of the electric field and the magnitude of the dipole moment.